(8x – 20)-16 =10x+51

Mathematics

1 answer:

Lemur [1.5K]2 years ago

6 0

When solving this equation, you get the answer as -43.5.

To solve, we simply need to collect like-terms and then rearrange to get x on one side and numbers on the other:

(8x - 20) - 16 = 10x + 51

8x - 36 = 10x + 51

- 8x

-36 = 2x + 51

- 51

-87 = 2x

÷ 2

-43.5 = x

I hope this helps!

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Tickets for the baseball games were $2.50 for general admission and 50 cents for kids. If there were six times as many general a

Makovka662 [10]

There was 3000 general admission tickets sold and 500 kid ticket sold.

How did I get this?

First, we need to see what information we have.
$2.50 = General admission tickets = (G)
$0.50 = kids tickets = (K)
There were 6x as many general admission tickets sold as kids. G = 6K

We need two equations:
G = 6K
$2.50G + $.50K = $7750
Since, G = 6K we can substitute that into the 2nd equation.

2.50(6K) + .50K = 7750
Distribute 2.50 into the parenthesis

15K + .50K = 7750
combine like terms

15.50K = 7750
Divide both sides by 15.50, the left side will cancel out.

K = 7750/15.50
K = 500 tickets
So, 500 kid tickets were sold.

Plug K into our first equation (G = 6k)

G = 6*500
G = 3000 tickets

So, 3000 general admission tickets were sold,

Let's check this:

$2.50(3000 tickets) = $7500 (cost of general admission tickets)
$.50(500 tickets) = $250 (cost of general admission tickets)
$7500 + $250 = $7750 (total cost of tickets)

4 0

2 years ago

Solve the equation x² = 1.21 *

gogolik [260]

Answer:

x = 1.1, -1.1

Step-by-step explanation:

This is the answer because:

1) Take the root of both sides and solve.

$\sqrt{1.21}$ = 1.1

2) x can also equal -1.1 because -1.1 times -1.1 gives us a positive 1.21 because negative x negative = positive

Hope this helps!

3 0

2 years ago

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

melamori03 [73]

Answer:

$6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}$

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs $AB=3$ , $BC=4$ , and $AC=5$ . The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base $b$ and height $h$ is given by $A=\frac{1}{2}bh$ . Therefore, the area of this right triangle is:

$A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}$

The other triangle is a bit trickier. Triangle $\triangle ADC$ is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by: