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Bad White [126]
3 years ago
5

What is b x (5x 10) x to y

Mathematics
1 answer:
tankabanditka [31]3 years ago
3 0

b^5 x b^10 = b^5+10

b^15

b^15 to y

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Kryger [21]

Answer:22,069

Step-by-step explanation:

6 0
3 years ago
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Jean and Jericho who are playing in the school grounds decided to sit on a
Natalija [7]

We want to see underline the correct part in each statement.

  • 1) This situation illustrates (direct, <u>inverse</u>) variation.
  • 2) The two quantities that must vary in this situation are (<u>weight and </u>
  • <u>distance from the cente</u><u>r</u>, height and weight).
  • 3) The heavier the kid, the (<u>closer</u>, farther) he/she should be at the
  • center to balance the seesaw?
  • 4) When Jean moves farther from the center, Jericho tends to go (up,
  • <u>down</u>).
  • 5) If Jericho moves closer to the center, Jean tends to go (up, <u>down</u>).

So, Jean and Jericho are playing on a seesaw.

Jean is heavier than Jericho.

Now, notice that a seesaw is a lever. So it "amplificates" the force that you apply in one end to lift the weight that is on the other end. Depending on the form of the lever and the weights, the force that you need to do changes.

If we define:

  • W₁ = Jean's weight.
  • d₁ = Jean's distance to the center.
  • W₂ = Jericho's weight.
  • d₂ = Jericho's distance to the center.

We must have:

W₁*d₁ = W₂*d₂

Then:

1) This situation illustrates (direct, <u>inverse</u>) variation.

d₁, the position of Jean, varies inversely with respect to Jean's weight.

2) The two quantities that must vary in this situation are (<u>weight and </u>

<u>distance from the center</u>, height and weight).

(by the equation above)

3) The heavier the kid, the (<u>closer</u>, farther) he/she should be at the

center to balance the seesaw?

By the given equation, we see that d₁ must be smaller than d₂.

And the last two are trivial:

4) When Jean moves farther from the center, Jericho tends to go (up,

<u>down</u>).

5) If Jericho moves closer to the center, Jean tends to go (up, <u>down</u>).

If you want to learn more, you can read:

brainly.com/question/18320907

7 0
2 years ago
Solve for the given length. The length of a rectangle is 3m longer than its width. The area of the rectangle is 154^2. What is t
Andre45 [30]
If the length of a rectangle is 3m longer than its width, then:
L=W+3
Is the area really 154^2? Or is it 154m^2? If yes, then:
A=LW
154=(W+3)(W)
154=(W^2+3W)
0=W^2+3W-154
0=(W-11)(W+14)
This means either (W-11) or (W+14) is equal to zero so:
W=11 and W=-14
To find out let's substitute the numbers:
154=(11+3)(11)
154=154
Therefore, the width of the rectangle is 11m
7 0
3 years ago
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(07.03)
Sloan [31]
The answer would be log base 2 of 6 over log base 2 of 3

8 0
3 years ago
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I'll give u a brainliest to answer this&lt;3
Andre45 [30]

Answer:

762

Step-by-step explanation:

3 0
3 years ago
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