3 years ago

Is it possible to have a solution that has 1 real and 1complex root?

Mathematics

1 answer:

Volgvan3 years ago

8 0

Answer:

Step-by-step explanation:

It is not possibe! you have to have a soulution that has 3 real and 1 complex root!

It does not make sence!

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Sweet T is making a double batch of cupcakes. He puts them in the oven at 6:50 that needs to bake for 25 minutes. When should sw

Serhud [2]

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3 years ago

Solve for X<br> (5x-2)^2=17<br> please help me everyone.

beks73 [17]

Answer:

$\frac{2}{5}±\frac{\sqrt{17}}{5}$

Step-by-step explanation:

$(5x-2)^2=17 \\ 5x-2=±\sqrt{17} \\ 5x = 2 ± \sqrt{17} \\ x=\frac{2}{5}± \frac{\sqrt{17}}{5}$

Don't mind the $±$

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3 years ago

Simplify the expression to one number using order of operations:

geniusboy [140]

<h2><u>COMBINED OPERATIONS</u></h2><h3>Exercise</h3>

First, let's solve the subtraction (in parentheses):

$\mathsf{\dfrac{-8\left(2-6\right)}{2}}$

$\mathsf{\dfrac{-8(-4)}{2}}$

Now, let's solve the multiplication. Remember: in multiplication, two negative numbers will give a positive number:

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$\mathsf{\dfrac{32}{2}}$

Finally, let's solve the division:

$\mathsf{\dfrac{32}{2} = 16}$

<h3><u>Answer</u>: 16</h3>

5 0

3 years ago

How do i solve this inequality? Number 46

myrzilka [38]

6(x-2.5) ≥ 8-6(3.5+x)

A number is before a parenthesis. The first step is to multiply.

6(x-2.5) = 6x -15

8-6(3.5+x)
2(3.5+x)= 7+2x

6x-15≥7+2x

Now solve

6x-15≥7+2x
-2x -2x

(-2x+2x)= They cancel out

(6x-2x)= 4x

4x-15≥7
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You get= x≥5.5

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3 years ago

(5+4-2)×(-2)=idfkadtsphm

lbvjy [14]

5+4=9 9-2=7

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your answer is -14

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3 years ago