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2 years ago

How can i simplify this 32/3-3

2 answers:

6 0

You can’t simplify it.
32/(3-3)
3 subtracted by 3 is 0.
But 32/0 will only be undefined because no number multiplied by 0 will get you 32.
So there is no actual number for it.

Answer: Unable to simplify. (Undefined)

Nikolay [14]2 years ago

3 0

Answer:

Step-by-step explanation:

well, if 3-3 is 0, then it is just 32.

I hope I got it right!

☾Lalita

p.s, it might be wrong.....

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Please help ( if you can then please help with <br> both questions )

tatuchka [14]

Answer:

For number 1, -1.125 and -9/8. For number 2, -46.

Step-by-step explanation:

6 0

2 years ago

The area of a trapezoid is 39 square millimeters. The height of the trapezoid is 6 millimeters. One of the base legths of the tr

Andrej [43]

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2 years ago

ASAP HELP FOR NUMBER 12, Thank you so much!!! :)

qaws [65]

12a)
5x + 45 + 3x - 25 = 180
8x + 20 = 180
8x = 160
x = 20

12b)
2x - 5 = 2(20) - 5 = 40 - 5 = 35
3x - 25 = 3(20) - 25 = 60 - 25 = 35

y = 180 - (35+35)
y = 180 - 70
y = 110

5 0

2 years ago

Which of the is not a function

Lina20 [59]

Answer:

Step-by-step explanation:

For a function the value of the variable "x" must correspond to a single value of "y"

Hope this helps

5 0

2 years ago

The quadratic function g(x) = a.ca + bx+c has the

Mumz [18]

<em>The value of b is 14 and the value of c is 65</em>

<h2>Explanation:</h2>

The quadratic function is a function of the form:

$f(x)=ax^2+bx+c$

Here we know that the leading coefficient $a=1$ so we reduce our equation to:

$g(x)=x^2+bx+c$

The roots are those values at which $g(x)=0$

So:

$x^2+bx+c=0 \\ \\ First \ root: \\ \\ (-7+4i)^2+b(-7+4i)+c=0 \\ \\ (-7)^2-2(7)(4i)+(4i)^2-7b+4bi+c=0 \\ \\ 49-56i+16i^2-7b+4bi+c=0 \\ \\ \\ Simplifying: \\ \\ 49-56i+16(-1)-7b+4bi+c=0 \\ \\ 49-56i-16-7b+4bi+c=0 \\ \\ 33-56i-7b+4bi+c=0 \\ \\ \\$

$Second \ root: \\ \\ (-7-4i)^2+b(-7-4i)+c=0 \\ \\ (-1)^2(7+4i)^2+b(-7-4i)+c=0 \\ \\ (7)^2+2(7)(4i)+(4i)^2-7b-4bi+c=0 \\ \\ 49+56i+16i^2-7b-4bi+c=0 \\ \\ \\ Simplifying: \\ \\ 49+56i+16(-1)-7b-4bi+c=0 \\ \\ 49+56i-16-7b-4bi+c=0 \\ \\ 33+56i-7b-4bi+c=0$

So we have:

$(1) \ 33-56i-7b+4bi+c=0 \\ \\ (2) \ 33+56i-7b-4bi+c=0 \\ \\ \\ Subtract \ 2 \ from: \\ \\ 33-56i-7b+4bi+c-(33+56i-7b-4bi+c)=0 \\ \\ 33-56i-7b+4bi+c-33-56i+7b+4bi-c=0 \\ \\ \\ Combine \ like \ terms: \\ \\ 33-33-56i-56i-7b+7b+4bi+4bi+c-c=0 \\ \\ -112i+8bi=0 \\ \\ Isolating \ b: \\ \\ b=\frac{112i}{8i} \\ \\ \boxed{b=14}$

Finding c from (1):

$33-56i-7b+4bi+c=0 \\ \\ \\ Substituting \ b: \\ \\ 33-56i-7(14)+4(14)i+c=0 \\ \\ 33-56i-98+56i+c=0 \\ \\ -65+c=0 \\ \\ \boxed{c=65}$

<h2>Learn more:</h2>

Complex conjugate: brainly.com/question/2137496

#LearnWithBrainly

5 0

3 years ago