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VashaNatasha [74]

3 years ago

13

Juan is training for the Great Bike

1 answer:

lina2011 [118]3 years ago

7 0

10 Miles Because 25-15=10

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Graph the inequality 4x -2y < 8

iris [78.8K]

Mine is pretty rough but hope it helps look at the attached image.

8 0

4 years ago

I NEED HELPPPP PLEASE

9966 [12]

Answer:

B

Step-by-step explanation:

x=-2, f(-2)=-7

x=0, f(0)=5

x=1, f(1)=2

8 0

3 years ago

Read 2 more answers

the volume V of a right circular cylinder of height 3 feet and radius r feet is V=V(r)=3(pi)(r^2). find the instantaneous rate o

Aleksandr-060686 [28]

When I see the words "instantaneous rate of change", I have to assume that you're in some stage of pre-calculus in your math class.

The instantaneous rate of change of a function is just its first derivative.

We have the function

V(r) = 3 π r²

and we need its first derivative with respect to ' r '. That shouldn't be
too hard, because the ' 3 π ' is nothing but constants.

Watch me while I do it slowly for you:

-- The derivative of ' r² ' with respect to ' r ' is ' 2r '.

-- The derivative of V(r) with respect to ' r ' is (3 π) times the derivative of ' r² '.

-- The derivative of V(r) with respect to ' r ' is (3 π) times (2r) = <u>6 π r</u> .

The value of the derivative when r=3 is (6 π 3) = 18π = about <em>56.5 feet³/foot .</em>

3 0

3 years ago

What is the solution to 2(3x+5)=5(2x-4)-4x??

Dmitrij [34]

First distribute
combine like terms
subtract 6x from both sides
then you get no solutions which is a special case

8 0

3 years ago

Given that k-5, k+7, k+55 are three consecutive terms in a geometric sequence, solve for k

choli [55]

Answer:

$k = 9$

Step-by-step explanation:

Given:

Consecutive Terms: k-5, k+7, k+55

Required:

Determine the value of k

To do this, we make use of the concept of common ratio.

The common ratio (r) of a geometric sequence is:

$r = \frac{T_{n}}{T_{n-1}}$

In other words:

$r = \frac{T_3}{T_2} = \frac{T_2}{T_1}$

Where 1, 2 and 3 represents the terms of the progression/sequence

So:

$r = \frac{T_3}{T_2} = \frac{T_2}{T_1}$ becomes

$\frac{k+55}{k+7} = \frac{k+7}{k-5}$

Cross Multiply:

$(k+55)(k-5) = (k+7)(k+7)$

Open Brackets

$k^2 + 55k - 5k - 275 = k^2 + 7k + 7k + 49$

$k^2 + 50k- 275 = k^2 + 14k + 49$

Collect Like Terms

$k^2 - k^2 + 50k-14k = 275 + 49$

$36k = 324$

Solve for k

$k = 324/36$

$k = 9$

5 0

4 years ago