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2 years ago

A parade traveled 12 miles in 3 hours. How far did the parade travel per minute?

Mathematics

1 answer:

Ratling [72]2 years ago

7 0

Answer:

176 feet per minute

Step-by-step explanation:

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Determine the direction that is parábola opens: y=4x^2+8x+2

marissa [1.9K]

It should open up because the a which is 4 is positive

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3 years ago

PLEASE ANSWER ASAP FOR BRANLEST!!!!!!!!!!!!!!!<br> Solve the equation<br> b/4 +2 = -1<br> what is b?

BARSIC [14]

The answer is b=-12

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3 years ago

Find the sum of 12 out of 7 plus 8 out of 7

Viktor [21]

12/7 + 8/7

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3 years ago

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A ladder is leaning against a vertical wall. The distance from the top of

4vir4ik [10]

Answer:

23.5(to the nearest tenth)

Step-by-step explanation:

pythagoras theorem states

c^2= a^2+ b^2

a= 23

b=5

c^2= 529+25

c^2= 554

c= square root of 554

c = 23.5 (approximately)

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3 years ago

A juggler tosses a ball into the air . The balls height, h and time t seconds can be represented by the equation h(t)= -16t^2+40

malfutka [58]

PART A

The given equation is

$h(t) = - 16 {t}^{2} + 40t + 4$

In order to find the maximum height, we write the function in the vertex form.

We factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + 4$

We add and subtract

$- 16(- \frac{5}{4} )^{2}$

to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + - 16( - \frac{5}{4})^{2} - -16( - \frac{5}{4})^{2} + 4$

We again factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4})^{2} ) - -16( - \frac{5}{4})^{2} + 4$

This implies that,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4}) ^{2} ) + 16( \frac{25}{16}) + 4$

The quadratic trinomial above is a perfect square.

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +25+ 4$

This finally simplifies to,

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +29$

The vertex of this function is

$V( \frac{5}{4} ,29)$

The y-value of the vertex is the maximum value.

Therefore the maximum value is,

$29$

PART B

When the ball hits the ground,

$h(t) = 0$

This implies that,

$- 16 ( t- \frac{5}{4}) ^{2} +29 = 0$

We add -29 to both sides to get,

$- 16 ( t- \frac{5}{4}) ^{2} = - 29$

This implies that,

$( t- \frac{5}{4}) ^{2} = \frac{29}{16}$

$t- \frac{5}{4} = \pm \sqrt{ \frac{29}{16} }$

$t = \frac{5}{4} \pm \frac{ \sqrt{29} }{4}$

$t = \frac{ 5 + \sqrt{29} }{4} = 2.60$

or

$t = \frac{ 5 - \sqrt{29} }{4} = - 0.10$

Since time cannot be negative, we discard the negative value and pick,

$t = 2.60s$

8 0

3 years ago