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3 years ago

Hcf of 72, 120, 192 with explanations

Mathematics

1 answer:

ryzh [129]3 years ago

5 0

Hcf means highest common factor
Which can be determined using many methods such as long division method and prime factorisation . Now let me solve your question . Hope u understand and mark as brainlist

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How long are the other sides

tester [92]

Answer:

10 inches.

Step-by-step explanation:

Because an isosceles has two equal sides and on different one

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3 years ago

The price, P, of a sweater was discounted by 10% in

aalyn [17]

Answer:

In terms of P, the price of the sweater after the discount is 0.72P.

Step-by-step explanation:

Since the price P would represent 100% and the price was discounted by 10%, you can calculate 10% of 100% and subtract that amount, which would be:

100*10%=10

100-10=90%

This means that the price paid would be 90% of P. Then, the statement says that the price was discounted by another 20%. So, you have to calculate 20% of the price and subtract that amount from the price:

90*20%=18

90-18=72%

According to this, this means that in terms of P, the price of the sweater after the discount is 0.72P.

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2 years ago

Ixllllllllllllllllll

marissa [1.9K]

Answer:

the answer would be t=3.59 if i am reading this correctly

Step-by-step explanation:

4 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

A recipe asks for 4 1/2 pounds of chicken. How many pounds of chicken are needed to make 12 of a recipe? Express the answer in s

jok3333 [9.3K]

At least 4 3/4 pounds

8 0

2 years ago

Read 2 more answers