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hram777 [196]
2 years ago
10

Can x = 2i be a "solution" for (x+3)^2 =5

Mathematics
1 answer:
Doss [256]2 years ago
4 0

Answer:

the x =2 cant be a solution for (x+3)^2 =5.

hope it helps.

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Step-by-step explanation:

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Read 2 more answers
If log(base4)x=2.5 and log(base Y)125=-3/2, find the numerical value of x/y, in simplest form.
Naya [18.7K]
X = (2^2)^(2.5) 
<span>x = 2^(2 * 2.5) </span>
<span>x = 2^5 </span>
<span>x = 32 
</span>y^(-3/2) = 125 
<span>y^(-3) = 125^2 </span>
<span>y^(-3) = (5^3)^2 </span>
<span>y^(-3) = (5^2)^3 </span>
<span>y^(-3) = 25^3 </span>
<span>y = 25^(-1) </span>
<span>y = 1/25 </span>


<span>x/y => </span>
<span>32 / (1/25) => </span>
<span>32 * 25 => </span>
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3 years ago
Which of the following statements about the sine, cosine, and tangent ratios is not correct?
torisob [31]
I think the answer is b
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3 years ago
Let g(x) = 4x - 20 and h(x) = -1/2x + k. For what value of k is the zero of h(x) equivalent to the zero of g(x)
uranmaximum [27]

Answer:

Step-by-step explanation:

The zero of a function is also known as the x-intercept of a function. The x-intercept of a function exists where y = 0. So we will begin by setting each of those lines = 0:

4x - 20 = 0     and     -1/2x + k = 0 and then solve each for x:

4x = 20 so

x = 5 and

-1/2x = -k and

1/2x = k so

x = 2k. Now we set the 2 x expressions equal to each other and solve for k:

5 = 2k so

k = 5/2

8 0
3 years ago
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