Rex paid $250 in taxes to the school district that he lives in this year. This year's taxes were a 12% increase from last year. What did Rex pay in school taxes last year?

Answers

Answer 1

Answer:

Answer:

$223.21

Step-by-step explanation:

Last year the amount of taxes was x, an unknown amount. That was 100% of last year's tax cost.

This year, taxes went up by 12%, so this year the taxes are 112% of what they were last year. This year the taxes were 112% of x, and they were $250.

112% of x = 250

1.12x = 250

x = 250/1.12

x = 223.21

Answer: $223.21

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Answers

Answer:

6,720

Step-by-step explanation:

B = 800(1.05)^8

B = 800(8.4)

B = 6,720

Answer:that's a proper mad question that

Step-by-step explanation:

Square QRST has vertices Q(2,8),R(-4,8),S(-4,2),and T(2,2) and triangle QTU has vertices Q(2,8),T(2,2) and U(6,2) What is the area of the quadrilateral formed by the square and the triangle?

Answers

Here is how you do it:
The answer is 48

What is the y-intercept of the equation: 3x + 6y = 42 *

Answers

Answer:

(0,7)

Step-by-step explanation:

Find the product mentally.
(2m - 9)^2

Answers

- - Solve:

$=2m+-9*2m+-9$
$2m*2m+2m*-9+-9*2m+-9*-9$
$4m^2-18m-18m+81$
$4m^2-36m+81$

Your final answer is 4m^2−36m+81

Use the formula: (a-b)² = a² -2ab + b²

(2m-9)² = (2m)² - 2(2m)(9) + 9²

= 4m² - 36m + 81

In right triangle LMN, L and M are complementary angles and sin(L) is 19/20. What is cos(M)?A.19/20
B.1/20
C.(square root of 39)/19
D.(square root of 39)/20

Answers

I believe the correct answer from the choices listed above is option A. In right triangle LMN, L and M are complementary angles and sin(L) is 19/20. Therefore, cosine of M would be 19/20 as well. It would be helpful if you draw a figure of the problem. Hope this answers the question.

Answer:

19/20

Step-by-step explanation:

Solve for x
logx+log(x-4)=2log5

Answers

$\log{x} + \log{(x-4)} = 2 \log{5}\n\n\log{\big(x(x-4)\big)} = \log{5^2}\n\n\log{(x^2-4x)} = \log{25} \n\nx^2-4x = 25\n\nx^2-4x-25=0 \n\nx = (4 \pm√(16+100))/(2) \n\nx = (4\pm2√(29))/(2) \n\nx = 2 \pm √(29) \n\n\text{But } x > 0 \implies x = 2 + √(29)$

$D:x>0\wedge x>4\nD:x>4\n\n\log x+\log(x-4)=2\log5\n\log x(x-4)=\log25\nx(x-4)=25\nx^2-4x-25=0\nx^2-4x+4-29=0\n(x-2)^2=29\nx-2=√(29) \vee x-2=-√(29)\nx=2+√(29) \vee x=2-√(29)\n2-√(29)\not>4\n\Downarrow\n\boxed{x=2+√(49)}$

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