Ask question

Ask question

All categories

3 years ago

7

The admission fee at a small fair is 1.50 for children and 4.00 for adults on a certain day 5,050 is collected if 1,000 adults a

ttended the fair write and solve a linear equation

1 answer:

blondinia [14]3 years ago

7 0

1.50x + 4.00(1000) =5050 would be linear equation.

1.50x + 4000=5050

1.50x=1050

x=1050/1.50

x=700

You might be interested in

What is the y-intercept of a line perpendicular to y= 1/2x + 6 that passes through the point (4,-3)

Finger [1]

Answer:

(0,5)

Step-by-step explanation:

Perpendicular lines have slopes which are negative reciprocals of each other. The slope of the line y=1/2x + 6 is 1/2. The slope perpendicular to it is -2.

Using m=-2, substitute it and (4,-3) into the point slope form and simplify to find the y-intercept.

$y-y_1 = m(x-x_1)\\y--3=-2(x-4)\\y+3=-2(x-4)\\y+3=-2x+8\\y=-2x+5$

The equation is now in slope intercept form y=mx+b and b is the y-intercept. The y-intercept of y=-2x+5 is (0,5).

4 0

3 years ago

Fill in the missing numbers to complete the pattern:<br> 2.65, 2.67, ______, ______, 2.73, 2.75

ludmilkaskok [199]

The answers are 2.69 and 2.71

4 0

3 years ago

Read 2 more answers

In a triangle ABC, if angle A = 53 degree and Angle C=45 degree then find the value of Angle B is:

murzikaleks [220]

Answer:

69

It wont always be easy, but always try to do what ' s right!

In a world of worriers, be the warrior!

When nothing goes right, go left.

4 0

2 years ago

Please click the photo and there are two questions to solve .Please help me guys

madreJ [45]

Answer:

Step-by-step explanation:

Distance formula:

$\sqrt{(X_2-X_1)^2+(Y_2-Y_1)^2}$

Distance for AB

$\sqrt{(4-2)^2+(1-(-5))^2}=\sqrt{4+36}=\sqrt{40}\\$

Distance for AC

$\sqrt{(8-2)^2+(-3-(-5))^2}=\sqrt{36+4}=\sqrt{40}$

2.

same formula solve when positive

$5=\sqrt{(6-3)^2+(y-2)^2}\\25=9+(y-2)^2\\16=(y-2)^2\\4=y-2\\6=y$

Now when it's negative

$5=\sqrt{(6-3)^2+(y-2)^2}\\25=9+(y-2)^2\\16=(y-2)^2\\-4=y-2\\y=-2$

so y can equal 6 or -2

3 0

3 years ago

At a campground, a rectangular fire pit is 3 feet by 2 feet. What is the area of the largest circular fire that

AysviL [449]

The area of the largest circular fire pit is 452 square inches.

Explanation:

Given that at a campground, a rectangular fire pit is 3 feet by 2 feet

The radius is given by 1 feet.

<u>Area:</u>

The area of the largest circular fire is given by

$A=\pi r^2$

Substituting the values in the formula, we get,

$A=(3.14)1^2$

$A=3.14 \ ft^2$

Thus, the area of the largest circular fire is 3.14 square feet.

<u>To convert feet to inches:</u>

The feet can be converted into inches by multiplying by 12.

Thus, we have,

$1 \ft= 12 \ inches$

Hence, we have,

$Area= 3.14\times 12^2$

Simplifying,we get,

$Area= 3.14\times144$

$Area=452.16 \ in^2$

Rounding off to the nearest square inch.

Thus, we have,

$Area= 452 \ in^2$

Thus, the area of the largest circular fire pit is 452 square inches.

5 0

3 years ago