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3 years ago

Go Which operation would be completed second in the following expression? 7² -3+9+8 dividend by two

2 answers:

5 0

First step: simplify exponents
second step: 49 -3
but the thing is when it’s both addition and subtraction u just go from left to right
hope it helps

Lera25 [3.4K]3 years ago

5 0

Answer:

Divide

Step-by-step explanation:

Using PEMDAS, the first would be parenthesis, but there are none, so we do not use P. Next is E which is exponents, which would be the first thing completed. Then is M and D which is multiplication and division. The division is used which is "dividend by two" as it's the second thing completed.

Hope this helped.

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Zoey made 6 3/4 cups of fruit salad for a picnic. At the picnic, they ate 1/3 of the fruit salad. How much of the fruit salad di

Aleks04 [339]

Answer:

9/4 cups

Step-by-step explanation:

6 3/4 = 27/4

27/4 * 1/3 = 27/12

27/12 = 9/4

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3 years ago

How to find imaginary zeros anf real zeros of F(x)=-4x^5-8x^3+12x

Fofino [41]

Answer:

$x=\{0, -1, 1, -i\sqrt{3}, i\sqrt{3}\}$

Step-by-step explanation:

We are given the function:

$f(x)=-4x^5-8x^3+12x$

And we want to finds its zeros.

Therefore:

$0=-4x^5-8x^3+12x$

Firstly, we can divide everything by -4:

$0=x^5+2x^3-3x$

Factor out an x:

$0=x(x^4+2x^2-3)$

This is in quadratic form. For simplicity, we can let:

$u=x^2$

Then by substitution:

$0=x(u^2+2u-3)$

Factor:

$0=x(u+3)(u-1)$

Substitute back:

$0=x(x^2+3)(x^2-1)$

By the Zero Product Property:

$x=0\text{ and } x^2+3=0\text{ and } x^2-1=0$

Solving for each case:

$x=0\text{ and } x=\pm\sqrt{-3}\text{ and } x=\pm\sqrt{1}$

Therefore, our real and complex zeros are:

$x=\{0, -1, 1, -i\sqrt{3}, i\sqrt{3}\}$

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3 years ago

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Answer:

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Kamila [148]

Here, this website explains it perfectly

Step-by-step explanation:

https://www.k5learning.com/blog/subtracting-positive-and-negative-numbers

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3 years ago

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25 POINTS AND BRAINLIEST PLEASE HELP ASAP

dexar [7]

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$MA=\sqrt{\left(\dfrac{\boxed{3}a}{2}\boxed{-}\boxed{0}\right)^2+\left(\dfrac{\boxed{b}}{2}\boxed{-}\boxed{0}\right)^2}=\\\\\\=
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$NB=\sqrt{\left(\dfrac{a}{2}\boxed{-}\boxed{2}a\right)^2+\left(\dfrac{b}{2}\boxed{-}\boxed{0}\right)^2}=\\\\\\=\sqrt{\left(\dfrac{a}{2}\boxed{-}\dfrac{\boxed{4}\boxed{a}}{2}\right)^2+\left(\dfrac{b}{2}-\boxed{0}\right)^2}=\\\\\\
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3 years ago

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