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Contact [7]
3 years ago
11

Can some one help me with this question its hard :( down below ill give brainliest

Mathematics
1 answer:
elixir [45]3 years ago
5 0

Answer:

C. 30f^5

Step-by-step explanation:

5.6 f^7+(-2)

30f^5

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PLEASE HELP!<br> WILL MARK BRAINLIEST!
RideAnS [48]

Answer:

1) 7m

2) 13

3) 4x^2y

4) 6s^3t^4

Step-by-step explanation:

gcf means the the greatest factor that works for both terms

1) 21m^3 and 28m

    let's first find the gcf for the coefficient

    21 and 28, the gcf is 7. (hopefully thats easy to explain)

    the gcf for the variables is the same mindset as the numbers

    m^3 and m, both can be factored by m

    so the gcf is 7 * m which is 7m

2)

  13x and 26

  let's first find the gcf for the coefficient

  13 and 26, gcf is 13

  there's only one variable, x, so there's not gcf for that

 so the gcf is just 13

3)

  8x^2y and -12x^3y^2

  let's first find the gcf for the coefficient

  8 and -12, the gcf is 4 (you can also say -4)

  for the variables , same mindset

 x^2y and x^3y^2

 the gcf for x is x^2, and for y it is just y

 so 4 * x^2 * y = 4x^2y

by understanding 3) hopefully you can understand 4)

pls comment for any questions

7 0
4 years ago
Noah is planning his birthday party. He is deciding between three themes: space, safari, and comics. He also has to determine wh
mixas84 [53]

Answer:

the answer is 18.l hope is good

8 0
3 years ago
Read 2 more answers
Inequalities I need help!​
olasank [31]

Answer:

\large\boxed{b

Step-by-step explanation:

-b-2>8\qquad\text{add 2 to both sides}\\\\-b-2+2>8+2\\\\-b>10\qquad\text{change the signs}\\\\b

4 0
3 years ago
Suppose a random variable x is best described by a uniform probability distribution with range 22 to 55. Find the value of a tha
const2013 [10]

Answer:

(a) The value of <em>a</em> is 53.35.

(b) The value of <em>a</em> is 38.17.

(c) The value of <em>a</em> is 26.95.

(d) The value of <em>a</em> is 25.63.

(e) The value of <em>a</em> is 12.06.

Step-by-step explanation:

The probability density function of <em>X</em> is:

f_{X}(x)=\frac{1}{55-22}=\frac{1}{33}

Here, 22 < X < 55.

(a)

Compute the value of <em>a</em> as follows:

P(X\leq a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.95\times 33=[x]^{a}_{22}\\\\31.35=a-22\\\\a=31.35+22\\\\a=53.35

Thus, the value of <em>a</em> is 53.35.

(b)

Compute the value of <em>a</em> as follows:

P(X< a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.49\times 33=[x]^{a}_{22}\\\\16.17=a-22\\\\a=16.17+22\\\\a=38.17

Thus, the value of <em>a</em> is 38.17.

(c)

Compute the value of <em>a</em> as follows:

P(X\geq  a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.85=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.85\times 33=[x]^{55}_{a}\\\\28.05=55-a\\\\a=55-28.05\\\\a=26.95

Thus, the value of <em>a</em> is 26.95.

(d)

Compute the value of <em>a</em> as follows:

P(X\geq  a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.89=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.89\times 33=[x]^{55}_{a}\\\\29.37=55-a\\\\a=55-29.37\\\\a=25.63

Thus, the value of <em>a</em> is 25.63.

(e)

Compute the value of <em>a</em> as follows:

P(1.83\leq X\leq  a)=\int\limits^{a}_{1.83} {\frac{1}{33}} \, dx \\\\0.31=\frac{1}{33}\cdot \int\limits^{a}_{1.83} {1} \, dx \\\\0.31\times 33=[x]^{a}_{1.83}\\\\10.23=a-1.83\\\\a=10.23+1.83\\\\a=12.06

Thus, the value of <em>a</em> is 12.06.

7 0
3 years ago
Solve this system of equations.<br><br><br><br> {3x+4y=36y=−12x+8
SIZIF [17.4K]

Step-by-step explanation:

(1 point)

No, there isn't a more efficient way to solve this system.

Yes, a more efficient way is to multiply the first equation by 4, add to eliminate y, then solve for x.

Yes, a more efficient way is to multiply the first equation by −4, add to eliminate y, then solve for x.

Yes, a more efficient way is to multiply the first equation by −4, add to eliminate x, then solve for y.

3 0
3 years ago
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