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2 years ago

8

Completely simplified the expression for x plus 10 plus 6 x plus 2 x by combining like terms into your answer in the box.

2 answers:

siniylev [52]2 years ago

7 0

We can first turn the text into a proper formula:

x plus 10 plus 6 x plus 2 x

x+10+6x+2x

The like terms are the numbers with

We can then put the like terms together to continue the calculation:

x+6x+2x+10

=9x + 10

Therefore, the answer is 9x + 10.

Hope it helps!

DiKsa [7]2 years ago

5 0

Answer:

9x+10

Step-by-step explanation:

x+10 +6x+2x

Combine like terms

x+6x+2x +10

9x+10

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A carpenter purchased 50 ft of redwood and 80 ft of pine for a total cost of $285. A second purchase, at the same prices, includ

Marizza181 [45]

Hey there!!

Let us take the price of the redwood as ' x ' and pine as ' y '

Then , we take it into an equation. We get,

50x + 80y = 285 -------------- ( 1 )

80x + 50y = 339 -------------- ( 2 )

Now, multiply the first equation with 8 and the second equation with 5

400x + 640y = 2280

400x + 250y = 1695

Now subtract the second equation from the first

390y = 585

y = 585 / 390

y = $1.5

Now substitute this into any equation

50x + ( 80 ) ( 1.5 ) = 285

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50x = 165

x = 165 / 50

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Pine = $1.5

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g100num [7]

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2 years ago

Read 2 more answers

PLZ HELP!!! Use limits to evaluate the integral.

Marrrta [24]

Split up the interval [0, 2] into <em>n</em> equally spaced subintervals:

$\left[0,\dfrac2n\right],\left[\dfrac2n,\dfrac4n\right],\left[\dfrac4n,\dfrac6n\right],\ldots,\left[\dfrac{2(n-1)}n,2\right]$

Let's use the right endpoints as our sampling points; they are given by the arithmetic sequence,

$r_i=\dfrac{2i}n$

where $1\le i\le n$ . Each interval has length $\Delta x_i=\frac{2-0}n=\frac2n$ .

At these sampling points, the function takes on values of

$f(r_i)=7{r_i}^3=7\left(\dfrac{2i}n\right)^3=\dfrac{56i^3}{n^3}$

We approximate the integral with the Riemann sum:

$\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{112}n\sum_{i=1}^ni^3$

Recall that

$\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4$

so that the sum reduces to

$\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{28n^2(n+1)^2}{n^4}$

Take the limit as <em>n</em> approaches infinity, and the Riemann sum converges to the value of the integral:

$\displaystyle\int_0^27x^3\,\mathrm dx=\lim_{n\to\infty}\frac{28n^2(n+1)^2}{n^4}=\boxed{28}$

Just to check:

$\displaystyle\int_0^27x^3\,\mathrm dx=\frac{7x^4}4\bigg|_0^2=\frac{7\cdot2^4}4=28$

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