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Mandarinka [93]
3 years ago
14

Pls answer math need help quick!! look at the attached file below for question

Mathematics
1 answer:
harkovskaia [24]3 years ago
7 0
Remark
I didn't know this was a calculus question. How dumb of me.

f(x + h) - f(x)
==========
x + h  - h

50/6 - 51/6 =( -1/6+ h)/h 
The h's will disappear when the limit is taken.

answer - 1/6 which is A. If anyone else answers take their answer. 


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What is the equation of a line that passes through the point (2, 7) and is perpendicular to the line whose equation is y=x4+5 ?
Valentin [98]
Attached a solution and showed work.

8 0
3 years ago
What is the fourth term of the sequence? <br> a1 = m<br> an = 2an-1<br><br> 2m<br> 4m<br> 6m<br> 8m
harkovskaia [24]

Answer:

8m.

Step-by-step explanation:

an = 2 an-1 means that each term in the sequence( except the first) is obtained by multiplying the last term by 2, so the first 4 terms are m, 2m, 4m, 8m.

3 0
3 years ago
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Please help!<br> This is mathematics!
Elanso [62]

Answer:

30, 30, 32, 32, 34, 38, 38, 38. Thats how you first need to set it up then what you do is add them all up which would equal to 272. Then you take 272 dived by how many numbers there are (theres 8 numbers) so you do 272/8 which would be 34! So 34 would be your answer.

Step-by-step explanation:

7 0
4 years ago
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Write the sentences as proportion. Twelve hundredths is to three as twenty-four hundredths is to eight
ValentinkaMS [17]

Answer:

see the explanation

Step-by-step explanation:

Remember that

A proportion is the ratio (quotient) between two numbers

so

Twelve hundredths to three is ----> \frac{0.12}{3}

Twenty-four hundredths to eight ----> \frac{0.24}{8}

therefore

Twelve hundredths is to three as twenty-four hundredths is to eight

\frac{0.12}{3}=\frac{0.24}{8}

Note The proportion is not true

3 0
3 years ago
Please prove this........​
Crazy boy [7]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = π    →     C = π - (A + B)

                                    → sin C = sin(π - (A + B))       cos C = sin(π - (A + B))

                                    → sin C = sin (A + B)              cos C = - cos(A + B)

Use the following Sum to Product Identity:

sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]

cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]

Use the following Double Angle Identity:

sin 2A = 2 sin A · cos A

<u>Proof LHS → RHS</u>

LHS:                        (sin 2A + sin 2B) + sin 2C

\text{Sum to Product:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-\sin 2C

\text{Double Angle:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-2\sin C\cdot \cos C

\text{Simplify:}\qquad \qquad 2\sin (A + B)\cdot \cos (A - B)-2\sin C\cdot \cos C

\text{Given:}\qquad \qquad \quad 2\sin C\cdot \cos (A - B)+2\sin C\cdot \cos (A+B)

\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]

\text{Sum to Product:}\qquad 2\sin C\cdot 2\cos A\cdot \cos B

\text{Simplify:}\qquad \qquad 4\cos A\cdot \cos B \cdot \sin C

LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C    \checkmark

7 0
3 years ago
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