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4 years ago

What is the solution for -8p - 12 = 5p + 14

1 answer:

7 0

Your trying to move -8p to the right side , so first add 12 to 14 the -12 + 12 is zero and add 12 to plus then it would end up to -8p=5p+ whatever your number is the you look for p

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On a road trip in Motanna, USA, Elise sees a road sign that says Helena 87. Elise tests the accuracy of her cars odometer and tr

mote1985 [20]

Answer:

No i dont think so but im not sure

Step-by-step explanation:

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3 years ago

How do you answer these two questions?

-BARSIC- [3]

I think you could multiply or divide. Hope this helps:)

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3 years ago

Please help!

crimeas [40]

Answer:

The zeros are:

$x =4, x=2, x = 5$

The function has three distinct real zeros.

Hence, option (B) is true.

Step-by-step explanation:

Given the expression

$h\left(x\right)=\left(x-4\right)^2\left(x^2-7x+\:10\right)$

Let us determine the zeros of the function by putting h(x) = 0 and solving the expression

$0=\left(x-4\right)^2\left(x^2-7x+10\right)$

switch sides

$\left(x-4\right)^2\left(x^2-7x+10\right)=0$

$x^2-7x+\:10=\left(x-2\right)\left(x-5\right)$

$\left(x-4\right)^2\left(x-2\right)\left(x-5\right)=0$

Using the zero factor principle

$\mathrm{If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

$x-4=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0$

$x =4, x=2, x = 5$

Thus, the zeros are:

$x =4, x=2, x = 5$

It is clear that there are three zeros and all the zeros are distinct real numbers.

Therefore,

The function has three distinct real zeros.

Hence, option (B) is true.

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3 years ago

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Lady bird [3.3K]

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Step-by-step explanation:

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3 years ago

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mariarad [96]

Answer:

5 $\sqrt{2}$

Step-by-step explanation:

Special triangle. 1:1: $\sqrt{2}$

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3 years ago

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