Solve:

$$(1/2)b+4=(1/8)b+88$$

$$(1/2)b+4=(1/8)b+88$$

Kendrick Finley
2022-10-27
Answered

Solve:

$$(1/2)b+4=(1/8)b+88$$

$$(1/2)b+4=(1/8)b+88$$

You can still ask an expert for help

Kamden Simmons

Answered 2022-10-28
Author has **10** answers

Subtract b/8 to both sides:

$$\frac{b}{2}+4-\frac{b}{8}=88$$

Subtract 4 to both sides:

$$\frac{b}{2}-\frac{b}{8}=88-4\phantom{\rule{0ex}{0ex}}\frac{b}{2}-\frac{b}{8}=84$$

Multiply b/2 by 4 to obtain a common denominator and subtract the fraction:

$$\frac{4b}{2\times 4}-\frac{b}{8}=84\phantom{\rule{0ex}{0ex}}\frac{3b}{8}=84$$

Multiply both sides by 8.

3b=(84)(8)

Divide both sides by 3.

b=672/3=224

Therefore b=224

Verify:

$$\frac{224}{2}+4=\frac{224}{8}+88\phantom{\rule{0ex}{0ex}}116=116$$

$$\frac{b}{2}+4-\frac{b}{8}=88$$

Subtract 4 to both sides:

$$\frac{b}{2}-\frac{b}{8}=88-4\phantom{\rule{0ex}{0ex}}\frac{b}{2}-\frac{b}{8}=84$$

Multiply b/2 by 4 to obtain a common denominator and subtract the fraction:

$$\frac{4b}{2\times 4}-\frac{b}{8}=84\phantom{\rule{0ex}{0ex}}\frac{3b}{8}=84$$

Multiply both sides by 8.

3b=(84)(8)

Divide both sides by 3.

b=672/3=224

Therefore b=224

Verify:

$$\frac{224}{2}+4=\frac{224}{8}+88\phantom{\rule{0ex}{0ex}}116=116$$

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I keep having trouble with one exercise. The questions are like this: how many $1/4$cm cubes does it take to fill a prism if the prism is $1$cm $\times $ $7/4$cm $\times $$2$cm?

I've tried converting all the side lengths to multiples(?) of $1/4$, such as $4/4\times 7/4\times 8/4$ and multiplying but I still get the wrong answer. How do I calculate this?

I keep having trouble with one exercise. The questions are like this: how many $1/4$cm cubes does it take to fill a prism if the prism is $1$cm $\times $ $7/4$cm $\times $$2$cm?

I've tried converting all the side lengths to multiples(?) of $1/4$, such as $4/4\times 7/4\times 8/4$ and multiplying but I still get the wrong answer. How do I calculate this?

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You see, there are many $-x$ terms. How can I make this shorter? Can I write:

$x\phantom{\rule{thinmathspace}{0ex}}\frac{-{a}_{1}-{a}_{2}-{a}_{3}}{{a}_{4}}\cdot l-u\phantom{\rule{thinmathspace}{0ex}}\frac{l}{{a}_{4}}$

Is this right?

Regards.

You see, there are many $-x$ terms. How can I make this shorter? Can I write:

$x\phantom{\rule{thinmathspace}{0ex}}\frac{-{a}_{1}-{a}_{2}-{a}_{3}}{{a}_{4}}\cdot l-u\phantom{\rule{thinmathspace}{0ex}}\frac{l}{{a}_{4}}$

Is this right?

Regards.

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Let $P$ be a polynomial of degree $n$ , $n\ge 2$, $P(X)=\prod _{i=1}^{n}{(X-{z}_{i})}^{{m}_{i}}$

with zeros ${z}_{1},{z}_{2},\dots ,{z}_{n}$

Decompose the rational fraction $\frac{{P}^{\prime}}{P}$

My thoughts:

$P(X)=\prod _{i=1}^{n}{(X-{z}_{i})}^{{m}_{i}}$

${P}^{\prime}(X)=\sum _{i=1}^{n}{m}_{i}{(X-{z}_{i})}^{{m}_{i}-1}\prod _{?}^{?}{(X-{z}_{i})}^{{m}_{?}}$

I don't know with what i should fill the ?

Then:

$\begin{array}{rl}{\displaystyle \frac{{P}^{\prime}}{P}}& ={\displaystyle \frac{\sum _{i=1}^{n}{m}_{i}{(X-{z}_{i})}^{{m}_{i}-1}\prod _{?}^{?}{(X-{z}_{i})}^{{m}_{i}}}{\prod _{i=1}^{n}{(X-{z}_{i})}^{{m}_{i}}}}\\ & =?\end{array}$

I'm stuck here

with zeros ${z}_{1},{z}_{2},\dots ,{z}_{n}$

Decompose the rational fraction $\frac{{P}^{\prime}}{P}$

My thoughts:

$P(X)=\prod _{i=1}^{n}{(X-{z}_{i})}^{{m}_{i}}$

${P}^{\prime}(X)=\sum _{i=1}^{n}{m}_{i}{(X-{z}_{i})}^{{m}_{i}-1}\prod _{?}^{?}{(X-{z}_{i})}^{{m}_{?}}$

I don't know with what i should fill the ?

Then:

$\begin{array}{rl}{\displaystyle \frac{{P}^{\prime}}{P}}& ={\displaystyle \frac{\sum _{i=1}^{n}{m}_{i}{(X-{z}_{i})}^{{m}_{i}-1}\prod _{?}^{?}{(X-{z}_{i})}^{{m}_{i}}}{\prod _{i=1}^{n}{(X-{z}_{i})}^{{m}_{i}}}}\\ & =?\end{array}$

I'm stuck here