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3 years ago

15

I need help with math please! :)

1 answer:

miss Akunina [59]3 years ago

4 0

Answer:

ok but heres the graph i will put more in a sec

Step-by-step explanation:

OML MY FINGER SLIPPED--

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Match each equation to a solution.<br> 2X=64

Marysya12 [62]

X=32

2x=64 divide each side by 2

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3 years ago

Select the correct answer.

Fofino [41]

Answer:

hihigigigigigigigi

4 0

3 years ago

How do you solve for ? 1/5(x-4)=5

zepelin [54]

$\frac{1}{5} *(x-4) =5$
$(x-4) = \frac{5}{ \frac{1}{5}}$ (take 1/5 to right hand side, thus it will divide by 5)
$(x-4) = 25$ (simplify)
$x = 29$

6 0

3 years ago

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Which is the following is the set of real zeros of the function f(x) = (x3 + 1000)(x4 - 160,000)?

marusya05 [52]

Answer:

A. { -20, -10, 20 }

Step-by-step explanation:

Given:

The function is given as:

$f(x)=(x^3+1000)(x^4-160000)$

Let us simplify the function.

First, we use the identity $a^3+b^3=(a+b)(a^2-ab+b^2)$

$x^2+1000= x^3+10^3=(x+10)(x^2-10x+10^2)\\x^3+1000=(x+10)(x^2-10x+100)$

Next, we use the identity $a^4-b^4=(a-b)(a+b)(a^2+b^2)$

$x^4-160000=x^4-20^4=(x-20)(x+20)(x^2+20^2)=(x-20)(x+20)(x^2+400)$

Now, the function can be rewritten as:

$f(x)=(x+10)(x^2-10x+100)(x-20)(x+20)(x^2+400)$

Now, the zeros are those values of $x$ for which $f(x)=0$

Now, for $f(x)=0$ , we must have either of the factors 0.

$x+10=0\\x=-10$

$x-20=0\\x=20\\\\x+20=0\\x=-20$

The factors $x^2-10x+100$ and $x^2+400$ can have no zeros as the first one has imaginary roots and second one is always greater than 0 irrespective of the $x$ values.

So, the possible set of zeros are { -20, 10, 20 }.

6 0

4 years ago

Is (-1,2) a solution of y=-5x+7

Arlecino [84]

No, it is not. I will show you:

-1 = x and 2 = y

2 = -5(-1) + 7

2 is not equal to 12. I hope this is right!

6 0

4 years ago

Read 2 more answers