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2 years ago

Help plz will give brainliest

Mathematics

2 answers:

lyudmila [28]2 years ago

8 0

There are 5 campgrounds
Belleville is farthest south
The volcanoes name is sue

grin007 [14]2 years ago

7 0

Answer:t the farthest city south is bellville and the volcano name is sue

Step-by-step explanation:

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If f(x) = -7x + 5x2 + 10, what does f(-2) equal?

Iteru [2.4K]

Answer:

Step-by-step explanation:

f(x) = -7x + 5x^2 + 10

f(-2) = -7(-2) + 5(-2)^2 + 10

= 14 + 20 + 10

= 44

6 0

2 years ago

A post 8 feet tall casts a shadow 12 feet long at the same time that a flagpole casts a shadow 96 feet long, how tall is the fla

lara [203]

Answer:

64 ft

Step-by-step explanation:

The shadow of the post is 12 feet long. The shadow of the flagpole is 96 feet long.

The shadow of the flagpole is 8 times longer than the shadow of the post.

8 x 8 = 64

8 0

2 years ago

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I fr need help I don’t know the answer

EastWind [94]

Answer:The answer is D

Step-by-step explanation:

3 0

3 years ago

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Solve for n.<br> 4 x n= 21.6

Kay [80]

Answer:

n = 5.4

Step-by-step explanation:

Equation: 4n = 21.6

To solve this equation for n, divide both sides by 4 to get n by itself.

<u />

<u>Step 1: Divide both sides by 4</u>

4 * n / 4 = 21.6 / 4

n = 5.4

Answer: n = 5.4

3 0

2 years ago

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A projectile is fired with muzzle speed 250 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe

qaws [65]

The projectile's horizontal and vertical positions at time $t$ are given by

$x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ\,t$

$y=30\,\mathrm m+\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ\,t-\dfrac g2t^2$

where $g=9.8\dfrac{\rm m}{\mathrm s^2}$ . Solve $y=0$ for the time $t$ it takes for the projectile to reach the ground:

$30+\dfrac{250}{\sqrt2}t-4.9t^2=0\implies t\approx36.2458\,\mathrm s$

In this time, the projectile will have traveled horizontally a distance of

$x=\dfrac{250\frac{\rm m}{\rm s}}{\sqrt2}(36.2458\,\mathrm s)\approx6400\,\mathrm m$

The projectile's horizontal and vertical velocities are given by

$v_x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ$

$v_y=\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ-gt$

At the time the projectile hits the ground, its velocity vector has horizontal component approx. 176.77 m/s and vertical component approx. -178.43 m/s, which corresponds to a speed of about $\sqrt{176.77^2+(-178.43)^2}\dfrac{\rm m}{\rm s}\approx250\dfrac{\rm m}{\rm s}$ .

7 0

2 years ago