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Yuri [45]
3 years ago
6

Factor 3h2 – 11h - 42

Mathematics
2 answers:
Zina [86]3 years ago
7 0
The answer is D
3*-42=-126
The factors of -126 that add to -11 are 7 and -18
replace -11h with 7h-18h
3h^2 + 7h - 18h - 42
Factor the first two terms and last two terms
h(3h+7) - 6(3h+7)
See how the two parentheses are the same?
THE ANSWER IS BELOW!!
(h-6)(3h+7)
the first parentheses has the terms that were outside, and the second has the original bracketed terms
This matches option D (it just multiplies the terms backwards, which gives the same result)
sattari [20]3 years ago
7 0

Answer:

The correct answer is option D. (3h + 7)(h – 6)

Step-by-step explanation:

It is given a quadratic function ,

3h² - 11h - 42

<u>To find the factors of given function</u>

3h² - 11h - 42  = 3h² - 18h  + 7h - 42

 = 3h(h -6) + 7(h - 6)

 =(h - 6)(3h + 7)

 = (3h + 7)(h – 6)

The correct answer is option D. (3h + 7)(h – 6)

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FIND THE UNIT OF X 5X+17=7 WHEN YOU FIND IT SIOLVE THE EQUATION WITH THE NEW NUMBER OF X
UkoKoshka [18]

Answer:

x = -2

Step-by-step explanation:

5x + 17 = 7

5x = -10

5x/5 = -10/5

x = -2

i could be wrong but i hope i helped

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3 years ago
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Maureen is taking an antibiotic. The table below shows the amount of antibiotic f(t), in mg, that was present in her body after
MariettaO [177]
One way to test this is by using the equation for Ken on Maureen. Let's say in the first hour,

f(t) = 200(0.976)^1 = 195.2 mg

In the second hour,

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In the third hour,

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If you compare this with Maureen's data which is 150, 90 and 54 for the first, second and third hour, respectively, you will see that Maureen's rate is much faster. However, you cannot tell by what factor because the function is exponential, not a multiple. There is no constant difference between their rates. Therefore, we only know that Maureen's rate is much faster.

The answer is <span>Maureen's body eliminated the antibiotic faster than Ken's body.</span>
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3 years ago
Best way to tell someone no without being mean
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Answer:

Try to search up euphemisms to help you in an argument where your forced to say no

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Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th
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Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

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S = \{1,1,2,2,2,3\}

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P(1) = \frac{n(1)}{n(s)}

P(1) = \frac{2}{6}

P(1) = \frac{1}{3}

P(2) = \frac{n(2)}{n(s)}

P(2) = \frac{3}{6}

P(2) = \frac{1}{2}

P(3) = \frac{n(3)}{n(s)}

P(3) = \frac{1}{6}

P(Odd Number) is then calculated as:

P(Odd\ Number) =  P(1) + P(3)

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Take LCM

P(Odd\ Number) = \frac{2+1}{6}

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P(Odd\ Number) =  \frac{1}{2}

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

S = \{1,1,2,2,2,3,3\}

The probability of each is:

P(1) = \frac{n(1)}{n(s)}

P(1) = \frac{2}{7}

P(2) = \frac{n(2)}{n(s)}

P(2) = \frac{3}{7}

P(3) = \frac{n(3)}{n(s)}

P(3) = \frac{1}{7}

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Take LCM

P(Odd\ Number) = \frac{2+1}{7}

P(Odd\ Number) = \frac{3}{7}

Comparing P(Odd Number) before and after

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P(Odd\ Number) = \frac{3}{7} --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

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