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andrezito [222]
4 years ago
8

Which of the following equations describes the line shown below? Check all

Mathematics
1 answer:
Brilliant_brown [7]4 years ago
6 0

Answer:

y = 7 x + 15

Step-by-step explanation:

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A bag of grass seed weighs 8 pounds.
lisabon 2012 [21]
The answer is a hope this helps
7 0
3 years ago
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Solve the system by substitution.<br> 2.5x-3y=-13<br> 3.25x-y=-14
MrMuchimi

Answer:

x=-4

y=1

Step-by-step explanation:

For the given system of equations,

2.5x-3y=-13.....eqn(1)

3.25x-y=-14.....eqn(2)

We first make y the subject of eqn (2)

This implies that,

y=3.25x+14.....eqn(3)

We substitute eqn (3) into eqn (1)

2.5x-3(3.25x+14)=-13

2.5x-9.75x-42=-13

-42+13=9.75x-2.5x

-29=7.25x

Dividing through by 7.25, we obtain

x=-4

We now put the value of x into eqn (2)

3.25(-4)-y=-14

This implies that,

-13+14=y

y=1

4 0
3 years ago
HELP FOR POINTSS
brilliants [131]
The answer would be
C = 10
P = 20
3 0
4 years ago
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Helppppppppppp me please
Scilla [17]

Answer:

18.85 rounded

Step-by-step explanation:

The length of an arc depends on the radius of a circle and the central angle Θ. We know that for the angle equal to 360 degrees (2π), the arc length is equal to circumference. Hence, as the proportion between angle and arc length is constant, we can say that:

L / Θ = C / 2π

As circumference C = 2πr,

L / Θ = 2πr / 2π

L / Θ = r

We find out the arc length formula when multiplying this equation by Θ:

L = r * Θ

Hence, the arc length is equal to radius multiplied by the central angle (in radians).

7 0
3 years ago
The radii of Earth and Pluto are 6,371 kilometers and 1,161 kilometers, respectively. Approximately how many spheres the size of
rosijanka [135]
\bf \qquad \qquad \textit{ratio relations}&#10;\\\\&#10;\begin{array}{ccccllll}&#10;&Sides&Area&Volume\\&#10;&-----&-----&-----\\&#10;\cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3}&#10;\end{array} \\\\&#10;-----------------------------\\\\&#10;\cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\&#10;-------------------------------\\\\

\bf \cfrac{\textit{earth's volume}}{\textit{pluto's volume}}\qquad \cfrac{s^3}{s^3}\implies \cfrac{6371^3}{1161^3}\implies \cfrac{258596602811}{1564936281}\approx 165.24417
8 0
4 years ago
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