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2 years ago

36 pictures for $ or 24 pictures for $5

Mathematics

2 answers:

Oksi-84 [34.3K]2 years ago

4 0

Answer:

I think it is 24 cus it's more bigger and cooler and hotter and shorter and butter

Serga [27]2 years ago

4 0

Answer:

$7.5

Step-by-step explanation:

24 pictures for $5

Multiply this by 2

48 pictures for $10

Then think about the number between 5 and 10

36 pictures for $7.5

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What is 9.286 as a mixed number

marshall27 [118]

9.286

[To change to mixed number, count the number of digits after the decimal point. There are 3 digits, so the denominator will be 1000]

$9.286 = 9 \frac{286}{1000}$

Then we simplify the fraction
(286 ÷ 2)/(1000÷2) = 143/500

Therefore,
$9.286 = 9 \frac{286}{1000}$
$9.286 = 9 \frac{143}{500}$

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3 years ago

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Find y for x=0,1,2, and 3.<br><br> y=4-x

LenKa [72]

4 - (0) = y
4 - 0 = 4 y = 4

4 - (1) = y
4 - 1 = 3 y = 3

4 - (2) = y
4 - 2 = 2 y = 2

4 - (3) = y
4 - 3 = 1 y = 1

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3 years ago

Solve for equation x. <br> 3In(x)+2In(4)=In(128)

Serggg [28]

$Domain:\ x > 0\\\\3\ln(x)+2\ln(4)=\ln(128)\qquad\text{use}\ \lag_ab^n=n\log_ab\\\\\ln(x^3)+\ln(4^2)=\ln(128)\qquad\text{use}\ \log_ab+\ln_ac=\ln_a(bc)\\\\\ln(16x^3)=\ln(128)\iff16x^3=128\qquad\text{divide both sides by 16}\\\\x^3=8\to x=\sqrt[3]8\\\\\boxed{x=2}\in D$

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3 years ago

How do I find the value of x?

Rzqust [24]

To find the value of x , bring the variable to the left side and bring all the remaining values to the right side. simplify the values to find the result.

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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

36 pictures for $ or 24 pictures for $5​

36 pictures for $ or 24 pictures for $5