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Oliga [24]
3 years ago
15

Solve each equation. Showing your reasoning. X^2-6x-20=7

Mathematics
1 answer:
scoundrel [369]3 years ago
4 0

Answer:

x=-3,x=9

Step-by-step explanation:

x^2-6x-20-7=0

x^2-6x-27=0

x^2-(9-3)x-27=0

x^2-9x+3x-27=0

x(x-9)+3(x-9)=0

(x+3)(x-9)=0

x=-3,x=9

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Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s). Consider the gi
STatiana [176]

Answer:

To determine the inverse of the given function, change f(x) to y, switch x and y and solve for y and f^{-1}(x)= \frac{ln\ x+4}{2}

Step-by-step explanation:

Data provided in the question

f(x) = e^2x - 4

Now

to find the inverse let

So,

y = e^2x - 4

Now

Replace x and y

Therefore

x = e^2y - 4

Now compute the value of y

So,

x + 4 = e^2y

Now take ln on both sides:

The equation is

ln(x+4) = ln(e^2y)

ln(x+4) = 2y

y = ln(x+4) ÷ 2

f^{-1}(x)= \frac{ln\ x+4}{2}

Therefore,  To determine the inverse of the given function, change f(x) to y, switch x and y and solve for y and f^{-1}(x)= \frac{ln\ x+4}{2}

5 0
3 years ago
Pls help and do quick!
motikmotik

Answer:

If 50.24cm = Circumference

Radius = 8cm

If 50.24cm^2 = Area

Radius = 4cm

Step-by-step explanation:

50.24cm what? Circumference? Area?

π = 3.14

2πr = Circumference

50.24 / 2π = r

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7 0
2 years ago
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Scilla [17]

Answer: 4.5 looks the most reasonable to me.

Step-by-step explanation:

3 0
2 years ago
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erica [24]
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8 0
3 years ago
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What are the real zeros of the function g(x) = x3 + 2x2 − x − 2? 1, –1, 2 1, –1, –2 1, –1 2, –2, 1
geniusboy [140]
 Hello there!

How are you doing today? I hope you are doing at least "okay".

g(x) = x³ + 2x² - x - 2

To find the zeros or the roots of a function, you need to set the function equal to 0 then solve for x.

x³ + 2x² - x - 2 = 0

Factor left side of the equation

(x + 1)(x - 1)(x + 2) = 0

Now we can set factors equal to 0

x + 1= 0 or x - 1 = 0 or x + 2 = 0

x= 0 - 1 or x = 0 + 1 or x= 0 - 2

x= -1 or x = 1 or x = -2

Thus,

The roots or the zeros of the function are:

-2, -1, 1

As always, it is my pleasure to help students like you!
8 0
2 years ago
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