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xenn [34]
4 years ago
10

HELP YOU WILL GET BRAINLIEST

Mathematics
1 answer:
dezoksy [38]4 years ago
7 0

Answer:

60π

Step-by-step explanation:

2π(3 × 7 ) + 2π (3)^2

Distribute: 2x3x7 and 2x3^2

42π + 18π

That gives 60π.

You might be interested in
CD is the segment bisector of AB at D. If AD = 4x-1 and DB = 9x-21. what is the length of AD and AB
Artemon [7]

Answer:

AD= 15

AB=30

Step-by-step explanation:

We are given that the Point D bisect the line segment AB

Hence AD = DB

AD= 4x-1

BD=9x-21

Hence

4x-1=9x-21

adding 21 on both sides and subtracting 4x from both sides

4x-4x-1+21=9x-4x-21+21

-1+21=5x

5x=20

dividing both sides by x we get

x=4

Hence

AD = 4x-1

= 4(4)-1= 16-1=15

AD= 15

AB = 2 x AD

     = 2 x 15

  = 30

3 0
4 years ago
Help me plzz!<br> Exponential word problems
slega [8]

Answer: (7)

 

Step-by-step explanation:

5 0
4 years ago
Find the slope and the y-intercept of the line.<br> y=2x-4
tatuchka [14]
Answer:
Slope: 2x
Y-intercept: -4
3 0
3 years ago
Read 2 more answers
Target Costing
Lynna [10]

The target cost for Model J20 is $2000.

The required cost reduction is $30.

The total saving will be $30.3.

<h3>How to compute the cost?</h3>

The target cost will be:

= Sales price - (100% × Target cost)

= 400 - (100% × Y)

Y + Y = 400

2Y = 400

Y = 400/2 = 200

Target cost = $200

Therefore, the target cost is $200.

The required cost reduction will be:

= $230 - $200

= $30

The required cost reduction is $30.

The three engineering improvements together will be:

Direct labor reduction = 7.5

Additional inspection = 17

Injection molding = 5.8

Total savings = 30.3

The total savings is $30.3.

Learn more about cost on:

brainly.com/question/10705492

#SPJ1

6 0
2 years ago
Find the solution to dy/dt = 7 y<br><br> satisfying<br> y(9) = 5
AVprozaik [17]
\rm \dfrac{dy}{dt}=7y,\qquad\qquad\qquad y(9)=5

I'm not sure what methods you've learned up to this point but one option is to apply separation of variables:

\rm \dfrac{dy}{y}=7dt

and then integrate from there,

\rm ln|y|=7t+c

exponentiate to isolate y,

\rm |y|=e^{7t+c}

Apply exponent rule,

\rm |y|=e^c e^{7t}

rename this e^c as some new constant, perhaps A,

\rm \pm y=A e^{7t}\qquad\qquad A>0

This A can only be positive, non-zero, but absorbing the plus/minus fixes that restriction,

\rm y=A e^{7t}\qquad\qquad A\ne0

Use your initial information to solve for this unknown value A,

\rm y(9)=A e^{7\cdot9}=5

solving for A, dividing by the exponential,

\rm A=5e^{-63}

So we get a final result of

\rm y(t)=5e^{-63}e^{7t}

apply exponent rule again to get a better looking answer, and factor,

\rm y(t)=5e^{7(t-9)}

Lemme know if too confusing.
3 0
3 years ago
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