30 students in a class took an algebra test.

Write the equation of the line that passes through the points (7,-6) and (0,2). Put

Alex

Answer:

$y+6=\frac{-8}{7} (x-7)$

Step-by-step explanation:

To find the slope of the equation use $\frac{y_2-y_1}{x_2-x_1}$ .

So, $\frac{-6-2}{7-0}$ = $\frac{-8}{7}$ . Now we can use our slope and any of the two points to write in point-slope form, which is $y-y_1=m(x-x_1)$ . Using the point (7,-6), the formula will give $y+6=\frac{-8}{7} (x-7)$ .

To check, you can plug in this equation and the points into a calculator to graph. The line passes through both points.

6 0

3 years ago

The geometric mean of 4 and 16 is? A. 64 B. 16 C. 12 D. 8

guajiro [1.7K]

The correct answer is D.8

5 0

3 years ago

Read 2 more answers

A dealer sold 200 tennis racquets. Some were sold for $33 each, and the rest were sold on sale for $18. The total receipts form

Mumz [18]

120 tennis racquets were sold for $ 18 each

Solution:

Let x = number that sell for $18 each

Then, 200 - x = number that sell for $33 each

The total receipts form these sales were 4800 dollars

Thus we frame a equation as:

number that sell for $18 each x 18 + number that sell for $33 each x 33 = 4800

$x \times 18 + (200-x) \times 33 = 4800\\\\Solve\ the\ above\ equation\ for\ x\\\\18x + 6600 - 33x = 4800\\\\Move\ the\ variables\ to\ one\ side\\\\18x-33x = 4800-6600\\\\-15x = -1800\\\\Divide\ both\ sides\ of\ equation\ by\ -15\\\\x = 120$

Thus 120 tennis racquets were sold for $ 18 each

6 0

3 years ago

a 400 g of a liquid of density 1.6g/cm3 is made from mixing liquids of density 1.2g/cm3 and 1.8g/cm3 find the mass of the liquid

gogolik [260]

Answer:

133.33g

Step-by-step explanation:

Let the:

Mass of 1.2g/cm³ of liquid = x

Mass of 1.8g/cm³ of liquid = y

From our Question above, our system of equations is given as:

x + y = 400........ Equation 1

x = 400 - y

1.2 × x + 1.8 × y = 1.6 × 400

1.2x + 1.8y = 640..... Equation 2

We substitute, 400 - y for x in Equation 2

1.2(400 - y) + 1.8y = 640

480 - 1.2y + 1.8y = 640

- 1.2y + 1.8y = 640 - 480

0.6y = 160

y = 160/0.6y

y = 266.67 g

Solving for x

x = 400 - y

x = 400 - 266.67g

x = 133.33g

Therefore, the mass of the liquid of density 1.2g/cm³ is 133.33g

7 0

3 years ago

Find constants a and b such that the function y = a sin(x) + b cos(x) satisfies the differential equation y'' + y' − 5y = sin(x)

vichka [17]

Answers:

a = -6/37

b = -1/37

============================================================

Explanation:

Let's start things off by computing the derivatives we'll need

$y = a\sin(x) + b\cos(x)\\\\y' = a\cos(x) - b\sin(x)\\\\y'' = -a\sin(x) - b\cos(x)\\\\$

Apply substitution to get

$y'' + y' - 5y = \sin(x)\\\\\left(-a\sin(x) - b\cos(x)\right) + \left(a\cos(x) - b\sin(x)\right) - 5\left(a\sin(x) + b\cos(x)\right) = \sin(x)\\\\-a\sin(x) - b\cos(x) + a\cos(x) - b\sin(x) - 5a\sin(x) - 5b\cos(x) = \sin(x)\\\\\left(-a\sin(x) - b\sin(x) - 5a\sin(x)\right) + \left(- b\cos(x) + a\cos(x) - 5b\cos(x)\right) = \sin(x)\\\\\left(-a - b - 5a\right)\sin(x) + \left(- b + a - 5b\right)\cos(x) = \sin(x)\\\\\left(-6a - b\right)\sin(x) + \left(a - 6b\right)\cos(x) = \sin(x)\\\\$

I've factored things in such a way that we have something in the form Msin(x) + Ncos(x), where M and N are coefficients based on the constants a,b.

The right hand side is simply sin(x). So we want that cos(x) term to go away. To do so, we need the coefficient (a-6b) in front of that cosine to be zero

a-6b = 0

a = 6b

At the same time, we want the (-6a-b)sin(x) term to have its coefficient be 1. That way we simplify the left hand side to sin(x)

-6a -b = 1

-6(6b) - b = 1 .... plug in a = 6b

-36b - b = 1

-37b = 1

b = -1/37

Use this to find 'a'

a = 6b

a = 6(-1/37)

a = -6/37

8 0

2 years ago