Which pair of equations represents two perpendicular lines?A. y=-7/8x+3 and -7y=-8x
B. 8y=3x+40 and y=8/2x-1
C. 5y=15-2x and 2/5x-4=y
D. y=9x+3 and y=9x-1/3

Answers

Answer 1

Answer: I am pretty sure that the answer is A. because the slopes are reciprocals of each other, and so they are perpendicular

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G is the incenter, or point of concurrency, of the angle bisectors of ΔACE.Triangle A C E has point G as its incenter. Lines are drawn from the points of the triangle to point G. Lines are drawn from point G to the sides of the triangle to form right angles. Line segments G B, G D, and G F are formed. Angle F E G is (3 n minus 1) degrees and angle D E G is 20 degrees.
What is the value of n?

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59G is the incenter, or point of concurrency, of the angle bisectors of ΔACE.

Triangle A C E has point G as its incenter. Lines are drawn from the points of the triangle to point G. Lines are drawn from point G to the sides of the triangle to form right angles. Line segments G B, G D, and G F are formed. Angle F E G is (3 n minus 1) degrees and angle D E G is 20 degrees.
What is the value of n?

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59

Answers

Answer:7

Step-by-step explanation:

Answer: 7

Step-by-step explanation:

if the radius of a circle is 3.5 inches its diameter is ____ inches and its approximate area is____ square inches

Answers

Answer:

Diameter = 7 inches

Area of circle is 38.46 inches²

Step-by-step explanation:

Given :The radius of a circle is 3.5 inches.

We have to find its diameter and area.

Since diameter is twice of radius.

So radius = 3.5 inches

Diameter = 2× 3.5 inches = 7 inches

Area of circle = $\pi r^2$

Substitute, r = 3.5

Area of circle = $\pi (3.5)^2$

Simplify , we have,

Area of circle = $\pi 12.25$

Using π = 3.14

We have,

Area of circle = $38.465$

Thus, Area of circle is 38.46 inches²

The diameter is 7 inches and the area is 38.465. I got the answer by taking the area of a circle's formula which is

pi r ^2

Kathy lives directly east of the park. The football field is directly south of the park. The library sits on the line formed between Kathy's home and the football field at the exact point where an altitude to the right triangle formed by her home, the park, and the football field could be drawn. The library is 9 miles from her home. The football field is 12 miles from the library.a. How far is the library from the park?
b. How far is the park from the football field?

Answers

see the attached figure to better understand the problem

Let

z---------> distance from the library to the park in miles

x-------> distance from the park to the to the football field in miles

y-------> distance from the park to Kathy's home in miles

we know that

In the right triangle ABC

Applying the Pythagorean Theorem

$x^(2) +y^(2) =(12+9)^(2) \n x^(2) +y^(2)=441$ -----> equation $1$

In the right triangle ABD

Applying the Pythagorean Theorem

$12^(2) +z^(2) =x^(2) \n 144 +z^(2)=x^(2)$ -----> equation $2$

In the right triangle BCD

Applying the Pythagorean Theorem

$9^(2) +z^(2) =y^(2) \n 81 +z^(2)=y^(2)$ -----> equation $3$

Add equation $2$ and equation $3$

$144 +z^(2)=x^(2)$

$81 +z^(2)=y^(2)\n ------$

$144+81+2z^(2) =x^(2) +y^(2)$ -----> equation $4$

Substitute equation $1$ in equation $4$

$144+81+2z^(2)=441\n 2z^(2) =441-225\n 2z^(2)=216\n z^(2)=108\n z=√(108) miles\n z=10.39 miles$

Find the value of x

$144 +z^(2)=x^(2)\n 144 +√(108)^(2)=x^(2) \n x^(2) =144+108\n x^(2) =252\n x=√(252) miles\n x=15.87 miles$

therefore

the answer is

Part a) The distance from the library to the park is equal to $10.39 miles$

Part b) The distance from the park to the to the football field is equal to $15.87 miles$

Look at the picture in the attachment.

Using the Pythagorean theorem, set up a system of three equations:
$x^2+y^2=(12+9)^2 \n12^2+z^2=x^2 \n9^2+z^2=y^2 \n \nx^2+y^2=441 \n144+z^2=x^2 \n81+z^2=y^2$

$\hbox{substitute } 144+z^2 \hbox{ for } x^2 \hbox{ and } 81+z^2 \hbox{ for } y^2 \hbox{ in the first equation:} \n144+z^2+81+z^2=441 \n225+2z^2=441 \n2z^2=441-225 \n2z^2=216 \nz^2=(216)/(2) \nz^2=108 \nz=√(108) \nz=√(36 * 3) \nz=6√(3) \n z \approx 10.39$

$81+z^2=y^2 \n81+108=y^2 \n189=y^2 \n√(189)=y \n√(9 * 21)=y \ny=3√(21) \n y \approx 13.75$

a. The library is approximately 10.39 miles (exactly: 6√3 miles) from the park.
b. The park is approximately 13.75 miles (exactly: 3√21 miles) from the football field.

What are the zeros of the polynomial function f (x)=x^2-12x+20

Answers

I hope this helps you

f (x)=(x-10)(x-2)

x=10

x=2

Answer:

the answer is 10 and 2

What is the reading in kwhr of the electric meter shown in the exam figure below?

Answers

We can't tell, because you won't let us SEE
the electric meter shown in the exam figure.

a new gym is being built at greenfield middle school. The length of the Gym is 12 inches on the builders blueprint. Find the actual length of the gym if the scale is 3/4 in = 5ft

Answers

Using scale conversion in mathematics, we set up a proportion based on the blueprint scale of 3/4 in equals 5 ft. Solving this proportion, we find the actual length of the gym is 80 feet.

In mathematics, this problem is about scale conversion.

The blueprint scale in this case is 3/4 in equals 5 ft.

To find out the actual length of the gym, we need to set up a proportion and solve for the missing value.

We know that 3/4 inches represents 5 feet on the actual gym.

According to the blueprint, the length of the gym is 12 inches.

First, let's set up the proportion: 3/4 in: 5 ft = 12 in: x ft.

Now we can cross multiply and solve for 'x'. (3/4) * x = 12 * 5, which simplifies to (3x/4) = 60.

Multiply both sides by 4 to isolate 'x': 3x = 240.

Finally, divide both sides by 3, we get x = 80.

Therefore, the actual length of the gym is 80 feet.

For more such questions on scale conversion, click on:

brainly.com/question/33515865