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lesya692 [45]
3 years ago
13

Which is the graph of f(x) = 4"?

Mathematics
1 answer:
Ugo [173]3 years ago
3 0

Answer: yee

Step-by-step explanation:

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Need help for this Algebra question plz!
Vikentia [17]

Answer:

2x+5y=-15

ax+bx=c

slope=-a/b so -2/5 matches up with point slope form

then to get y intercept substitite x in point slope for 0 and solve for Y to get -3

then do c/b to get y intercept, so -15/5, which equals -3.

5 0
3 years ago
Can someone please show me how<br><br> 19169000*e^(0.15)= 222712201<br><br> When e= 2.7182818284?
Ierofanga [76]

Answer:

It doesn't. x = 22 271 201

Step-by-step explanation:

You are going to need a calculator no matter how you do it.

x =19 169 000 \times e^{0.15}

(a) The direct method

x = 19 169 000\times 2.718 281 828^{0.15} = 19 169 000 \times 1.161 834 243 = 22 271 201

(b) The indirect method

\ln \left (19 169 000\times 2.718 281 828^{0.15} \right ) = \ln(19 169 000) + 0.15 = 16.768 805 + 0.15 = 16.918 804\\\\e^{16.918 804} = 22 271 201

8 0
3 years ago
What is the area of this figure?​
iVinArrow [24]

Answer: To find the area of a rectangle, multiply its height by its width. For a square you only need to find the length of one of the sides (as each side is the same length) and then multiply this by itself to find the area.

6 0
2 years ago
1 + 5sin^2x= 7sin^2x solve trig equation. Can you show me how to solve this trig equation?
pantera1 [17]

Answer: x = \frac{\pi}{4}\pm k\frac{\pi}{2}\,\,\,\,\,\,k=\{0,1,2,...\}

Step by step:

1 + 5\sin^2x= 7\sin^2x\\\sin^2 x\rightarrow z\\1 + 5z = 7z\\2z = 1\\z = \frac{1}{2}\implies \sin^2 x = \frac{1}{2}\\|\sin x| = \frac{1}{\sqrt{2}}\\\sin x = \pm\frac{1}{\sqrt{2}}=\pm\frac{\sqrt{2}}{2}\\\implies x = \frac{\pi}{4}\pm k\frac{\pi}{2}\,\,\,\,\,\,k=\{0,1,2,...\}

(Used a table of common angles)

8 0
3 years ago
Solve for xif mZRQS = 2x+4 and mZTQS = 6x+ 20.<br> 180<br> 19.5<br> -4<br> 90
Likurg_2 [28]

Answer:

x= -4

Step-by-step explanation:

2x+4=6x+20

so 2x-6x=20-4

divide both sides by -4

-4x/-4 = 16/4

=-4

7 0
3 years ago
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