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3 years ago

Find the zeros for (x+7) (x+5)+0

Mathematics

1 answer:

JulijaS [17]3 years ago

3 0

Steps to solve for the zero:

(x + 7)(x + 5) = 0

~Set both factors to equal zero

x + 7 = 0

x + 5 = 0

~Solve for x in [ x + 7 = 0 ]

x + 7 = 0

x + 7 - 7 = 0 - 7

x = -7

~Solve for x in [ x + 5 = 0 ]

x + 5 = 0

x + 5 - 5 = 0 - 5

x = -5

Therefore, the solutions are x = -7 and x = -5

Best of Luck!

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Answer:

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Step-by-step explanation:

1.8 added 8 times = 1.8 * 8

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1.8 = 1 + 0.8

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3 years ago

A mixture contains forty ounces of glycol and water and is ten percent glycol. If the mixture is to be strengthened to twenty-fi

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Read 2 more answers

The question is Stephen read for 34 minutes on Monday and 45 minutes on Tuesday. Use breaking apart to find how many minutes Ste

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Breaking apart means that you just take the tens from each number, and the ones from each number, then add them up.

The tens from your numbers are:
30 from 34
40 from 45

The ones from your numbers are:
4 from 34
5 from 45

Now, you have to add up the tens and ones separately.
30 + 40 = 70
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Finally, you add up those 2 numbers that you got- 70 and 9

The answer: Stephen read for 79 minutes total.

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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,

$t = 2.60s$

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3 years ago