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3 years ago

John and alisha randomly met at the local movie theater. what will affect their first impression of each other?

Mathematics

1 answer:

Vesna [10]3 years ago

7 0

How they dress, What movie they buy, and what snacks you buy

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Lines need to be painted on the field at a football stadium. The painter estimated that one can of paint would last 11 yards. Ho

zlopas [31]

Answer:

area = pi*24^2/4 = 452 yd^2

so that means you will need 452/59 = 7.6 (or 8) gallons

8 0

2 years ago

(02.05 MC)

Amiraneli [1.4K]

Table for f(x)

$\boxed{\begin{array}{c|c}\boxed{\bf x}&\boxed{\sf f(x)}\\ \sf 0 &\sf 1 \\ \sf 2&\sf 9\\ \sf 4 &\sf 17\end{array}}$

Table for g(x)

$\boxed{\begin{array}{c|c}\boxed{\bf x}&\boxed{\sf g(x)}\\ \sf 0 &\sf 1 \\ \sf 2&\sf 7\\ \sf 4 &\sf 13\end{array}}$

Option C is correct

3 0

2 years ago

Read 2 more answers

what is the slope of the line that passes through the points (10,2) and (19,14) write your answear in simplist form

irakobra [83]

Answer: 1 1/3

Step-by-step explanation: slope is the difference in y values over the difference in the c values

4 0

3 years ago

Read 2 more answers

Visa charges 24.5% on the balance each month. Find the new balance for $2000 after I month

Oliga [24]

Answer:

I think $490

Step-by-step explanation:

Don't quote me on that

8 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