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kherson [118]
3 years ago
5

True or false? (Please help me)

Mathematics
1 answer:
gayaneshka [121]3 years ago
6 0

Answer:

False

Explanation:

5 is 1/5 of 25

2 is 1/10 of 20

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The square root of -12 × the square root of -9 (THIS IS AN ALGEBRA 2 QUESTION)
nlexa [21]

Answer:

\sqrt{-12} = 2\sqrt{3} i\\\sqrt{-9}= 3i\\3i * 2\sqrt{3}i = -6\sqrt{3

Step-by-step explanation:

5 0
3 years ago
Nadia will be x years old y years from now. How old was she z years ago?
Vsevolod [243]
A years old. Na but are you needing an equation ?
8 0
3 years ago
Given r(x) = 11/ (x - 42)
julsineya [31]

For a given function f(x) we define the domain restrictions as values of x that we can not use in our function. Also, for a function f(x) we define the inverse g(x) as a function such that:

g(f(x)) = x = f(g(x))

<u>The restriction is:</u>

x ≠ 4

<u>The inverse is:</u>

y = 4 + \sqrt{\frac{11}{x} }

Here our function is:

f(x) = \frac{11}{(x - 4)^2}

We know that we can not divide by zero, so the only restriction in this function will be the one that makes the denominator equal to zero.

(x - 4)^2 = 0

x - 4 = 0

x = 4

So the only value of x that we need to remove from the domain is x = 4.

To find the inverse we try with the general form:

g(x) = a + \sqrt{\frac{b}{x} }

Evaluating this in our function we get:

g(f(x)) = a + \sqrt{\frac{b}{f(x)} }  = a + \sqrt{\frac{b*(x - 4)^2}{11 }}\\\\g(f(x)) = a + \sqrt{\frac{b}{11 }}*(x - 4)

Remember that the thing above must be equal to x, so we get:

g(f(x)) = a + \sqrt{\frac{b}{11 }}*(x - 4) = x\\\\{\frac{b}{11 }} = 1\\{\frac{b}{11 }}*4 - a = 0

From the two above equations we find:

b = 11

a = 4

Thus the inverse equation is:

y = 4 + \sqrt{\frac{11}{x} }

If you want to learn more, you can read:

brainly.com/question/10300045

3 0
2 years ago
HELP ME I NEED SOMEHELP<br> PLZ<br> PLZ<br> PLZ
otez555 [7]

Step-by-step explanation:

2/7

2×2 = 4

___

7×2 = 14

2×3 = 6

___

7×3 = 21

2×4 = 8

___

7×4 = 12

8 0
3 years ago
Please help expand the expression using pascal's triangle<br><br> (1-2x)^4
aliya0001 [1]
(1 - 2x)⁴
(1 - 2x)(1 - 2x)(1 - 2x)(1 - 2x)
[1(1 - 2x) - 2x(1 - 2x)][1(1 - 2x) - 2x(1 - 2x)]
[1(1) - 1(2x) - 2x(1) - 2x(-2x)][1(1) - 1(2x) - 2x(1) - 2x(-2x)]
(1 - 2x - 2x + 4x²)(1 - 2x - 2x + 4x²)
(1 - 4x + 4x²)(1 - 4x + 4x²)
1(1 - 4x + 4x²) - 4x(1 - 4x + 4x²) + 4x²(1 - 4x + 4x²)
1(1) - 1(4x) + 1(4x²) - 4x(1) - 4x(-4x) - 4x(4x²) + 4x²(1) - 4x²(4x) + 4x²(4x²)
1 - 4x + 4x² - 4x + 16x² - 16x³ + 4x² - 16x³ + 16x⁴
1 - 4x - 4x + 4x² + 16x² + 4x² - 16x³ - 16x³ + 16x⁴
1 - 8x + 24x² - 32x³ + 16x⁴
5 0
3 years ago
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