<img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%201%20-%202x%20%20%5C%5C%20g%28x%29%20%3D%20x%20%2B%204%20%5C%5C%20h%28x%29%20%

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3 years ago

3D%20%20%7Bx%7D%5E%7B2%7D%20%2B%201" id="TexFormula1" title="f(x) = 1 - 2x \\ g(x) = x + 4 \\ h(x) = {x}^{2} + 1" alt="f(x) = 1 - 2x \\ g(x) = x + 4 \\ h(x) = {x}^{2} + 1" align="absmiddle" class="latex-formula">
$hgf(x) = 4 {x}^{2} + px + q$
Find the value of p and the value of q

Mathematics

1 answer:

antiseptic1488 [7]3 years ago

7 0

Answer:

(x-3)(x+2) i.e. x2 - x -6 i.e. p = -1 and q = -6

Step-by-step explanation:

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Anna35 [415]

Answer:

C) p(x) has a zero between x=2 and x=4

Step-by-step explanation:

For a polynomial p(x), $p(a)=R$ means that, when p(x) is divided by $x-a$ , the remainder is $R$ .

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Therefore the correct answer is C

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5 times as many stakes

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Simplify the next two expressions

prisoha [69]

1. 7x-40
2. 8x-30y-10

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3 years ago

An object is dropped from a small plane. As the object falls, its distance, d, above the ground after t seconds, is given by the

DedPeter [7]

The inequality can be used to find the interval of time taken by the object to reach the height greater than 300 feet above the ground is:

$d = -16t^2 + 1000$

Solution:

The object falls, its distance, d, above the ground after t seconds, is given by the formula:

$d = -16t^2 + 1000$

To find the time interval in which the object is at a height greater than 300 ft

Frame a inequality,

$-16t^2 + 1000 > 300$

Solve the inequality

Subtract 1000 from both sides

$-16t^2 + 1000 - 1000 > 300 - 1000\\\\-16t^2 > -700$

$16t^2 < 700\\\\Divide\ both\ sides\ by\ 16\\\\t^2 < \frac{700}{16}\\\\Take\ square\ root\ on\ both\ sides\\\\t < \sqrt{\frac{700}{16}}\\\\t < \pm 6.61$

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Therefore,

t < 6.61

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3 years ago