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stepan [7]
3 years ago
10

Serena has attached a 10 inch ribbon to the corners of a frame to hang it on the wall. The frame 9 inches wide. How far above th

e top of the frame will the hook need to be? Round

Mathematics
1 answer:
kherson [118]3 years ago
4 0

Answer: The hook would be 2.2 inches (approximately) above the top of the frame

Step-by-step explanation: Please refer to the picture attached for further details.

The top of the picture frame has been given as 9 inches and a 10 inch ribbon has been attached in order to hang it on a wall. The ribbon at the point of being hung up would be divided into 5 inches on either side (as shown in the picture). The line from the tip/hook down to the frame would divide the length of the frame into two equal lengths, that is 4.5 inches on either side of the hook. This would effectively give us two similar right angled triangles with sides 5 inches, 4.5 inches and a third side yet unknown. That third side is the distance from the hook to the top of the frame. The distance is calculated by using the Pythagoras theorem which states as follows;

AC^2 = AB^2 + BC^2

Where AC is the hypotenuse (longest side) and AB and BC are the other two sides

5^2 = 4.5^2 + BC^2

25 = 20.25 + BC^2

Subtract 20.25 from both sides of the equation

4.75 = BC^2

Add the square root sign to both sides of the equation

2.1794 = BC

Rounded up to the nearest tenth, the distance from the hook to the top of the frame will be 2.2 inches

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Marlene rides her bike at a rate of 16 miles per hour. The time in hours that she rides is represented by the variable t, and th
Vesna [10]

Question is Incomplete; Complete question is given below;

Marlene rides her bike at a rate of 16 miles per hour. The time in hours that she rides is represented by the variable t, and the distance she rides is represented by the variable d. Which statements are true of the scenario? Check all that apply.

a. The independent variable, the input, is the variable d, representing distance.

b. The distance traveled depends on the amount of time Marlene rides her bike.

c. The initial value of the scenario is 16 miles per hour.

d. The equation t = d + 16 represents the scenario.

e. The function f(t) = 16t represents the scenario.

Answer:

OPTION (b) and (e) are the true statements.

Step-by-step explanation:

Given:

t=> time in hours that she rides

d=>  distance she rides

We have to find the statements that are true of the scenario.

Solution:

d=16t

The equation itself is a linear equation which presents an equation's scenario and time (t) is a independent variable and distance (d) is a dependent variable.

Equation for distance according to function is  f(t)=16t

Here, we can use graph tool, you can see the attested image below.

Function's domain is the interval⇒ [0,∞)

t\geq0

Function's range is the interval⇒ [0,∞)

f(t)\geq0

The statements which are TRUE are explained below :

b) The distance traveled depends on the amount of time Marlene rides her bike

It is true because as explained above that equation for distance according to function is  f(t)=16t

e) The function f(t) = 16t represents the scenario.

It is true because as explained that the scenario has been represented by the function above that f(t)=16t

* The statement (a) is false because the variable t is the independent variable.

* The statement (c) is false because 16 is been represented as the slope or the rate of the linear equation.

* The statement (d) is false because t = d + 16 donot represents the scenario.

8 0
3 years ago
A basketball hoop has a circumference of 56.5 cm.what is its diameter
Bingel [31]
It's diameter is 17.98. (I'm pretty sure I'm right I even used a calculator, if not sorry)
4 0
3 years ago
Read 2 more answers
Can someone please answer 8|x+2|-6=5|x+2|+3
Anettt [7]

Step-by-step explanation:

8|x+2|-6=5|x+2|+3

|8x+16|-6=|5x+10|+3

I've been thinking... Is that | or (? Are you sure it's |?

Oh well, I'm continuing...

8x+16-6=5x+10+3

8x+10=5x+13

3x=3

x=1

4 0
4 years ago
The points (0, 3), (4, 0), (9, 0), (0, 11), and (14, 0) are plotted on the graph. On a coordinate plane, points (0, 3), (4, 0),
Advocard [28]

Answer:

can you take a picture it might make it easier for me

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3 years ago
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Suppose that you had the following data set. 500 200 250 275 300 Suppose that the value 500 was a typo, and it was suppose to be
hodyreva [135]

Answer:

\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305

s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108

\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105

s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221

The absolute difference is:

Abs = |340.221-115.108|= 225.113

If we find the % of change respect the before case we have this:

\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%

So then is a big change.

Step-by-step explanation:

The subindex B is for the before case and the subindex A is for the after case

Before case (with 500)

For this case we have the following dataset:

500 200 250 275 300

We can calculate the mean with the following formula:

\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305

And the sample deviation with the following formula:

s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108

After case (With -500 instead of 500)

For this case we have the following dataset:

-500 200 250 275 300

We can calculate the mean with the following formula:

\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105

And the sample deviation with the following formula:

s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221

And as we can see we have a significant change between the two values for the two cases.

The absolute difference is:

Abs = |340.221-115.108|= 225.113

If we find the % of change respect the before case we have this:

\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%

So then is a big change.

8 0
3 years ago
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