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2 years ago

What is the area of rectangle FGHI

Mathematics

1 answer:

Veseljchak [2.6K]2 years ago

5 0

Answer:

A=wl

Step-by-step explanation:

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System AAA \text{\quad}start text, end text System BBB

lisabon 2012 [21]

Answer:

replace one question with itself where the same quantity is added to both sides and yes

Step-by-step explanation:

The other answer is wrong :)

4 0

3 years ago

Read 2 more answers

Okay here's the first few questions

Marina86 [1]

Answer:

1,yes

2,no

3,x=√117

4,x=12 y=6

5,x=12, y=15.75

6 0

3 years ago

let x represent one number and let y represent the other number. use the given conditions to write a system of equations. solve

kiruha [24]

9514 1404 393

Answer:

(x, y) = (8, 2)

Step-by-step explanation:

The relevant equations are ...

3x -y = 22

x +2y = 12

We can eliminate y by adding twice the first equation to the second.

2(3x -y) +(x +2y) = 2(22) +(12)

7x = 56

x = 8

Substituting into the first equation gives ...

3(8) -y = 22

y = 24 -22 = 2

The first number is 8; the second number is 2.

7 0

3 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

3 years ago

2x-3y> or equal to 12

Drupady [299]

Answer:

it's not greater than it would be 2x-3y because you can add then together because the not the same variable

Step-by-step explanation:

5 0

3 years ago