Answer:
use desmos its a graphing caculator that gives you exaxt graphing answers
on the map, 2 cm equals 1 km. what is the actual length of a street that is 10 cm long on the map?
Answer: 5km
Step-by-step explanation: 2cm=1 km
10cm divided by 2 is 5km
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The area of the figure is ______ square units.
Answer:
8 units
Step-by-step explanation:
Area of square:
length * width
2 * 3
= 6 units
Area of triangle:
1/2 * base * height
1/2 * 2 * 2
= 2 units
Area of entire shape:
6 + 2
= 8 units
So, the area of the shape is 8 units.
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Xan and Katie are twins. Xan weighs 666 pounds 999 ounces, and Katie weighs 888 pounds 111 ounce.
What is the difference, in ounces, between Xan and Katie's weight?
Answer:
Difference in weight is 24 once
Step-by-step explanation:
got it right
-(-x) when x=-17
Is the answer 17 or -17 ?
Answer:
-17
Step-by-step explanation:
-(-x)
This simplifies to positive x
x
Let x = -17
-17
Answer:
-17
Step-by-step explanation:
-(-x)= +x therefore
-{-(-17)}= -17
An instructor gives an exam with sixteen questions. Students are allowed to choose any eleven to answer. (Hint: To answer the following questions, these examples may be helpful: Examples 9.5.4, 9.5.5, 9.5.6, and 9.5.7.) (a) How many different choices of eleven questions are there?
The number of different choices of eleven questions that a student can make on an exam with sixteen questions is 4,096.
This number can be calculated by the combination formula. The combination formula is used to calculate the number of combinations of a certain size that can be made from a larger set. In this case, the larger set is sixteen questions and the smaller set is eleven questions. The combination formula, written as C (n,r) is equal to n! / (r! (n-r)!), where n is the size of the larger set and r is the size of the smaller set. Applying this formula to the problem at hand, 16! / (11! (16-11)!), yields 4,096.
There are sixteen cards, and eleven of them must be chosen, so the number of different choices for eleven questions is 4,096.
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Suppose that y varies directly with x and y = 10 when x = 20. What is y when x = 15?
Given that,
y varies directly with x and y = 10 when x = 20.
To find,
The value of y when x = 15.
Solution,
\(y\propto x\\\\y=kx\)
Put x = 20 and y = 10
\(k=\dfrac{y}{x}\\\\k=\dfrac{10}{20}\\\\=\dfrac{1}{2}\)
Put k = 1/2 and x = 15 to find the value of y.
\(y=\dfrac{1}{2}\times 15\\\\=7.5\)
So, the value of y is 7.5
Absolute vale of x+2 if x is less than 2
Answer: The expression for the absolute value of x+2 when x is less than 2 is -(x+2).
Step-by-step explanation: When x is less than 2, x+2 is a negative number. The absolute value of a negative number is its opposite with the negative sign, so the absolute value of x+2 is -(x+2). Therefore, when x is less than 2, the expression for the absolute value of x+2 is -(x+2).
PLZ HELP PLEASE ITS A BIG GRADE PLEASE
Answer:
A: 4 Units Perimeter Is All The Sides not That Hard
Step-by-step explanation:
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the graph of the parabola y=3(x+5)2-2 has vertex (-5,-2). if this parabola is shifted 1 unit down and 6 units to the right, what is the equation of the new parabola?
Answer:
y = 3(x-1)^2 -3
Step-by-step explanation:
y = 3(x+5)^2 - 2
vertex: (-5, -2)
1 unit down :
-2 - 1 = -3
6 units to the right:
3(x+5-6)^2 -2
3(x-1)^2 - 2
make what's in the parenthesis equal to 0:
(x-1)^2 = 0
x = 1
or
-5 + 6 = 1
new vertex : (1, -3)
equation : y = 3(x-1)^2 -3
Translation involves shifting of points from one position to another. The equation of the new parabola is: \(y = 3(x - 1)^2 - 3\)
Given that:
\(y = 3(x + 5)^2 - 2\)
\((h,k) = (-5,-2)\) --- vertex
The general equation of a parabola is:
\(y = a(x - h)^2 + k\)
By comparison:
\(a = 3\\ h = -5 \\ k= -2\)
When the vertex is shifted 1 unit down, the rule is:
\((x,y) \to (x,y-1)\)
So, we have:
\((x,y) \to (-5,-2-1)\)
\((x,y) \to (-5,-3)\)
When the vertex is shifted 6 unit right, the rule is:
\((x,y) \to (x + 6,y)\)
So, we have:
\((h,k) \to (-5 + 6,-3)\)
\((h,k) \to (1,-3)\)
This means that:
\(h = 1\\ k =-3\)
Recall that:
\(a =3\)
Substitute these values in:
\(y = a(x - h)^2 + k\)
\(y = 3(x - 1)^2 - 3\)
Hence, the equation of the new parabola is: \(y = 3(x - 1)^2 - 3\)
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At midnight the temperature was ➖ 5 degrees Celsius. At noon, the temperature was 19 degrees Celsius. Which expression represents the change in temperature?
Answer:
cold and hot Is the answer
Referring to the figure, find the unknown
length. Write your answer in simplest form.
The length of unknown side c is 30 units.
Given,
The figure is a right angled triangle.
Altitude of the right angled triangle is given as 24 units.
Base of the right angled triangle is given as 18 units.
We have to find the unknown side of the right angled triangle which is its hypotenuse.
According to Pythagorean theorem:
Hypotenuse² = Altitude² + Base²
c² = 24² + 18²
c² = 576 + 324
c² = 900
c = 30
Therefore, the length of the unknown side c, which is the hypotenuse of the right angled triangle is 30 units.
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Tell whether each equation is true, false, or open. Explain.
2.) 5x+7 = 17
1.) 85+ (-10) = 95
3.) -8(-2) - 7 = 14 – 5
Answer:
1.) 85+ (-10) = 95 is false.
2.) 5x+7 = 17 is false.
3.) -8(-2) - 7 = 14 – 5 is false.
Step-by-step explanation:
1. is false because since 85 ≠ 95, then 85 + ( - 10) = 95 is false.
2. is false because since 12 ≠ 17, then 5x + 7 = 17 is false.
3. is false because since -15 ≠ 14, then -8( - 2) - 7 = 14–5 is false.
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A researcher is interested in the effects of sun exposure on mood in men and women. She splits her subjects up by gender, then randomly assigns them to high and low-exposure conditions (resulting in four total conditions: high-exposure men, high-exposure women, low-exposure men, and low-exposure women). This is an example of a:
One-group-pretest-protest
Mixed factorial design
\(\huge\underline\mathtt\colorbox{cyan}{One-group pretest-posttest}\)
Explanation:The one-group pretest-posttest design is a type of quasi-experiment in which the outcome of interest is measured 2 times: once before and once after exposing a non-random group of participants to a certain intervention/treatment.
The objective is to evaluate the effect of that intervention which can be:
A training program
A policy change
A medical treatment, etc.
54 is 150 of what number ?
Answer:
x=36
Step-by-step explanation:
If you are using a calculator, simply enter 54×100÷150, which will give you the answer.
Two horses start a race at the same time. Horse A gallops at a steady rate of 32 feet per second and Horse B gallops at a steady rate of 28 feet per second.
After 5 seconds, how much farther will Horse A have traveled?
Answer:
Answer: 20 feet further
Step-by-step explanation:
Answer:
horse A will have travel 20 feet farther than horse B
Please help! I'm stuck on this one :(
Answer:
[900, 1300]
Step-by-step explanation:
According to the empirical rule, 95% is within ±2 standard deviations.
1100 − 2(100) = 900
1100 + 2(100) = 1300
What is this area of this rectangle
Step-by-step explanation:
To find the area of a rectangle, we simply need to multiply its length by its width. The area is simply the width of a rectangle times height. Area = h × h, Area = h2. Hence for a square, the area is the square of its sides.
The average (A) of two numbers, m and n, is given by the formula A = m+t/2 . Find
the average of the two numbers 36 and 72.
The solution is
Answer:
\(54\)
Step-by-step explanation:
Formula for average A of two numbers m and n is:
\(A=\frac{m+n}{2}\)
Substitute the value m=36 and n=72
\(A=\frac{36+72}{2}\)
This is equivalent to:
\(A=\frac{108}{2}\)
The average is:
\(A=54\)
The price of an item has been reduced by $3.49. The new sale price is $31.38. What was the original price?
Add the amount it was reduced to the new price:
3.49 + 31.38 = 34.87
Original price = $34.87
Answer:
When finding the unknown original price, add the sale reduction to the new sale price to get the old sale price.
\(3.49 + 31.38 = 34.87\)
Therefore, the original sales price was 34.87
when the stretched string of the apparatus represented below is made to vibrate, point p does not move. point p is most probably at the location of
A node is a point of no displacement in a standing wave. Therefore, if point P does not move, it is most likely located at a node. the points along the wave that experience maximum displacement are called antinodes.
A node is a point of no displacement in a standing wave, meaning that if the stretched string of the apparatus represented is made to vibrate, point P will not move. Point P is most likely located at a node as it experiences no displacement. A standing wave is created when two waves combine and the resulting wave is stationary. The points along the wave that experience no displacement are called nodes, and the points along the wave that experience maximum displacement are called antinodes. Nodes can be found at points that are integral multiples of half the wavelength of the wave. Therefore, it can be concluded that point P is at a node since it does not move when the string is made to vibrate.
The complete question is :
When the stretched string of the apparatus represented below is made to vibrate, point p does not move. Point p is most probably at the location of _____.
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What is x equal to in the equation 10+x=5(1/5x+2)
Answer:
30
Step-by-step explanation:
Theorm
5. Write the equation of the line represented by the following graph (2 points):
Answer:
y=2/3x-2
Step-by-step explanation:
We can first find the y-intercept of the graph.
The y-intercept is the value of y when x is 0.
So, based on the graph, we can see that the y-intercept is at (0, -2).
Now, we need to find another point in order to find the slope of the line.
Another point would be (3,0).
This is also the x-intercept.
We can put these points into the formula for slope to find the slope of the given line.
The formula for slope is: y1-y2/x1-x2
In the formula:
y1 is the y coordinate of the first point
y2 is the y coordinate of the second point
x1 is the x coordinate of the first point
x2 is the x coordinate of the second point
We can plug our points in.
-2-0/0-3
Simplify.
-2/-3
2/3
The slope is 2/3.
Put it into an equation.
We can use the form y=mx+b
This is where m is the slope and b is the y-intercept.
We can put our corresponding values in.
The final equation would therefore be: y=2/3x-2.
Which distance formula or formulas show(s) a joint variation?
d=r•t , formula show a joint variation , Option C is the answer.
What is a Joint Variation ?The variation when a variable is dependent upon two or more variables and varies directly as either of them are taken constant is called Joint Variation.
According to the question , which among the options shows a Joint Variation.
Option 1 d = 50 t
It shows direct variation , as increase in t will directly increase d 50 times * the increase
Option 2 d = 16t²
It also shows direct square variation .
Option 3 d=r•t
In this variation d is dependent on two variables and so it is a joint variation.
Therefore Option C is the answer.
The complete question is ?
Which distance formula or formulas show a joint variation? d=50t, d=16t^2, d=r•t
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Let f left parenthesis x right parenthesis equals x cubed minus x squared minus 1 and x subscript 0 equals 1 . To the nearest three decimal places, find x subscript 5 using Newton's method of approximation.
The value of \(x_5\), rounded to the nearest decimal place using Newton's method of approximation is 1.466.
The correct answer is option B.
To find the value of \(x_5\)using Newton's method of approximation, we need to iterate the following formula:
\(x_(_n_+_1_) = x_n - f(x_n) / f'(x_n)\)
where f'(x) represents the derivative of the function f(x). Let's calculate the values step by step.
Given function: f(x) = \(x^3 - x^2 - 1\)
Step 1: Find the derivative of f(x)
f'(x) = \(3x^2 - 2x\)
Step 2: Initialize the starting value
\(x_0\)= 1
Step 3: Calculate \(x_1\)
\(f(x_0) = f(1) = (1^3) - (1^2) - 1 = 1 - 1 - 1 = -1\)
\(f'(x_0) = f'(1) = (3(1^2)) - (2(1)) = 3 - 2 = 1\)
\(x_1 = x_0 - f(x_0) / f'(x_0)\)
= 1 - (-1) / 1
= 2
Step 4: Calculate \(x_2\)
\(f(x_1) = f(2) = (2^3) - (2^2) - 1 = 8 - 4 - 1 = 3\)
\(f'(x_1) = f'(2) = (3(2^2)) - (2(2)) = 12 - 4 = 8\)
\(x_2 = x_1 - f(x_1) / f'(x_1)\)
= 2 - 3 / 8
= 1.625
Step 5: Repeat the process until we reach \(x_5\)
Performing the calculations for \(x_3, x_4, andx_5\), we find:
\(x_3\) ≈ 1.465
\(x_4\) ≈ 1.466
\(x_5\)≈ 1.466
After applying Newton's method of approximation to the function f(x) = \(x^3 - x^2 - 1,\)starting with,\(x_0 = 1,\) we iteratively calculated the values of \(x_1, x_2, x_3, x_4, and x_5\). The final approximation, \(x_5\), is approximately 1.466 when rounded to the nearest decimal place. This aligns with option B as the correct answer.
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What’s the total distance he travels
Answer:
10
Step-by-step explanation:
find the length of each leg, being 6 and 8. Then use the pythagorean theorem a^2+b^2=c^2
6^2+8^2=c^2
36+64=c^2
100=c^2
c=10
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Write as many equations as possible that could represent the relationship between
the ages of the two children in each family described. Be prepared to explain what each part of your equation represents.
a.) In Family A, the youngest child is 7 years younger than the oldest, who is 18.
b.) In Family B, the middle child is 5 years older than the youngest child.
Answer:
a) Here we have:
"In Family A, the youngest child is 7 years younger than the oldest, who is 18"
Let's define:
Y = age of the youngest child.
O = age of the oldest child.
Then we know that:
Y = O - 7
O = 18
Then we can replace the second equation into the first one:
Y = 18 - 7 = 11.
b) Here we have:
"In Family B, the middle child is 5 years older than the youngest child."
Let's define:
Y = age of the youngest child.
M = age of the middle child.
Here we have only one equation:
M = Y + 5.
Answer:
your answer is....
Step-by-step explanation:M = Y + 5.
A) Find an equation for the line perpendicular to the tangent line to the curve y=x^3-4x+6 at the point (2,6)
-The equation is y=
b) What is the smallest slope on the curve? At what point on the curve does the curve have this slope?
-The smallest slope on the curve is
-The curve has the smallest slope at the point
c) Find equations for the tangent lines to the curve at the points where the slope of the curve is 8.
Answer:
f(x) = x³ - 4x + 6
f'(x) = 3x² - 4
a) f'(2) = 3(2²) - 4 = 12 - 4 = 8
6 = 8(2) + b
6 = 16 + b
b = -10
y = 8x - 10
b) 3x² - 4 = 0
3x² = 4, so x = ±2/√3 = ±(2/3)√3
= ±1.1547
f(-(2/3)√3) = 9.0792
f((2/3)√3) = 2.9208
c) 3x² - 4 = 8
3x² = 12
x² = 4, so x = ±2
f(-2) = (-2)³ - 4(-2) + 6 = -8 + 8 + 6 = 6
6 = -2(8) + b
6 = -16 + b
b = 22
y = 8x + 22
f(2) = 6
y = 8x - 10
The equation perpendicular to the tangent is y = -1/8x + 25/4
-The smallest slope on the curve is 2.92
The curve has the smallest slope at the point (1.15, 2.92)
The equations at tangent points are y = 8x + 16 and y = 8x - 16
Finding the equation perpendicular to the tangentFrom the question, we have the following parameters that can be used in our computation:
y = x³ - 4x + 6
Differentiate
So, we have
f'(x) = 3x² - 4
The point is (2, 6)
So, we have
f'(2) = 3(2)² - 4
f'(2) = 8
The slope of the perpendicular line is
Slope = -1/8
So, we have
y = -1/8(x - 2) + 6
y = -1/8x + 25/4
The smallest slope on the curveWe have
f'(x) = 3x² - 4
Set to 0
3x² - 4 = 0
Solve for x
x = √[4/3]
x = 1.15
So, we have
Smallest slope = (√[4/3])³ - 4(√[4/3]) + 6
Smallest slope = 2.92
So, the smallest slope is 2.92 at (1.15, 2.92)
The equation of the tangent lineHere, we set f'(x) to 8
3x² - 4 = 8
Solve for x
x = ±2
Calculate y at x = ±2
y = (-2)³ - 4(-2) + 6 = 6: (-2, 0)
y = (2)³ - 4(2) + 6 = 6: (2, 0)
The equations at these points are
y = 8x + 16
y = 8x - 16
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1.7% as a fraction in simplest form
Answer:
\( \sf1.7\% \\ = \sf \frac{1.7}{100} \\ = \boxed{ \bold{ \red{\frac{17}{1000} }}}\)
Hope it helps!!
Let register x5 hold the hex number 01a74cd0hex and let x6 hold the hex number 00467b44 hex. If this information is enough, then answer the following three questions, else explain what extra information you may need to proceed.a) (1.5 pts) What are the contents of register x31 after the following instruction is executed? addi x31, x5, 32
b) (1.5pts) What would be the contents of register x28 after the following instruction is executed? add x28, x5, x6
c) (1.5 pts) What would be the contents of register x29 after the following instruction is executed? ld x29, 1056decimal (x30)
d) (1.5 pts) What would be the contents of register x29 after the following instruction is executed? sd x28, 16 decimal (x29)
a) The contents of register x31 after the following instruction is executed? addi x31, x5, 32 is 277968880 decimal.
b) The contents of register x28 after the following instruction is executed is 278427652 decimal.
c)The contents of register x29 after the following instruction is executed is 1056 decimal
d) The contents of register x29 after the following instruction is executed is 278427652 decimal
In computer architecture, registers are special storage locations within the CPU that hold data temporarily. These registers can hold values in binary, decimal, or hexadecimal formats. In this scenario, we have two registers x5 and x6, holding hexadecimal values. We will use these values to answer the given questions about register manipulation.
a) To determine the contents of register x31 after the instruction "addi x31, x5, 32", we need to add 32 to the value in x5 and store the result in x31. The value in x5 is 01a74cd0hex, which is equivalent to 277968848 in decimal. Adding 32 to this value, we get 277968880 decimal. Therefore, the contents of register x31 after the instruction is executed would be 277968880 decimal.
b) To determine the contents of register x28 after the instruction "add x28, x5, x6", we need to add the values in x5 and x6 and store the result in x28. The value in x5 is 01a74cd0hex, which is equivalent to 277968848 decimal. The value in x6 is 00467b44hex, which is equivalent to 458804 decimal. Adding these two values, we get 278427652 decimal. Therefore, the contents of register x28 after the instruction is executed would be 278427652 decimal.
c) To determine the contents of register x29 after the instruction "ld x29, 1056decimal (x30)", we need to load the value at memory location 1056decimal + the value in register x30 into register x29. The value in register x30 is not given in this scenario, so we cannot determine the contents of register x29.
d) To determine the contents of register x29 after the instruction "sd x28, 16 decimal (x29)", we need to store the value in register x28 at memory location 16 decimal + the value in register x29. The contents of register x28 is 278427652 decimal
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A soccer ball travels upward from a height of 11 feet with an initial velocity of 20
feet per second. The quadratic function h (t) = -16t² + 20t+11 models the height
of the ball, where h (t) is the height, in feet, of the soccer ball and t is the time that
ball has been in the air, in seconds. When is the soccer ball above 15 feet?
A. The soccer ball is above 15 feet between 0 seconds and 0.5 second.
B. The soccer ball is above 15 feet between 0.5 second and 1 second.
C. The soccer ball is above 15 feet between 0:25 second and 1
second.
D. The soccer ball is above 15 feet between 0 seconds and 0.25 second.
The soccer ball is above 15 feet between 0.25 seconds and 1 second.
To determine when the soccer ball is above 15 feet, we need to find the values of t that satisfy the inequality h(t) > 15.
Given the quadratic function h(t) = -16t² + 20t + 11, we can rewrite the inequality as follows:
-16t² + 20t + 11 > 15
Subtracting 15 from both sides:
-16t² + 20t - 4 > 0
Simplifying further:
-16t² + 20t - 4 = -4(4t² - 5t + 1) = -4(t - 1)(4t - 1) > 0
Now, we can solve for t by finding the values that make the inequality true. We have two factors: (t - 1) and (4t - 1).
Setting each factor greater than zero and solving for t:
t - 1 > 0 => t > 1
4t - 1 > 0 => 4t > 1 => t > 1/4
So, we have t > 1 and t > 1/4. To satisfy both conditions, t must be greater than the maximum of 1 and 1/4, which is 1.
Therefore, the soccer ball is above 15 feet for t > 1 second.
The correct answer is:
C. The soccer ball is above 15 feet between 0.25 seconds and 1 second.
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