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3 years ago

The width of a soccer goal is 7 m. how wide is the goal in centimeters ?

1 answer:

4 0

We know, 1 m = 100 cm
Multiply both sides by 7,
7 m = 100 * 7 = 700 cm

In short, Your Answer would be 700 cm

Hope this helps!

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Find the length of the line segment that connects the points (7, 16) and (47, 140)

tresset_1 [31]

Answer:

$length = \sqrt{ {(47 - 7)}^{2} + {(140 - 16)}^{2} } \\ = \sqrt{ {40}^{2} + {124}^{2} } \\ = \sqrt{16976} \\ = 130.29 \: units$

6 0

2 years ago

BRAINLIEST Look at the pattern and decide what number comes next.<br> 2 7 3 8 4 9 5 ?

baherus [9]

Answer:

Step-by-step explanation:

2 + 5 = 7

3 + 5 = 8

4 + 5 = 9

5 + 5 = 10

4 0

2 years ago

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What is the equation of the function shown in the graph?

Zanzabum

Answer:

Step-by-step explanation:

I do not now what you mean?

5 0

2 years ago

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$

8 0

2 years ago

If f(x) =2x²-x, find f(-4)-f(9)

sergeinik [125]

F(-4) = 36
f(9) = 153

3 0

2 years ago