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kondor19780726 [428]

2 years ago

7

Simplify the expression 2a + 5b² - a - b - 3b²

2 answers:

Margarita [4]2 years ago

8 0

Answer:

a-b+2b^2

Step-by-step explanation:

collect like terms

2a-a-b+5b^2-3b^2

a-b+2b^2

Vikki [24]2 years ago

6 0

Answer:

a+2b²-b (correct order is 2b²+a-b)

Step-by-step explanation:

combine like terms ( terms with the same variable and are to the same power) so thats 5b² and -3b² and 2a and -a. the -b doesnt have any like terms so leave that alone for now. combine like terms 5b²-3b² is 2b² and 2a-a is a. so you have 2b²+a and plug in the -b so you get 2b²+a-b.

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Step-by-step explanation:

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2 years ago

Factor? for 6xy−3x−8y+4

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6xy -3x -8y +4
3x (2y - 1 ) - 4 (2y -1)
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3 years ago

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tekilochka [14]

Answer:

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Step-by-step explanation:

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2 years ago

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Elena L [17]

Answer:

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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