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2 years ago

9

Brittany is making hair bows to sell at a craft show. Each hair bow requires 3/4 yard of ribbon. Brittany plans to make 50 bows.

How many yards of ribbon does Brittany need to purchase?
A. 37 1/2 yards

B. 49 1/4 yards

C. 50 3/4 yards

D. 66 2/3 yards

1 answer:

uranmaximum [27]2 years ago

6 0

A. 50 x (3/4) = 37.5

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Given a function I and a subset A of its domain, let I(A) represent the range of lover the set A; that is, I(A) = {I(x) : x E A}

Ede4ka [16]

Step-by-step explanation:

(a)

Using the definition given from the problem

$f(A) = \{x^2 \, : \, x \in [0,2]\} = [0,4]\\f(B) = \{x^2 \, : \, x \in [1,4]\} = [1,16]\\f(A) \cap f(B) = [1,4] = f(A \cap B)\\$

Therefore it is true for intersection. Now for union, we have that

$A \cup B = [0,4]\\f(A\cup B ) = [0,16]\\f(A) = [0,4]\\f(B)= [1,16]\\f(A) \cup f(B) = [0,16]$

Therefore, for this case, it would be true that $f(A\cup B) = f(A)\cup f(B)$ .

(b)

1 is not a set.

(c)

To begin with

$A\cap B \subset A,B$

Therefore

$g(A\cap B) \subset g(A) \cap g(B)$

Now, given an element of $g(A) \cap g(B)$ it will belong to both sets, therefore it also belongs to $g(A\cap B)$ , and you would have that

$g(A)\cap g(B) \subset g(A \cap B)$ , therefore $g(A)\cap g(B) = g(A \cap B)$ .

(d)

To begin with $A,B \subset A \cup B$ , therefore

$g(A) \cup g(b) \subset g(A\cup B)$

7 0

2 years ago

Ok guys i have only 1 hr to get this. Determine the y-intercept for the equation 3x + 2y = 12. HELP

sukhopar [10]

Answer:

x= -2/3 y+4

Step-by-step explanation:

Let's solve for x.

3x+2y=12

Step 1: Add -2y to both sides.

3x+2y+−2y=12+−2y

3x=−2y+12

Step 2: Divide both sides by 3.

3x

3

=

−2y+12

3

5 0

1 year ago

Fond the area of the shaded sector of circle C to the nearest tenth

Lesechka [4]

Solution

Step 1

Write an expression for the area of sector

$Area\text{ (A) = }\frac{\theta}{360}\times\pi\times r^2$

Where

θ = 52°

π = 3.14

r = 8cm

Step 2

Find the area of the sector after substitution of values into the formula above.

$\begin{gathered} A\text{ =}\frac{52}{360}\times3.14\times8^2 \\ A=29.028cm^2 \end{gathered}$

To the nearest tenth the area of the shaded sector of circle C = 29.0cm²

5 0

9 months ago

What is the domain of the function shown in thr table

Reil [10]

Option A ( 2 , 4 , 6 , 8) because domain are x values

4 0

2 years ago

CALC- limits<br> please show your method

gladu [14]

A. Factor the numerator as a difference of squares:

$\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt x-3}=\lim_{x\to9}\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}=\lim_{x\to9}(\sqrt x+3)=6$

c. As $x\to\infty$ , the contribution of the terms of degree less than 2 becomes negligible, which means we can write

$\displaystyle\lim_{x\to\infty}\frac{4x^2-4x-8}{x^2-9}=\lim_{x\to\infty}\frac{4x^2}{x^2}=\lim_{x\to\infty}4=4$

e. Let's first rewrite the root terms with rational exponents:

$\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}$

Next we rationalize the numerator and denominator. We do so by recalling

$(a-b)(a+b)=a^2-b^2$
$(a-b)(a^2+ab+b^2)=a^3-b^3$

In particular,

$(x^{1/3}-x)(x^{2/3}+x^{4/3}+x^2)=x-x^3$
$(x^{1/2}-x)(x^{1/2}+x)=x-x^2$

so we have

$\displaystyle\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}\cdot\frac{x^{2/3}+x^{4/3}+x^2}{x^{2/3}+x^{4/3}+x^2}\cdot\frac{x^{1/2}+x}{x^{1/2}+x}=\lim_{x\to1}\frac{x-x^3}{x-x^2}\cdot\frac{x^{1/2}+x}{x^{2/3}+x^{4/3}+x^2}$

For $x\neq0$ and $x\neq1$ , we can simplify the first term:

$\dfrac{x-x^3}{x-x^2}=\dfrac{x(1-x^2)}{x(1-x)}=\dfrac{x(1-x)(1+x)}{x(1-x)}=1+x$

So our limit becomes

$\displaystyle\lim_{x\to1}\frac{(1+x)(x^{1/2}+x)}{x^{2/3}+x^{4/3}+x^2}=\frac{(1+1)(1+1)}{1+1+1}=\frac43$

3 0

2 years ago