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3 years ago

You earn $596.92 per week.

Mathematics

1 answer:

mash [69]3 years ago

5 0

Answer:

your net pay is $441.84

Step-by-step explanation:

596.92 - 155.08 = 441.84

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ABCD is an isosceles trapezoid.

Eduardwww [97]

Answer:

∡b = 53°

∡d = 127°

∡c = 127°

Step-by-step explanation:

in trapezoids, the two top angles are equal and the two bottom are equal

also, there are 360 degrees in any 4-sided polygon

53 + 53 + x + x = 360

2x + 106 = 360

2x = 254

x = 127

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3 years ago

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A pair of parallel lines is cut by a transversal as shown in the figure. Match the angles with their corresponding angles.

garri49 [273]

D and H are corresponding
G and C are corresponding
E and A are corresponding
B and F are corresponding

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Which equation demonstrates the distributive property?

PIT_PIT [208]

B because distributive property goes like this:
a(b+c)
a(b)+a(c)
That's just an example though

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Show the ratios are equivalent by simplifying any 4 of them.

aliya0001 [1]

When simplified all of them equal 3/1. They are asking you to just chose 4 of them. Hope that helps!

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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