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2 years ago

The admission cost to enter a fair is $5. The cost to ride each ride at the fair is

Mathematics

1 answer:

Llana [10]2 years ago

6 0

Answer:

Step-by-step explanation:

ne spent $42 for shoes. This was $14 less than twice what she spent for a blouse. How much was the blouse?

First, identify what the question is asking for.

Jane spent $42 for shoes. This was $14 less than twice what she spent for a blouse . How much was the blouse?

Next, identify the numbers.

Jane spent $42 for shoes. This was $14 less than twice what she spent for a b

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Use the Distributive Property to find the Product 4 and 2/5 × 10

arlik [135]

Distributive Property ( A number, n, is multiplied to the expression in parentheses)

$4( \frac{2}{5} \times 10) = \\ 4( \frac{20}{5} ) = \\ 4(4) = 16$

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3 years ago

Can anyone rewrite the quadratic function in standard form?

ycow [4]

Answer:

So we first open multiply the parenthesis by 9. 9x+5 is what it is. Now we need to simply it even more with the 2 parenthesis

(9x+5)(x+1) = 9x^2+9x+5x+5 = 9x^2+14x+5

<h2>Answer: 9x^2+14x+5</h2>

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2 years ago

How to tell if a number is prime or composite

kobusy [5.1K]

A prime number has only two factors: 1 and itself. Acomposite number has more than two factors. Thenumber 1 is neither prime nor composite. The prime numbers between 2 and 31 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31 since each of these numbers has only two factors, itself and 1.

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3 years ago

Read 2 more answers

Sheri is making 4 pounds of mixed nuts. She will use One-third the amount of macadamia nuts as she does almonds. A graph titled

notsponge [240]

1 pound of macadamia nuts; 3 pounds of almonds of each nut will she use.

Given that,

Sheri is making 4 pounds of mixed nuts.

She will use One-third the number of macadamia nuts as she does almonds.

A graph titled Sheri's mixed nut recipe has pounds of macadamia nuts on the x-axis, and pounds of almonds on the y-axis. 2 lines intersect at (1, 3).

We have to find,

How many pounds of each nut will she use?

According to the question,

Total mixed nuts are 4 pounds,

She will use One-third the number of macadamia nuts as she does almonds.

2 lines intersect at (1, 3).

Therefore,

Sheri's mixed nut recipe has pounds of macadamia nuts on the x-axis,

At x-axis, the point 1 lies.

So, macadamia nuts have 1 pound.

And pounds of almonds on the y-axis,

At y-axis, the point 3 lies.

So, almond nuts have 3 pounds.

Hence, 1 pound of macadamia nuts; 3 pounds of almonds of each nut will she use.

For more details about Graphs refer to the link given below.

brainly.com/question/9351049

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2 years ago

Read 2 more answers

Perform the following operations s and prove closure. Show your work.

nadezda [96]

Answer:

1. $\frac{x}{x+3}+\frac{x+2}{x+5}$ = $\frac{2x^2+10x+6}{(x+3)(x+5)}\\$

2. $\frac{x+4}{x^2+5x+6}*\frac{x+3}{x^2-16}$ = $\frac{1}{(x+2)(x-4)}$

3. $\frac{2}{x^2-9}-\frac{3x}{x^2-5x+6}$ = $\frac{-3x^2-7x-4}{(x+3)(x-3)(x-2)}$

4. $\frac{x+4}{x^2-5x+6}\div\frac{x^2-16}{x+3}$ = $\frac{1}{(x-2)(x-4)}$

Step-by-step explanation:

1. $\frac{x}{x+3}+\frac{x+2}{x+5}$

Taking LCM of (x+3) and (x+5) which is: (x+3)(x+5)

$=\frac{x(x+5)+(x+2)(x+3)}{(x+3)(x+5)}\\=\frac{x^2+5x+(x)(x+3)+2(x+3)}{(x+3)(x+5)} \\=\frac{x^2+5x+x^2+3x+2x+6}{(x+3)(x+5)} \\=\frac{x^2+x^2+5x+3x+2x+6}{(x+3)(x+5)} \\=\frac{2x^2+10x+6}{(x+3)(x+5)}\\$

Prove closure: The value of x≠-3 and x≠-5 because if there values are -3 and -5 then the denominator will be zero.

2. $\frac{x+4}{x^2+5x+6}*\frac{x+3}{x^2-16}$

Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)

Factors of x^2+5x+6 = x^2+3x+2x+6 = x(x+3)+2(x+3) =(x+2)(x+3)

Putting factors

$=\frac{x+4}{(x+3)(x+2)}*\frac{x+3}{(x-4)(x+4)}\\\\=\frac{1}{(x+2)(x-4)}$

Prove closure: The value of x≠-2 and x≠4 because if there values are -2 and 4 then the denominator will be zero.

3. $\frac{2}{x^2-9}-\frac{3x}{x^2-5x+6}$

Factors of x^2-9 = (x)^2-(3)^2 = (x-3)(x+3)

Factors of x^2-5x+6 = x^2-2x-3x+6 = x(x-2)+3(x-2) =(x-2)(x+3)

Putting factors

$\frac{2}{(x+3)(x-3)}-\frac{3x}{(x+3)(x-2)}$

Taking LCM of (x-3)(x+3) and (x-2)(x+3) we get (x-3)(x+3)(x-2)

$\frac{2(x-2)-3x(x+3)(x-3)}{(x+3)(x-3)(x-2)}$

$=\frac{2(x-2)-3x(x+3)}{(x+3)(x-3)(x-2)}\\=\frac{2x-4-3x^2-9x}{(x+3)(x-3)(x-2)}\\=\frac{-3x^2-9x+2x-4}{(x+3)(x-3)(x-2)}\\=\frac{-3x^2-7x-4}{(x+3)(x-3)(x-2)}$

Prove closure: The value of x≠3 and x≠-3 and x≠2 because if there values are -3,3 and 2 then the denominator will be zero.

4. $\frac{x+4}{x^2-5x+6}\div\frac{x^2-16}{x+3}$

Factors of x^2-5x+6 = x^2-3x-2x+6 = x(x-3)-2(x-3) = (x-2)(x-3)

Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)

$\frac{x+4}{(x-2)(x+3)}\div\frac{(x-4)(x+4)}{x+3}$

Converting ÷ sign into multiplication we will take reciprocal of the second term

$=\frac{x+4}{(x-2)(x+3)}*\frac{x+3}{(x-4)(x+4)}\\=\frac{1}{(x-2)(x-4)}$

Prove Closure: The value of x≠2 and x≠4 because if there values are 2 and 4 then the denominator will be zero.

5 0

3 years ago