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2 years ago

What is an equation of the line that passes through the points (−6,−2) and (6, 8)

Mathematics

1 answer:

-BARSIC- [3]2 years ago

5 0

Answer: $y=\frac{5}{6}x+3$

Step-by-step explanation:

To find the equation of the line with the given points, we want to first find the slope then the y-intercept. We use the formula $m=\frac{y_2-y_1}{x_2-x_1}$ .

$m=\frac{8-(-2)}{6-(-6)} =\frac{10}{12}=\frac{5}{6}$

Now that we know the slope, we can use one of the points to find the y-intercept.

$8=\frac{5}{6}(6)+b$ [multiply]

$8=5+b$ [subtract both sides by 5]

$b=3$

Now, we know the equation is $y=\frac{5}{6}x+3$ .

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Show whether or not y=x+3 is tangential to the curve y^2=x

wolverine [178]

The line y = x + 3 has slope 1, so we look for points on the curve where the tangent line, whose slope is dy/dx, is equal to 1.

y² = x

Take the derivative of both sides with respect to x, assuming y = y(x) :

2y dy/dx = 1

dy/dx = 1/(2y)

Solve for y when dy/dx = 1 :

1 = 1/(2y)

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y = 1/2

When y = 1/2, we have x = y² = (1/2)² = 1/4. However, for the given line, when y = 1/2, we have x = y - 3 = 1/2 - 3 = -5/2.

This means the line y = x + 3 is not a tangent to the curve y² = x. In fact, the line never even touches y² = x :

x = y² ⇒ y = y² + 3 ⇒ y² - y + 3 = 0

has no real solution for y.

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2 years ago

From a group of 12 students, we want to select a random sample of 4 students to serve on a university committee. How many combin

borishaifa [10]

Answer:

495 combinations of 4 students can be selected.

Step-by-step explanation:

The order of the students in the sample is not important. So we use the combinations formula to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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4 from a set of 12. So

$C_{n,x} = \frac{12!}{4!(8)!} = 495$

495 combinations of 4 students can be selected.

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3 years ago