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3 years ago

Is there a relationship between the distance and the sum? Is there a relationship between the distance and the difference

1 answer:

4 0

The distance is always the opposite of the difference. The distance is exactly the difference. The distance is the absolute value of the difference. The distance is not related to difference.

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A telephone pole casts a 10 foot shadow. A horse that stands 5.5 feet tall casts a 2.5 foot shadow. How tall is the pole?

Makovka662 [10]

Answer:

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6 0

3 years ago

Solve this inequality: 3b – 7 < 32

butalik [34]

It's a lot like solving equations.

3b-7 < 32
Add 7 to each side.
3b < 39
Divide by 3 on each side.
b < 13

Hope this helps :)

6 0

3 years ago

Read 2 more answers

Consider the original complex figure and the reduction.

Dmitriy789 [7]

Answer:

1/8

Step-by-step explanation:

I'm going to try to explain this as easy as possible. What I did was take the original shape and divide it by the new shape. For this question, I solved it by dividing 32(the original base) by 4(the new base) and got 8. So the scale factor of the reduction was 1/8.

8 0

2 years ago

a circle has a center (3,5) and a diameter AB. The coordinates of A are (-4,6). what are the coordinates of B?

Kruka [31]

Find the length of the radius.

$r= \sqrt{(3-(-4))^2+(5-6)^2}
\\r = \sqrt{7^2+(-1)^2}
\\r= \sqrt{50} 
\\ r=5 \sqrt{2}$

Find the length of the diameter.

d = 2r = 2 × 5√2 = 10√2

Point B must lie on a line AC, where C is a center of a circle.
Find equation of line AC.
A(–4, 6), C(3, 5)

$y-y_1= \frac{y_2-y_1}{x_2-x_1}(x-x_1)
\\y-6= \frac{5-6}{3-(-4)} (x-(-4))
\\y-6=- \frac{1}{7} (x+4)
\\7y-42=-x-4
\\x+7y-38=0$

The distance from B(x, y) to C(3, 5) is 5√2.
$\sqrt{(x-3)^2+(y-5)^2}=5 \sqrt{2} 
 \\(x-3)^2+(y-5)^2=(5 \sqrt{2})^2 
\\(x-3)^2+(y-5)^2=50$

Solve system of equations.
$x+7y-38=0
\\(x-3)^2+(y-5)^2=50
\\
\\x=38-7y
\\(38-7y-3)^2+(y-5)^2=50
\\(35-7y)^2+(y-5)^2=50
\\1225-490y+49y^2+y^2-10y+25-50=0
\\50y^2-500y+1200=0
\\y^2-10y+24=0
\\y^2-6y-4y+24=0
\\y(y-6)-4(y-6)=0 \\(y-6)(y-4)=0 \\y_1=6,y_2=4
$

$x_1=38-7y_1=38-7 \times 6 = 38-42=-4
\\x_2=38-7y_2=38-7 \times 4 = 38-28=10
$

Point B could have coordinates
$\\
\\(x_1,y_1)=(-4,6),(x_2,y_2)=(10,4)
$

But (–4, 6) are the coordinates of point A.
Therefore, point B has coordinates (10,4).

8 0

3 years ago

Can someone please help me on this question I'm already having a bad day and I just want to get this stupid math over with

balu736 [363]

Answer:

Option A

Step-by-step explanation:

To check which function satisfies the graph, substitute the points. Points: $(1, 12)$ is on the graph. Substitute it on the functions given in the options.

Option A: $f(1) = 20 { (\ \frac{3}{5} )\ }^1$

$\implies 4 \times 3 = 12$

Other options do not fit this $(1,12)$ . Hence Option A is our answer.

5 0

3 years ago