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3 years ago

5

Mr. and Mrs. Jones wanted to plan a fun day with their two children. An adult ticket to the local amusement park is $20. A child

’s ticket is 40 percent less than the adult ticket. What is the total amount for a family of four to enter the park?

2 answers:

Nata [24]3 years ago

4 0

Answer: It would cost $56

Step-by-step explanation:

2 adults = 20 + 20 = 40

20 * 0.40 = 8

2 children = 8 + 8 = 16

16 + 40 = 56

nika2105 [10]3 years ago

4 0

Answer:

$64

Step-by-step explanation:

Adult ticket = $20 (base cost, which we take is 100% of the total)

100% - 40% = 60%

Child ticket = $20 x 60% = $12

2 Adult + 2 Child Tickets = $20 + $20 + $12 + $12 = $64

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If 2(4x + 3)/(x - 3)(x + 7) = a/x - 3 + b/x + 7, find the values of a and b.

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Answer:

$a=3$ and $b=5$ .

Step-by-step explanation:

So I believe the problem is this:

$\frac{2(4x+3)}{x-3}(x+7)}=\frac{a}{x-3}+\frac{b}{x+7}$

where we are asked to find values for $a$ and $b$ such that the equation holds for any $x$ in the equation's domain.

So I'm actually going to get rid of any domain restrictions by multiplying both sides by (x-3)(x+7).

In other words this will clear the fractions.

$\frac{2(4x+3)}{x-3}(x+7)}\cdot(x-3)(x+7)=\frac{a}{x-3}\cdot(x-3)(x+7)+\frac{b}{x+7}(x-3)(x+7)$

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As you can see there was some cancellation.

I'm going to plug in -7 for x because x+7 becomes 0 then.

$2(4\cdot -7+3)=a(-7+7)+b(-7-3)$

$2(-28+3)=a(0)+b(-10)$

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Divide both sides by -10:

$\frac{-50}{-10}=b$

$5=b$

Now we have:

$2(4x+3)=a(x+7)+b(x-3)$ with $b=5$

I notice that x-3 is 0 when x=3. So I'm going to replace x with 3.

$2(4\cdot 3+3)=a(3+7)+b(3-3)$

$2(12+3)=a(10)+b(0)$

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Divide both sides by 10:

$\frac{30}{10}=a$

$3=a$

So $a=3$ and $b=5$ .

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3 years ago

Read 2 more answers

Circle E with center at (−5, 3) and radius of 8 is similar to circle F with center at (1, 2) and radius of 4. An incorrect infor

aalyn [17]

Answer: Step 2

Step-by-step explanation:

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2 years ago

Find the difference. (9/x^2-9x)-(6/x^2-81)

Sunny_sXe [5.5K]

The difference is $\frac{3 x+81}{x(x-9)(x+9)}$

Explanation:

The expression is $\left(\frac{9}{x^{2}-9 x}\right)-\left(\frac{6}{x^{2}-81}\right)$

Removing the parenthesis, we have,

$\left\frac{9}{x^{2}-9 x}\right-\left\frac{6}{x^{2}-81}\right$

Factoring the terms $x^{2}-9 x$ and $x^{2}-81$ , we get,

$\frac{9}{x(x-9)}-\frac{6}{(x+9)(x-9)}$

Taking LCM, we get,

$\frac{9(x+9)-6x}{x(x-9)(x+9)}}$

Simplifying the numerator, we get,

$\frac{9x+81-6x}{x(x-9)(x+9)}}$

Subtracting the numerator, we have,

$\frac{3 x+81}{x(x-9)(x+9)}$

Hence, the difference is $\frac{3 x+81}{x(x-9)(x+9)}$

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