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AlexFokin [52]
3 years ago
12

Can someone help me with this one

Mathematics
1 answer:
Nostrana [21]3 years ago
6 0
It’s name would be a monomial because it only has 1 term. (Terms are departed by addition or subtraction sign) The degree would be the highest exponent in the polynomial, in this case we only have 1 term so the degree of that term is 2
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There would be 60 medium shirts

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Justin Chan bought a Scion car for a price of $8,200, putting down $800 and financing the remainder with 60 monthly payments of
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What is the slope of the line (9,-6) and (-6- -9)?
Vlada [557]

Answer:

1/5

Step-by-step explanation:

Slope formula = (y2-y1)/(x2-x1)

((-6)- (-9))/ ((9)-(-6)

(-6 + 9)/ (9 + 6)

3/15

1/5

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Which of these answers is closest to zero-2, 0.2, 1, 1/2
Dmitry_Shevchenko [17]
0.2 is closest to zero
4 0
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
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