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Lynna [10]
3 years ago
13

Find the value of x that makes mln (8x - 22) (5x + 290°​

Mathematics
1 answer:
Anastasy [175]3 years ago
4 0

Answer:

give me brainliest now plsssss

Step-by-step explanation:

(8x-22)=5x+29

We move all terms to the left:

(8x-22)-(5x+29)=0

We get rid of parentheses

8x-5x-22-29=0

We add all the numbers together, and all the variables

3x-51=0

We move all terms containing x to the left, all other terms to the right

3x=51

x=51/3

x=17

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Which is the initial value that shrinks an exponential growth function by 50%?
ValentinkaMS [17]
Suppose that the exponential growth function is  y = 4^x.

If x = 1, then y = 4^1 = 4.
If x = 3, then y = 4^3 = 64.

Let the initial value be 1/2.

The growth function is now y = 1/2 * 4^x.

If x =1, then y = 2 which is half of the y value before adding the 1/2.

If x = 3, then y = 1/2 * 64 = 32.

Answer: 1/2
7 0
3 years ago
Read 2 more answers
The hypotenuse is 8 inches. The side opposite angle X is 7 inches. Angle X = ?
Alex73 [517]

Step-by-step explanation:

In the picture.

The answer is <em><u>X = 61°</u></em>

6 0
3 years ago
Over which interval is the graph of f(x) = one-halfx2 + 5x + 6 increasing?<br><br> (–6.5, ∞)
juin [17]

Using the first derivative test, it is found that the graph is increasing on the interval is (-5, ∞).

<h3>What is Function?</h3>

A function is a process or a relation that associates each element 'a' of a non-empty set A , at least to a single element 'b' of another non-empty set B.

Here, the function is given by:

        f(x)  =   x²/2 + 5x + 6

On differentiating both sides, with respect to x we get

      f'(x) = x + 5

Then, we test the signal of the derivative, which is the first derivative test.

      x + 5 ≥ 0

       x ≥ -5

Thus, It is found that the Function is increasing on the interval is (-5, ∞).

Learn more about Function from:

brainly.com/question/12431044

#SPJ1

7 0
2 years ago
Evaluate y = 9 * (5/2)^x for x = -3.
dem82 [27]
You can find your answer by just substituting -3 for x......
6 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5C%7B%20%5Cfrac%20%7B%20%28%20%5Csqrt%20%7B%203%20%7D%20%29%20%5Ctimes%203%20%5E%20%7B%20-%20
ZanzabumX [31]

Answer:

Step-by-step explanation:

Exponent law:

    \sf \bf a^m * a^n = a^{m+n}\\\\ (a^m)^n = a^{m*n}

    \sf a^{-m}=\dfrac{1}{a^m}

       First convert radical form to exponent form and then apply exponent law.

 \sf \sqrt{3}=3^{\frac{1}{2}}\\\\\sqrt{5}=5^{\frac{1}{2}}

\sf \left(\dfrac{(\sqrt{3}*3^{-2}}{(\sqrt{5})^2}\right)^{\frac{1}{2}}= \left(\dfrac{3^{\frac{1}{2}}*3^{-2}}{(5^{\frac{1}{2}})^2} \right )^{\frac{1}{2}}

                      = \left(\dfrac{3^{\frac{1}{2}-2}}{5^{\frac{1}{2}*2}}\right)^{\frac{1}{2}}\\\\=\left(\dfrac{3^{\frac{1-4}{2}}}{5}\right)^{\frac{1}{2}}\\\\=\left(\dfrac{3^{\frac{-3}{2}}}{5}\right)^{\frac{1}{2}}\\\\=\dfrac{3^{\frac{-3}{2}*{\frac{1}{2}}}}{5^{\frac{1}{2}}}\\\\ =\dfrac{3^{{\frac{-3}{4}}}}{5^{\frac{1}{2}}}

3 0
2 years ago
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