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4 years ago

HELP PLEASE PLEASE :’(

Mathematics

1 answer:

Svet_ta [14]4 years ago

8 0

Practice:

1. A.

2. D. y=5+0.84x-0.03x^2, x>=0

3. B. y=2/x, x>0

Quiz:

1. A C E. (-4,4) (0,0) (6,9)

2. A. y=-1/3x+19/3

3. D. x=2y^2-1, y>=0

4. B. y=3/x^2, x>0

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Answer:

There's two answers.

-5, -4, and -3

3, 4, 5

Step-by-step explanation:

$x^{2} +(x+1)^{2} +(x+2)^{2} =50\\x^{2} +x^{2} +2x+1+x^{2} +4x+4=50\\3x^{2} +6x+5=50\\3x^{2} +6x-45=0\\3(x^{2} +2x-15)=0\\(x+5)(x-3)=0\\x = -5\\x = 3\\$

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3 years ago

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Rufina [12.5K]

Answer:

Note that i= √-1 so you say √-25=√25i so the answer is 5i

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3 years ago

Simplify 6 to the 3rd

elixir [45]

Answer:

Isnt it 2?

Step-by-step explanation:

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3 years ago

What is the slope of the line that goes through points(-2,2),and(-4,-2)

Dahasolnce [82]

Slope = y2 - y1/x2 - x1
slope = -2 - 2/-4 - -2
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3 years ago

Find the six trigonometric values for this angle. Will mark Brainliest.

nlexa [21]

Step-by-step explanation:

Terminal arm of the angle passes through the point (-2, 11) = (x, y)

$r = \sqrt{ {x}^{2} + {y}^{2} } \\ = \sqrt{ {( - 2)}^{2} + {11}^{2} } \\ = \sqrt{4 + 121} \\ = \sqrt{125} \\ = 5 \sqrt{5} \\ \\ sin \beta = \frac{y}{r} = \frac{11}{5 \sqrt{5} } \\ \\ cos\beta = \frac{x}{r} = \frac{ - 2}{5 \sqrt{5} } = - \frac{ 2}{5 \sqrt{5} } \\ \\ tan \beta = \frac{y}{x} = \frac{11}{ - 2} = - \frac{11}{2} \\ \\ cosec \beta = \frac{r}{y} = \frac{5 \sqrt{5} }{11 } \\ \\ sec\beta = \frac{r}{x} = \frac{5 \sqrt{5}}{ - 2} = - \frac{5 \sqrt{5}}{ 2} \\ \\ cot \beta = \frac{x}{y} = \frac{ - 2}{ 11} = - \frac{2}{11}$

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3 years ago