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2 years ago

Been struggling with this question can someone help?

Mathematics

2 answers:

Genrish500 [490]2 years ago

4 0

Answer:

a and b

Step-by-step explanation:

............. ...... ..

natulia [17]2 years ago

4 0

Answer:

The answer is B

Step-by-step explanation:

$\frac{x}{4}=16\\\mathrm{Multiply\:both\:sides\:by\:}4\\\frac{4x}{4}=16\cdot \:4\\\mathrm{Simplify}\\x=64$

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Determine whether the relation is a function. (2, -3). (3, 2). (4, 7), (3, 14, (6, 23)

Vika [28.1K]

Answer:

A: Yes

Step-by-step explanation:

Each input has exactly one output! So each x value has exactly one y output. hope this helps!

5 0

2 years ago

I need help to see if their equivalent help fast pls

Taya2010 [7]

They are not equivalent. When you distribute the 1/4, the expressions are different.

8 0

3 years ago

Some students are making solid figures with 1-inch unit cubes. Brittany is making a cube. She builds the first two layers of the

Yuki888 [10]

Answer:

6³ = 216 cubes

Step-by-step explanation:

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4 0

3 years ago

CALC- limits 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

3 years ago

Jill bought oranges and bananas. she bought 12 pieces of fruit and spent $5. Oranges cost $0.50 each and bananas cost $0.25 each

inysia [295]

The first part is 0.5x+0.25y= $12.00

3 0

2 years ago