All categories

3 years ago

Tyler baked 702 cookies. He sold them in boxes of 18. After selling all of the boxes of cookies for the

Mathematics

2 answers:

maksim [4K]3 years ago

6 0

Answer: Each box was sold for $39

Step-by-step explanation:

If you divide 18 divided by 702 then you get 39

user100 [1]3 years ago

3 0

Answer:

3.50

Step-by-step explanation:

if you divide 18 by 702 you will get 39 and 136.50 divided by 39 is 3.50. and to make sure this is right you can multiply 3.50 by 39 and you will get 136.50

You might be interested in

Anya’s team won 5 out of every 7 games this season.

spin [16.1K]

Answer:

20 out of 28 games.

Step-by-step explanation:

28 divided by 7 is 4.

5 times 4 is 20.

20 out of 28 games.

(This is simple math honestly.)

7 0

2 years ago

Problem page three consecutive even integers have a sum of 36 . find the integers.

ANTONII [103]

3 consecutive even integers = 36 .
ANSWER 10+12+14= 36

6 0

3 years ago

Solve the equation.<br><br> −10x + 1 + 7x = 37

matrenka [14]

You combine like terms which are -10x and 7x and 1 and 37. Then it becomes -3x=36. x would equal to 36/-3, which is -12.

8 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

Use the rule x + 8 to write a pattern. Begin with x = 0.

Nutka1998 [239]

8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104

7 0

2 years ago