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Ahat [919]
4 years ago
11

Joe decides to take his six year old son to the planetarium. The price for the child's ticket is 5.75 dollars less than the pric

e for the adult's ticket. If you represent the price for the child's ticket using the variable "x," how would you write the algebraic expression for the adult's ticket price? 5.75x x over 5.75 x + 5.75 x5.75
Mathematics
2 answers:
zaharov [31]4 years ago
6 0
It would be x + 5.75 because the child ticket is 5.75 less than the adult ticket so to solve how to get the adult ticket you do the opposite
AleksandrR [38]4 years ago
5 0

To answer this item, we are instructed that the price of the ticket for the children is equal to x. Since, we are also given that this is 5.75 less than the price of tickets for the adults, we may express this as follows,

   x = n - 5.75

n is the price of ticket for the adults. This can be calculated by transposing 5.75 to the other side of the equation,

  n = x + 5.75

The answer is the third choice.

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(8x2 - 2x) - (x2 - x + 5)
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Answer:

7x^2 -x -5

Step-by-step explanation:

(8x^2 - 2x) - (x^2 - x + 5)

Distribute the minus sign

(8x^2 - 2x) - x^2 + x - 5

Combine like terms

7x^2 -x -5

7 0
4 years ago
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Rectangle QRST with vertices Q(-3,2), R(-1,4), S(2,1), and T(0,-1)) in the x-axis
charle [14.2K]

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6 0
3 years ago
ella is making 3 batches of banana milkshake she needs 3/4 gallon of milk for each bath.Her measuring cup holds 1/4 gallon.How m
Dmitry [639]
1 batch need 3/4 gallon of milk

She has 1/4 gallon cup

(3/4) divided by (1/4)

A technique in math is just flip the divider which is (1/4) to (4/1)
so it becomes (3/4) multiply (4/1) = (3x4) / (4x1) = 12/4 = 3 times for one batch

She needs three batches so 3 x 3 = 9 times to fill the measuring cup to make all 3 batches of milkshake.
3 0
4 years ago
Solve each equation <br> | n/5| +1=2
erastova [34]

Answer:

|n \div 5|  + 1 = 2 \\  |n \div 5|  = 1 \\ n = 5

4 0
3 years ago
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50 points + brainliest
xxTIMURxx [149]
Solving this problem involves repeated application of the distance formula. In order to figure out which vertices we need to connect to another vertex, we should first plot the points on the coordinate plane to get an idea of what the polygon looks like. To form the sides of this polygon (which is, in our case, a pentagon), we'll need to connect the points in the following pairs:

(-2, -2) and (3, -3)
(3, -3) and (4, -6)
(4, -6) and (1, -6)
(1, -6) and (-2, -4)
(-2, -4) and (-2, -2)

In case you forgot, the distance formula is simply an application of the Pythagorean Theorem that treats the x-distance and y-distance between two points as the "legs" of a right triangle, and the shortest distance between them as the "hypotenuse."

If a and b are the legs of a right triangle, and c is the hypotenuse, the Pythagorean Theorem can be written as:

a^2+b^2=c^2

Or, if we're just looking for the value of c:

c=\sqrt{a^2+b^2}

Since the hypotenuse in our case represents <em>distance</em>, it's more descriptive to rename that variable <em>d</em>. Also, the "legs" a and b in this problem represent the distances between the x and y components of the two points. If we take any two points (x_1,y_1) and (x_2,y_2), the distance between the x components of those points would be their difference, x_2-x_1, and the distance between the y components would be y_2-y_1. Substituting that all in, the distance formula becomes:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2

All that's left to do now is substitute our specific points into the formula for each side of the polygon:

(-2, -2) and (3, -3):
d=\sqrt{(3-(-2))^2+(-3-(-2))^2}\\ d=\sqrt{(3+2)^2+(-3+2)^2}\\ d=\sqrt{5^2+(-1)^2}\\ d=\sqrt{25+1}\\ d=\sqrt{26}\\ d\approx5.1

(3, -3) and (4, -6)
d=\sqrt{(4-3)^2+(-6-(-3))^2}\\ d=\sqrt{1^2+(-6+3)^2}\\ d=\sqrt{1+(-3)^2}\\d=\sqrt{1+9}\\ d=\sqrt{10}\\ d\approx3.2

(4, -6) and (1, -6)
d= \sqrt{(1-4)^2+(-6-(-6))^2} \\ d= \sqrt{(-3)^2+0^2} \\ d= \sqrt{9} \\ d=3

(1, -6) and (2, -4)
d= \sqrt{(2-1)^2+(-4-(-6))^2}\\ d= \sqrt{1^2+(-4+6)^2}\\ d= \sqrt{1+2^2}\\ d= \sqrt{1+4} \\ d= \sqrt{5}\\ d\approx2.2

(2, -4) and (-2, -2)
d= \sqrt{(-2-2)^2+(-2-(-4))^2}\\ d= \sqrt{(-4)^2+(-2+4)^2} \\ d= \sqrt{16+2^2}\\ d= \sqrt{16+4}\\ d= \sqrt{20} \\ d\approx4.5

Rounding beforehand and adding up all of the distances gives us a perimeter of 18 units, which is remarkably close to the more precise approximation of 17.96 units. Given your options, 17.9 units would be the closest to the result we obtained here.

6 0
4 years ago
Read 2 more answers
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