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3 years ago

I have no idea what to do with this someone help

Mathematics

2 answers:

joja [24]3 years ago

4 0

The given distance of 5.20 would be A.

Replace A with 5.20 to solve for T.

T = 5.20^3/2
T = 11.9 years.

Naddik [55]3 years ago

3 0

I think the answer is 11.8 years. Hope this helps.

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Write each fraction as a percent 4/100

nadezda [96]

Answer:

4%is the answer you are looking for

8 0

3 years ago

Read 2 more answers

Mrs Katie constructs a cube with 512 magnetic block students and her two classes will each make an identical cube. There are 28

devlian [24]

I take it each student is going to make a block.
She has 25 + 28 students all together which is 53 students total
Each one requires 512 magnetic blocks

Therefore she needs 53 * 512 = 27136 blocks.

6 0

3 years ago

Which equation has the solution x=-4? a 2(x-4)=16 b 3x+6-2x=-2 c 1/4x+4=3 d(x+6)3=-6 need help plz

xz_007 [3.2K]

Answer:

C: 1/4x + 4 = 3

Step-by-step explanation:

1/4x + 4 = 3

1/4x = 3 - 4

x = - 4

Thus, option c gives out the solution x = - 4.

4 0

3 years ago

OKAY i CANT EVEN FIGURE THIS ONE OUT NEED HELP.<br> -3 (5 – 2y)

inessss [21]

negative multiplied by positives = negative numbers

3 x 5 = -15

negative multiplied by a negative = a positive number

-3 x -2y = +6y

so combine both gets -15 +6y

or a better way to write it is 6y-15

5 0

3 years ago

Use the intermediate value theorem to find the value of c such that f(c) = M. f(x) = x^2 - x + 1 text( on ) [1,8]; M = 21 c =

DanielleElmas [232]

Answer:

$c = 5$

Step-by-step explanation:

Given

$f(c) = M$

$f(x) = x^2 - x + 1$

$Interval: [1,8]$

$M = 21$

Required

Find c using Intermediate Value theorem

First, check if the value of M is within the given range:

$f(x) = x^2 - x + 1$

$f(1) = 1^2 - 1 + 1$

$f(1) = 1$

$f(x) = 8^2 - 8 + 1$

$f(x) = 57$

$1 \le M \le 57$

$1 \le 21 \le 57$

M is within range.

Solving further:

We have:

$f(c) = f(x) = M$

$f(x) = 21$

Substitute 21 for f(x) in $f(x) = x^2 - x + 1$

$21 = x^2 - x + 1$

Express as quadratic function

$x^2 - x + 1 - 21 = 0$

$x^2 - x - 20 = 0$

Expand

$x^2 + 4x - 5x - 20$

$x(x+4)-5(x+4)=0$

$(x - 5)(x+4) = 0$

$x - 5 = 0$ or $x + 4= 0$

$x = 5$ or $x = -4$

The value of $x = -4$ is outside the $Interval: [1,8]$

So:

$x = 5$

$f(c) = f(x) = M$

$f(c) = f(5) = 21$

By comparison:

$c = 5$

4 0

3 years ago