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3 years ago

Question 14 (2 points) Compute: 2(8) + 5(6) = O 12

Mathematics

1 answer:

FrozenT [24]3 years ago

5 0

Answer:

2×8+5×6=16+30=46

Step-by-step explanation:

you multiplie 2×8

and 5×6

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Simplify this expression: 1/2(10x + 12) - (3.5x - 7) *

Anna71 [15]

Simplified Expression:

1.5x+13

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3 years ago

PLSSSS PLSSS HELPPPP Use a Calculator to find the equation which models the exponential which models the exponential data given

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Step-by-step explanation:

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3 years ago

Which could be the resulting equation when elimination is used to solve the given system of equations ?

WINSTONCH [101]

Answer:

10·b = 60

Step-by-step explanation:

The given system of equation is presented as follows;

5·a + 5·b = 25...(1)

-5·a + 5·b = 35...(2)

Given that the coefficient of a in equation (1) is equal in magnitude but opposite in sign to the coefficient of 'a' in equation (2), to eliminate the variable 'a' when using the elimination method, we add both equations as follows;

5·a + 5·b + (-5·a + 5·b) = 25 + 35 = 60

5·a - 5·a + 5·b + 5·b = 60

5·a - 5·a = 0

5·b + 5·b = 10·b

∴ 5·a - 5·a + 5·b + 5·b = 60 = 0 + 10·b = 10·b

∴ 10·b = 60

5 0

3 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

3 years ago

What is 6/9 in it's lowest terms??

gtnhenbr [62]

The simplified answer of 6/9 is 2/3.

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3 years ago