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3 years ago

I need some help! please help me if you know the answers! will give brainliest!

Mathematics

2 answers:

miv72 [106K]3 years ago

7 0

C because you have to round and it would be C

Katyanochek1 [597]3 years ago

6 0

What he said lol XD

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Please solve number 11:(

MrRa [10]

Answer:

Market price $400

Cost price $300

Step-by-step explanation:

MP: market price

CP: cost price

1. (MP x 0.9) - CP = 60

CP = 0.9MP - 60

2. MP - CP = 100

CP = MP - 100

0.9MP - 60 = MP - 100

MP - 0.9MP = 100 - 60

0.1MP = 40

MP = 400

Cost price = MP - 100 = 400 - 100 = 300

3 0

2 years ago

There are 3 alternative routes by which you may drive to work: Alabaster Street, Brillantine Street, and Clancy Street. It is th

Solnce55 [7]

Answer:

25%

Step-by-step explanation:

sorry i did it in my head

3 0

3 years ago

What is addition property

n200080 [17]

Answer:

When we add two or more whole numbers, their sum is the same regardless of the order of the addends. The sum of both 2 + 4 and 4 + 2 is 6. That means, we can add whole numbers in any order. When three or more numbers are added, the sum is the same regardless of the grouping of the addends.

Step-by-step explanation:

there

6 0

3 years ago

Read 2 more answers

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

A small company had a loss of $425 in January. If it continues to have the same loss each month for 4 months, what will be the c

Tresset [83]

425(4)= 1700. After 4 months they would have lost $1700

3 0

3 years ago