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IrinaK [193]
3 years ago
15

You increase the size of a page by 30%. Then you decrease it by 30%. What is the size of the page now?

Mathematics
2 answers:
kotegsom [21]3 years ago
8 0
Lets use 1 to solve this problem because it is easiest to show. 

Multiply 1 by 1.3 for a 30% increase (1*1.3=1.3). 

Then multiply 1.3 by .7 for a 30% decrease. (1.3*.7=.91)  

Therefore the page had a 9% reduction from its original size or is 91% of its original size. 
Mariana [72]3 years ago
7 0

Answer:

Smaller the size you started with

Step-by-step explanation:

Because for example 100 times .30 is 30 100+30=130 and 130 times .30 is 39 100-39=91. 100 > 91

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400 decimeters = 4,000 cm

4,000 cm = 40 m
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Raj’s bathtub is clogged and is draining at a rate of 1.5 gallons of water per minute. The table shows that the amount of water
timurjin [86]

Raj’s bathtub is clogged and is draining at a rate of 1.5 gallons of water per minute. The table shows that the amount of water remaining in the bathtub, y, is a function of the time in minutes, x, that it has been draining.

What is the range of this function?

all real numbers such that y ≤ 40

all real numbers such that y ≥ 0

all real numbers such that 0 ≤ y ≤ 40

all real numbers such that 37.75 ≤ y ≤ 40

5 0
3 years ago
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5x + 2x + 2 = 9x + 4​
bija089 [108]

Step-by-step explanation:

Given

5x + 2x + 2 = 9x + 4

9x - 5x - 2x = 2 - 4

9x - 7x = - 2

2x = - 2

x = - 2 / 2

Therefore x = -1.

5 0
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PLSS HELP IF YOU TURLY KNOW THISS
coldgirl [10]

Answer:

x = 7

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7x = 49

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\boxed{\green{x = 7}}

3 0
3 years ago
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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

4 0
3 years ago
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