Mrs. Hill went to get an oil change, because she has a 20% off coupon. F the oil change is $35.00, how much is the discount? In problem #1, how much does Mrs. Hill have to pay?

Answer:

domain is the lenght of the road (in miles)

range is the number of days

Step-by-step explanation:

domain is the input because there building the road that is the input because the number of days is the output (range) because that is what happens from working on the roads (the input) what come out is the days they worked on the road

Answer:

You would be able to afford 9 gallons of gas.

Step-by-step explanation:

34.83 divided by 3.87 is 9. Have a nice day!

Answer    g

Step-by-step explanation:

g

Answer:

don't know i tell true

Step-by-step explanation:

Answer:

I believe the answers are

B) 1

D) $1,000

Step-by-step explanation:

Answer:

[tex]\frac{11}{10}[/tex]

Step-by-step explanation:

[tex]\frac{8}{10}[/tex] - [tex]\frac{x}{10}[/tex] = [tex]\frac{3}{10}[/tex]

so [tex]\frac{8}{10}[/tex] +[tex]\frac{3}{10}[/tex] =[tex]\frac{x}{10}[/tex]

since all the denominators are the same we simply add the numerators to find x

x would be 8 +3 = 11

so [tex]\frac{11}{10}[/tex] is the answer

Answer:

D

Step-by-step explanation:

Process of Elimination:

A: 3.976 is greater than 3.97

B: 1.025 is lesser than 1.25

C: 9.145 is greater than 9.14

All of the options checked are true, leaving the answer only to be D.

Answer:

x = -5/2, (5+5i√3)/4, (5-5i√3)/4

Answer:

g=6

Step-by-step explanation:

6.4×6-1.4×6=

38.4-8.4=30

its true if it is not the result is incorrect

The quotient of 110 divided by 4 is 27 with a remainder of 2.

To solve 110 divided by 4 using the area model, we can represent the division problem visually by creating an area model.

Draw a rectangle to represent 110.

Divide the rectangle into four equal parts to represent dividing by 4.

Distribute the value of 110 evenly across the four parts. Since 110 is not divisible evenly by 4, distribute as much as possible to each part.

Determine the quotient. Count the number of equal parts, which in this case is 27.

Therefore, the quotient of 110 divided by 4 is 27 with a remainder of 2.

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I hope this helps. Comment if any doubt.

Answer:

Step-by-step explanation:

3(m-2)+2(m-2)

3m-6+2m-4

3m+2m-6-4

5m-6-4

5m-10

The expressions 3(m-2)+2(m-2) and 50m-2 are not equivalent expressions.

After simplifying the first expression 3(m-2)+2(m-2) in part (a), we got 5m-10, which is different from the second expression, 50m-2.

Answer:

regan ate more than maggie

Answer:

16x + 12

Step-by-step explanation:

20x + 12 - 4x

20x - 4x = 16x

16x + 12

Answer: 15x + 12

Step-by-step explanation:

(4 x 5x). (4 x 3). (20x - 4x). 15x + 12 is your answer.

Answer:

It is an identity, proved below.

Step-by-step explanation:

I assume you want to prove the identity. There are several ways to prove the identity but here I will prove using one of method.

First, we have to know what cot and cosec are. They both are the reciprocal of sin (cosec) and tan (cot).

[tex]\displaystyle \large{\cot x=\frac{1}{\tan x}}\\\displaystyle \large{\csc x=\frac{1}{\sin x}}[/tex]

csc is mostly written which is cosec, first we have to write in 1/tan and 1/sin form.

[tex]\displaystyle \large{1+(\frac{1}{\tan x})^2=(\frac{1}{\sin x})^2}\\\displaystyle \large{1+\frac{1}{\tan^2x}=\frac{1}{\sin^2x}}[/tex]

Another identity is:

[tex]\displaystyle \large{\tan x=\frac{\sin x}{\cos x}}[/tex]

Therefore:

[tex]\displaystyle \large{1+\frac{1}{(\frac{\sin x}{\cos x})^2}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{1}{\frac{\sin^2x}{\cos^2x}}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}[/tex]

Now this is easier to prove because of same denominator, next step is to multiply 1 by sin^2x with denominator and numerator.

[tex]\displaystyle \large{\frac{\sin^2x}{\sin^2x}+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}\\\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}[/tex]

Another identity:

[tex]\displaystyle \large{\sin^2x+\cos^2x=1}[/tex]

Therefore:

[tex]\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}\longrightarrow \boxed{ \frac{1}{\sin^2x}={\frac{1}{\sin^2x}}}[/tex]

Hence proved, this is proof by using identity helping to find the specific identity.

Answer: many solutions

Step-by-step explanation:

D10W has a concentration 10%

So

Dextrose amount=

[tex]\\ \sf\Rrightarrow 10\%\:of\:756[/tex]

[tex]\\ \sf\Rrightarrow 0.10(756)[/tex]

[tex]\\ \sf\Rrightarrow 75.6ml[/tex]

Answer:

The answer is 9 ( C on edge)

Step-by-step explanation:

Answer:

9

Step-by-step explanation:

C on edge

I took the test

Step-by-step explanation:

24 cans cost $8.

so, 1 can is 1/24 if the total, and that applies then also to the price.

so, the cost of 1 can is 8/24 = 1/3 ≈ $0.33

Answer:

Cost per can = $0.33.

Step-by-step explanation:

8 / 24

= 1/3

= 0.333....

Answer:

Step-by-step explanation:

Given:

s = b + 7

=> 11 = b + 7

=> b = 4

a = 15 + b

=> a = 15 + 4

=> a = 19

Therefore, the equation is 'a = 15 + 4' and the answer is 'a = 19'.

Hoped this helped.

I apologize if this was incorrect or un-helpful.

So

As z=11

[tex]\\ \rm\Rrightarrow y+7=11[/tex]

[tex]\\ \rm\Rrightarrow y=4[/tex]

My age=

Answer:

i got 47/88

Step-by-step explanation:

Divide each number by two

Hope you could get an idea from here.

Doubt clarification - use comment section.

The equation [tex]p^3+q^3+r^3=(\frac{1}{p^3}+\frac{1}{q^3}+\frac{1}{r^3})p^2q^2r^2[/tex] has been proved to be true

The equation is given as:

[tex]\frac pq = \frac qr[/tex]

Cross multiply

[tex]pr = q^2[/tex]

Rewrite as:

[tex]q^2 = pr[/tex]

Also, we have:

[tex]p^3+q^3+r^3=(\frac{1}{p^3}+\frac{1}{q^3}+\frac{1}{r^3})p^2q^2r^2[/tex]

Substitute [tex]q^2 = pr[/tex]

[tex]p^3+q^3+r^3=(\frac{1}{p^3}+\frac{1}{q^3}+\frac{1}{r^3})p^3r^3[/tex]

Expand

[tex]p^3+q^3+r^3=\frac{1}{p^3} \times p^3r^3 +\frac{1}{q^3} \times p^3r^3 +\frac{1}{r^3} \times p^3r^3[/tex]

Simplify

[tex]p3+q3+r3=r^3 +\frac{1}{q^3} \times p^3r^3 + p^3[/tex]

Rewrite as:

[tex]p3+q3+r3=r^3 +\frac{1}{q^3} \times (pr)^3 + p^3[/tex]

Substitute [tex]pr = q^2[/tex]

[tex]p3+q3+r3=r^3 +\frac{1}{q^3} \times (q^2)^3 + p^3[/tex]

[tex]p3+q3+r3=r^3 +\frac{1}{q^3} \times q^6 + p^3[/tex]

Divide q^6 by q^3

[tex]p3+q3+r3=r^3 +q^3 + p^3[/tex]

Rewrite the equation as:

[tex]p3+q3+r3=p^3 +q^3 + r^3[/tex]

Hence, the equation has been proved

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Alma was definitely correct to say that the tickets are $14 each.

The total amount in dollars the class have for tickets to the science museum

is $140.

The teacher writes as follows:

Alma says the tickets are $14 each.

Mohammed says that 14 students are going for the trip

The teacher divided 140 by 10. This means the number of student in the class is 10 not 14 as Mohammed enunciated.

The teacher divided the total amount available for the ticket($140) by the number of student in the class(10). Then the quotient ($14) is the amount in dollars which each student will pay for the ticket to the science museum.

Therefore, Alma was definitely correct.

learn more about algebra here:  https://brainly.com/question/11455464?referrer=searchResults

Answer:

y = 3.2

Step-by-step explanation:

8/y = 10/4

10y = 32

y = 32 / 10

y = 3.2

Answer:

-3

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(2-(-4))/(6-8)

m=(2+4)/-2

m=6/-2

m=-3

Answer:

- 1/4

Step-by-step explanation:

Where you see X, substitute with 1/3

Solve the power/exponent first:

6(1/3) = 2

That is ; - 4(1/4)^2

Solve the parenthesis:

(1/4)^2 = 1/16

-4(1/16)

= - 1/4