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3 years ago

Multiply -9w(-3w^8) Simplify

Mathematics

1 answer:

andrezito [222]3 years ago

6 0

3^2w^9

Step by step solution :

Step 1 :

Equation at the end of step 1 :

0 - (9w • (0 - 3w8))

Step 2 :

Multiplying exponential expressions :

2.1 w1 multiplied by w8 = w(1 + 8) = w9

Multiplying exponents :

2.2 32 multiplied by 31 = 3(2 + 1) = 33

Equation at the end of step 2 :

0 - -33w9

Step 3 :

Final result :

33w9

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Step-by-step explanation:

A goes up by 1

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How do I to problems like "From 45 ft to 92 ft"?

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3 years ago

Solve for x <br><br><br> 5(3+x)=5+7x

MakcuM [25]

The answered the question is X=5

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3 years ago

Read 2 more answers

A violin student records the number of hours she spends practicing during each of nine consecutive weeks: 6.2 5.0 4.3 7.4 5.8 7.

spin [16.1K]

Answer:

A) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the first quartile.

Correct we satisfy that 1.2 <1.7 the lower limit

Step-by-step explanation:

Assuming this complete question: A violin student records the number of hours she spends practicing during each of nine consecutive weeks:

6.2 5.0 4.3 7.4 5.8 7.2 8.4 1.2 6.3

For this case we need to sort the data first on increasing way and we got:

1.2, 4.3, 5.0, 5.8, 6.2, 6.3, 7.2, 7.4, 8.4

For this case we have 9 values the median would be on the 5 position:

$Median = 6.2$

The first quartile would be 5 since we analyze 1.2, 4.3, 5.0, 5.8, 6.2 and the middle point is 5.0

The third quartile would be 7.2 since we analyze 6.2, 6.3, 7.2, 7.4, 8.4 and the middle point is 7.2

Then the interquartile rnage would be:

$IQR = Q_3 -Q_1= 7.2-5=2.2$

And $1.5 IQR = 1.5*2.2=3.3$

The lower limit on this case would be:

$LL= Q_1 -1.5 IQR= 5-3.3=1.7$

And our value is 1.2 is lower than the lower limit 1.7

Considering the smallest data value (1.2) and using the 1.5 × IQR rule, we would:

A) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the first quartile.

Correct we satisfy that 1.2 <1.7 the lower limit

B) not classify the value 1.2 as an outlier, because it is not more than 1.5 × IQR below the first quartile.

False 1.2 is more than 1.5 IQR below the first quartile

C) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the median.

False, we analyze if is an outlier comparing with the Q1 or Q3 not with the median

D) classify the value 1.2 as an outlier, because it is more than 1.5 × IQR below the mean.

False we analyze if is an outlier comparing with the Q1 or Q3 not with the mean

5 0

3 years ago

dybincka [34]

Answer:

$57.2\ ft$

Step-by-step explanation:

we know that

The formula of slope is "rise over run", where the "rise" (means change in y, up or down) and the "run" (means change in x, left or right)

In this problem

The tangent of angle of 5 degrees is equal to the quotient of "rise over run"

Let

y ----> the rise of the ramp

x ----> the run of the ramp

$tan(5^o)=\frac{y}{x}$

we have

$y=5\ ft$

substitute and solve for x

$tan(5^o)=\frac{5}{x}$

$x=\frac{5}{tan(5^o)}$

$x=57.2\ ft$

5 0

3 years ago