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seropon [69]
3 years ago
11

Expand the following. In other words, remove the parentheses (brackets):

Mathematics
1 answer:
Nina [5.8K]3 years ago
3 0

Answer:

I believe that it would be A.

But answer A should be 3a + ab - a^{2}

Step-by-step explanation:

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Please i will give brainliest <br>​
igor_vitrenko [27]

Answer:

6✓2=8.49

✓8.49^2-6^2

✓ 36

6

✓24^2+6^2

✓612

=24.74=x

5 0
3 years ago
(b) The area of a rectangular pool is 5488 m\2 If the length of the pool is 98 m, what is its width?​
mrs_skeptik [129]

Answer:

56

Step-by-step explanation:

To find the area of a rectangle we have the foruma A=WxL.

But we already have the area and length so we can plug that in

5488=Wx98

Now its an algebreic expression.

SInce its multiplying we do the opposite, so we divide 98 on both sides.

98/98 crosses itself out so now its 5488/98. Which equals 56. So now our expression is W=56. To fact check we put the numbers 56 and 98 into the formula to see if we get 5488.

A=56x98

A=5488

7 0
1 year ago
Bob and Doug are both 100-meter sprinters. Bob's sprint time is normally distributed with a mean of 10.00 seconds and Doug's spr
alexdok [17]

Answer:

The standard deviation (σ) = 0.05

Step-by-step explanation:

The question is to find the standard deviation.

STEEP 1: FIND THE MEAN

(10+9.9) ÷ 2 = 9.95

STEP 2: SQUARE THE DIFFERENCE BETWEEN SPRINT TIME AND MEAN

10-9.95= 0.05

0.05^2 = 0.0025

9.9 - 9.95= -0.0025

-0.0025^2 = 0.0025

STEP 3: FIND THE VARIANCE

0.0025+0.0025= 0.005

0.005/2= 0.0025

STEP 4: FIND THE STANDARD DEVIATION (σ )

√variance

√0.0025 = 0.05

Therefore

σ = 0.05.

From the standard deviation, the percentage probability of the higher value to occurs is

0.05×100= 5%

That means Doug has 95%

And Bob has 5%

3 0
2 years ago
269 - 100 = 169 points left lets goooooooo get em now
Butoxors [25]

Answer:

yessss pointssssssssssssssssssss

5 0
2 years ago
Read 2 more answers
Please help me... I don’t understand this at all...
timama [110]
For the problem in the upper left corner, see "figure 1"

For the problem in the upper right corner, see "figure 2"

For the problem in the lower left corner, see "figure 3"

For the problem in the lower right corner, see "figure 4"

3 0
2 years ago
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