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2 years ago

Find the distance FG between the points F (0,-7) and G (-6,-3) . Round your answer to the nearest tenth, if necessary

Mathematics

1 answer:

Mrrafil [7]2 years ago

7 0

Answer:

7.2

Step-by-step explanation:

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50 POINTS !! PLEASE HELP !! ILL GIVE BRAINLIEST TO THE RIGHT ANSWERS.

Musya8 [376]

Answer:

The length of the missing leg is 56 meters.

Step-by-step explanation:

Use the Pythagorean theorem to figure out the leg of the missing leg.

Formula: a^2+b^2=c^2, given a=leg, b=leg, and c=hypotenuse

The hypotenuse is 65 meters, and one of the legs is 33 meters, so plug it into the formula and solve for a.

a^2+33^2=65^2

a^2+1089=4225

a^2=3136

Take the square root of both sides, and you get 56 meters.

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2 years ago

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The average distance from Mars to the sun is about one hundred forty- one million, six hundred twenty thousands miles. How is th

gavmur [86]

Answer: 141,620,000

Step-by-step explanation:

one hundred forty-one million, six hundred twenty, thousands

↓ ↓ ↓

141, 620, 000

5 0

3 years ago

What is the difference between gcf (greatest common factor) and lcm (least common multiple)?

WARRIOR [948]

The GCF is based upon what integer divides evenly into two numbers; the LCM is based upon what integer two numbers share in a list of multiples. The GCF must be a prime number; the LCM must be a composite number.

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2 years ago

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Drag the numbers shown to their approximate locations on the number line

pychu [463]

Answer:

Step-by-step explanation:

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