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3 years ago

Please help its a quiz

Mathematics

2 answers:

Inessa [10]3 years ago

6 0

Answer:

What are each of the drop downs love so i can help u?

Irina-Kira [14]3 years ago

6 0

-10+15=5. But I don’t know what the rest of the drop downs are.

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Everado paid $1.12 for a dozen eggs at his local grocery store. Determine the cost of 3 eggs.

Setler [38]

First you would divide 1.12 by 12 which is 0.093 repeating 3 times 3 to get the cost of three eggs.Therefore the answer would be he paid 28 cents for 3 eggs.

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4 years ago

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in a week karry ran 9.3 miles and Tricia ran 4.4 miles. Use mental math to find how much farther Karry ran than Tricia. Explain

Ede4ka [16]

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4 years ago

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Which equation is correct for the perpendicular bisector of the line segment whose endpoints are (−1,1) and (7,−5)?

kherson [118]

Slope = (-5-1)/(7+1) = -6/8 = -3/4
Perpendicular slope = 4/3

Midpoint = ((-1+7)/2 , (1+(-5))/2) = (3, 3)

equation:
y = mx + c
y = 4/3 x + c

at point (3, 3)
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6 0

3 years ago

A rock thrown vertically upward from the surface of the moon at a velocity of 36m/sec reaches a height of s = 36t - 0.8 t^2 met

Verdich [7]

Answer:

a. The rock's velocity is $v(t)=36-1.6t \:{(m/s)}$ and the acceleration is $a(t)=-1.6 \:{(m/s^2)}$

b. It takes 22.5 seconds to reach the highest point.

c. The rock goes up to 405 m.

d. It reach half its maximum height when time is 6.59 s or 38.41 s.

e. The rock is aloft for 45 seconds.

Step-by-step explanation:

Velocity is defined as the rate of change of position or the rate of displacement. $v(t)=\frac{ds}{dt}$
Acceleration is defined as the rate of change of velocity. $a(t)=\frac{dv}{dt}$

The rock's velocity is the derivative of the height function $s(t) = 36t - 0.8 t^2$

$v(t)=\frac{d}{dt}(36t - 0.8 t^2) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(36t\right)-\frac{d}{dt}\left(0.8t^2\right)\\\\v(t)=36-1.6t$

The rock's acceleration is the derivative of the velocity function $v(t)=36-1.6t$

$a(t)=\frac{d}{dt}(36-1.6t)\\\\a(t)=-1.6$

b. The rock will reach its highest point when the velocity becomes zero.

$v(t)=36-1.6t=0\\36\cdot \:10-1.6t\cdot \:10=0\cdot \:10\\360-16t=0\\360-16t-360=0-360\\-16t=-360\\t=\frac{45}{2}=22.5$

It takes 22.5 seconds to reach the highest point.

c. The rock reach its highest point when t = 22.5 s

Thus

$s(22.5) = 36(22.5) - 0.8 (22.5)^2\\s(22.5) =405$

So the rock goes up to 405 m.

d. The maximum height is 405 m. So the half of its maximum height = $\frac{405}{2} =202.5 \:m$

To find the time it reach half its maximum height, we need to solve

$36t - 0.8 t^2=202.5\\36t\cdot \:10-0.8t^2\cdot \:10=202.5\cdot \:10\\360t-8t^2=2025\\360t-8t^2-2025=2025-2025\\-8t^2+360t-2025=0$

For a quadratic equation of the form $ax^2+bx+c=0$ the solutions are

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=-8,\:b=360,\:c=-2025:\\\\t=\frac{-360+\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2-\sqrt{2}\right)}{4}\approx 6.59\\\\t=\frac{-360-\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2+\sqrt{2}\right)}{4}\approx 38.41$

It reach half its maximum height when time is 6.59 s or 38.41 s.

e. It is aloft until s(t) = 0 again

$36t - 0.8 t^2=0\\\\\mathrm{Factor\:}36t-0.8t^2\rightarrow -t\left(0.8t-36\right)\\\\\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\\\\t=0,\:t=45$

The rock is aloft for 45 seconds.

5 0

4 years ago

10 POINTS ALL FOR YOU IF YOU ANSWER IT

Gennadij [26K]

Answer:

x is equal to 30

<ABC is equal to 100°

<DBC is equal to 80°

Step-by-step explanation:

3x +10 +2x +20 =180

5x = 180 - 10- 20

5x = 150, divide both sides by 5

x is equal to 30

<ABC = 3x + 10 = 3 × 30 + 10 = 100°

<DBC = 180 - <ABC = 180 - 100 = 80°

3 0

2 years ago