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Alex777 [14]
4 years ago
6

G^2 +23 when. g=6 please help

Mathematics
1 answer:
lapo4ka [179]4 years ago
4 0

Answer:

59

Step-by-step explanation:

Substitute.

6 ^ 2 = 36

36 + 23 = 59

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A +-21 = -38<br><br>help me please​
ivann1987 [24]

Answer:

A= -17

Step-by-step explanation:

A +-21 = -38

    +21     +21

-38+21= -17

A= -17

5 0
3 years ago
Read 2 more answers
Suppose a certain drug test is 85% accurate, meaning that if a person is a user, the result is positive 85% of the time, and if
kakasveta [241]

Answer:

29.411 %  of the result is true.

Step-by-step explanation:

This test's sensitivity is low.

Suppose we have 5 drug users and 85% of the test is accurate meaning that 5* 0.85= 4.25 people are drug users which is true.

But if work for the non users than 5% people being drug users means 95 people are non drug users out of the hundred.

And 85 % accuracy would give  95* 0.85= 80.75 which is very high.

And out of the 80.75 * 15%= 12.11 only 12 are positive .

The results give us  5 out of 17  true possibilities

or 5/17 *100= 29.411 %  of the result is true.

8 0
3 years ago
(Im sure this is a easy answer but im just double checking)
IgorC [24]
This answer seems to be C
5 0
3 years ago
What is the quotient in simplified form? State any restrictions on the variable? \frac{x^2-16}{x^2+5x+6} /\frac{x^2+5x+4}{x^2-2x
lora16 [44]
\frac{x^2-16}{x^2+5x+6} / \frac{x^2+5x+4}{x^2-2x-8}

We can begin by rearranging this into multiplication:

\frac{x^2-16}{x^2+5x+6} * \frac{x^2-2x-8}{x^2+5x+4}

Now we can factor the numerators and denominators:

\frac{(x+4)(x-4)}{(x+3)(x+2)} * \frac{(x-4)(x+2)}{(x+4)(x+1)}

The factors (x+4) and (x+2) cancel out, leaving us with:

\frac{(x-4)}{(x+3)} * \frac{(x-4)}{(x+1)}

Our answer comes out to be:

\frac{(x-4)^{2} }{(x+3)(x+1)} or \frac{ x^{2} -8x+16}{ x^{2}+4x+3 }

Based on the numerator of the second fraction (since we used its inverse), the denominators of both, and the factors we canceled out earlier, the restrictions are x ≠ -4, -3, -2, -1, 4
4 0
3 years ago
The zeros of the function f(x)=(x+2)^2 - 25 are?
zloy xaker [14]
<u>Zeros of the function</u>
f(x) = (x + 2)² - 25
f(x) = (x + 2)(x + 2) - 25
f(x) = x(x + 2) + 2(x + 2) - 25
f(x) = x(x) + x(2) + 2(x) + 2(2) - 25
f(x) = x² + 2x + 2x + 4 - 25
f(x) = x² + 4x + 4 - 25
f(x) = x² + 4x - 21
x² + 4x - 21 = 0
x = <u>-(4) +/- √((4)² - 4(1)(-21))</u>
                      2(1)
x = <u>-4 +/- √(16 + 84)</u>
                   2
x = <u>-4 +/- √(100)
</u>               2<u>
</u>x = <u>-4 +/- 10
</u>            2<u>
</u>x = -2 <u>+</u> 5<u>
</u>x = -2 + 5    x = -2 - 5
x = 3           x = -7
f(x) = x² + 4x - 21
f(3) = (3)² + 4(3) - 21
f(3) = 9 + 12 - 21
f(3) = 21 - 21
f(3) = 0
(x, f(x)) = (3, 0)
or
f(x) = x² + 4x - 21
f(-7) = (-7)² + 4(-7) - 21
f(-7) = 49 - 28 - 21
f(-7) = 21 - 21
f(-7) = 0
(x, f(x)) = (-7, 0)

<u>Vertex</u>
<u>X - Intercept</u>
<u />-b/2a = -(4)/2(1) = -4/2 = -2

<u>Y - Intercept</u>
y = x² + 4x - 21
y = (-2)² + 4(-2) - 21
y = 4 - 8 - 21
y = -4 - 21
y = -25
(x, y) = (-2, -25)
<u />
5 0
3 years ago
Read 2 more answers
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