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2 years ago

5

Can any one pleasssssseeee help me with algebra home work? I need help on A, B, and C

1 answer:

Komok [63]2 years ago

7 0

By looking at this scatter plot you can see that 2 people bowled above their average, 1 person bowled the same number for their average and their highest score (this person bowled 200 as their highest score as well as their average), and the biggest difference between someones high score and their average is one person has 200 for their bowling average, but only 175 for their high score. Everyone else's high scores and averages are about the same.

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Write 2603094 using exponents?

Artemon [7]

2*10^6 + 6*10^5 + 3*10^3 +9*10^1 +4*10^0

see expanded form is adding the digits together. and it would be what ever one number is times 10 to the power of how many digits are after that number. you skip the 0's though. hope this helps

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2 years ago

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What is the area of the figure shown below? giving brainly!!

siniylev [52]

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4073.76

Step-by-step explanation:

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Type the correct answer in the box. Use numerals instead of words.

stiks02 [169]

Answer:

78

Step-by-step explanation:

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2 years ago

Determine the domain and range of the given function.

frez [133]

Answer: domain is all real numbers and range is all real numbers greater than or equal to -2

Step-by-step explanation:

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2 years ago

What is the positive slope of the asymptote of (y+11)^2/100-(x-6)^2/4=1? The positive slope of the asymptote is .

VladimirAG [237]

Answer:

the positive slope of the asymptote = 5

Step-by-step explanation:

Given that:

$\frac{(y+11)^2}{100} -\frac{(x-6)^2}{4} =1$

Using the standard form of the equation:

$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}= 1$

where:

(h,k) are the center of the hyperbola.

and the y term is in front of the x term indicating that the hyperbola opens up and down.

a = distance that indicates how far above and below of the center the vertices of the hyperbola are.

For the above standard equation; the equation for the asymptote is:

$y = \pm \frac{a}{b} (x-h)+k$

where;

$\frac{a}{b}$ is the slope

From above;

(h,k) = 11, 100

$a^2$ = 100

a = $\sqrt{100}$

a = 10

$b^2 =4$

b = $\sqrt{4}$

b = 2

$y = \pm \frac{10}{2} (x-11)+2$

$y = \pm 5 (x-11)+2$

y = 5x-53 , -5x -57

Since we are to find the positive slope of the asymptote: we have

$\frac{a}{b}$ to be the slope in the equation $y = \pm \frac{10}{2} (y-11)+2$

$\frac{a}{b}$ = $\frac{10}{2}$

$\frac{a}{b}$ = 5

Thus, the positive slope of the asymptote = 5

8 0

2 years ago