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3 years ago

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each Locker requires 3 ft of space how many lockers can the school build with 43 ft of space how much space will the school have

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german3 years ago

6 0

This should be the correct answer

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Directed line segment AB has endpoints A(2,3) and B(8,15). Point P has coordinated P(6,11). What ratio does P partition AB into?

Vladimir79 [104]

Refer to the graph shown below. It confirms that points A, B and P are
collinear.

Calculate the length of AP (as a).
a = √[(6 - 2)² + (11 - 3)²]
= 8.9443

Calculate the length of BP (as b).
b = √[(8 - 6)² + (15 - 11)²]
= 4.4721

Calculate ratio a/b.
a/b = 8.9443/4.4721 = 2
Therefore P partitions AB in a 2:1 ratio so that AP = 2*BP.

Answer: 2:1 ratio.

7 0

2 years ago

Un estanque tiene 13/2 litros de leche y se le agregan 87/10. ¿Cuánta leche quedó en el estanque? ¿Sí en el estanque caben 65/4

Katarina [22]

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3 0

2 years ago

Denise has a stool with a circle seat on top. The diameter of the seat is 14inches, and she wants to know the area of the top in

saul85 [17]

Answer:21?

Step-by-step explanation:

7 0

3 years ago

What is the midpoint of the segment shown below?

kolbaska11 [484]

The midpoint of the segment is (-15/2, -15/2)

<h3>How to determine the midpoint?</h3>

The complete question is in the attached image

The points are given as:

(-8, -7) and (-7, -8)

The midpoint is calculated as:

(x,y) = 1/2 * (x1 + x2, y1 + y2)

So, we have:

(x,y) = 1/2 * (-8 - 7, -7 - 8)

Evaluate the difference

(x,y) = 1/2 * (-15, -15)

Evaluate the product

(x,y) = (-15/2, -15/2)

Hence, the midpoint of the segment is (-15/2, -15/2)

Read more about midpoints at:

brainly.com/question/4747771

#SPJ1

8 0

1 year ago

The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find

Serga [27]

Answer:

Required solution is $y(x)=1+A\cos x+B\sin x$ where A and B are constants.

Step-by-step explanation:

Given nonhomogenous differential equation is,

$y''(x)+y'(x)=1\hfill (1)$ with $y_1(x)=1$

To find another solution, consider $m=\frac{\partial}{\partial x}$ be such that,

$m^2+1=0\implies m=\pm i$

Hence,

$C.F=A\cos x+B\sin x$ where A and B are constants.

Let $D=\frac{\partial}{\partial x}$

$P.I=\frac{1}{1+D^2}(1)$

$=(1+D^2)^{-1}(1)$

$=(1-D^2+.....)(1)=1$

Hence,

$y(x)=1+A\cos x+B\sin x$ where A and B are constants.

which is required solution.

6 0

3 years ago