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lorasvet [3.4K]
2 years ago
13

What is the value of x

Mathematics
1 answer:
Lostsunrise [7]2 years ago
7 0

Answer: your answer is 36.3 cause it's half of 72.6.

Step-by-step explanation:

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Find the shaded region of the figure below​
lilavasa [31]

Answer:

-x³ + 3x² - 14x + 12

Step-by-step explanation:

Area of outer rectangle = (x² + 3x - 4) * (2x - 3)

      = (x² + 3x - 4) * 2x  + (x² + 3x - 4) * (-3)

     =x²*2x + 3x *2x - 4*2x  + x² *(-3) + 3x *(-3)  - 4*(-3)

     =2x³ + 6x² - 8x - 3x² - 9x + 12

    = 2x³ + <u>6x² - 3x²</u>   <u>- 8x - 9x</u> + 12     {Combine like terms}

    = 2x³ + 3x² - 17x + 12

Area of inner rectangle = (x² - 1)* 3x

                                       = x² *3x - 1*3x

                                       = 3x³ - 3x

Area of shaded region = area of outer rectangle - area of inner rectangle

         = 2x³ + 3x² - 17x + 12 - (3x³ - 3x)

         = 2x³ + 3x² - 17x + 12 -3x³ + 3x

        = 2x³ - 3x³ + 3x² - 17x + 3x + 12

        = -x³ + 3x² - 14x + 12

4 0
3 years ago
8^0 is greater than 8^1 true or false
liq [111]

Answer:

False 8^0 is not greater than 8^1

Step-by-step explanation:

8^0 = 1

8^1= 8

8>1

3 0
3 years ago
W(x)=-3x-4 find w (7)
JulsSmile [24]

Answer:

w=-7

Step-by-step explanation:

Soooo-

W(x)=-3x-4

divide both sides by x

W=-3-4

combine like terms

W=-7

3 0
3 years ago
Read 2 more answers
This is a description of a(n)<br> A) asteroid. <br> B) comet. <br> C) meteor. <br> D) star.
sergey [27]
B. Comet


Sometimes called a dirty snowball


6 0
3 years ago
Assuming that the sample mean carapace length is greater than 3.39 inches, what is the probability that the sample mean carapace
joja [24]

Answer:

The answer is "".

Step-by-step explanation:

Please find the complete question in the attached file.

We select a sample size n from the confidence interval with the mean \muand default \sigma, then the mean take seriously given as the straight line with a z score given by the confidence interval

\mu=3.87\\\\\sigma=2.01\\\\n=110\\\\

Using formula:

z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}  

The probability that perhaps the mean shells length of the sample is over 4.03 pounds is

P(X>4.03)=P(z>\frac{4.03-3.87}{\frac{2.01}{\sqrt{110}}})=P(z>0.8349)

Now, we utilize z to get the likelihood, and we use the Excel function for a more exact distribution

=\textup{NORM.S.DIST(0.8349,TRUE)}\\\\P(z

the required probability: P(z>0.8349)=1-P(z

4 0
2 years ago
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