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4 years ago

6

Josie read 246 pages of a book last month .Her older brother says he read about 3 to 4 times as many pages .Why is 2500 not a re

asonable estimate for the number of pages that josie's brother read

1 answer:

adell [148]3 years ago

6 0

Answer:

Because 2500 would be around 10 times as many pages, 3-4 times as many would be around 700-900 more pages, not 2500.

Step-by-step explanation:

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Calculate the total area of the shaded region.

LUCKY_DIMON [66]

so hmmm seemingly the graphs meet at -2 and +2 and 0, let's check

$\stackrel{f(x)}{2x^3-x^2-5x}~~ = ~~\stackrel{g(x)}{-x^2+3x}\implies 2x^3-5x=3x\implies 2x^3-8x=0 \\\\\\ 2x(x^2-4)=0\implies x^2=4\implies x=\pm\sqrt{4}\implies x= \begin{cases} 0\\ \pm 2 \end{cases}$

so f(x) = g(x) at those points, so let's take the integral of the top - bottom functions for both intervals, namely f(x) - g(x) from -2 to 0 and g(x) - f(x) from 0 to +2.

$\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill$

$\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}$

7 0

2 years ago

7.89 greater or less than 7.189<br><br> 6.030 greater less or equal 6.03

NemiM [27]

7.89 is greater than 7.189, 6.03 is equal to 6.030, because that last zero doesn't really change anything it cancels itself out

5 0

3 years ago

Read 2 more answers

I need help please

frutty [35]

260 units2 is the answer

3 0

3 years ago

Read 2 more answers

Solve for n: c= 6n−5g /11t

qwelly [4]

Answer: $\frac{11tc+5g}{66t}=n$

Step-by-step explanation:

You have the equation $c=6n-\frac{5g}{11t}$ .

Then, to solve for the variable $n$ from the equation you need:

Make the subtraction of the right side of the equation:

(As the denominators are 1 and $11t$ , the least common denominator is $11t$ )

$c=\frac{(6n)(11t)-5g}{11t}\\\\c=\frac{66nt-5g}{11t}$

Multiply $11t$ to both sides:

$(11t)c=(\frac{66nt-5g}{11t})(11t)\\\\11tc=66nt-5g$

Add $5g$ to both sides:

$11tc+5g=66nt-5g+5g\\\\11tc+5g=66nt$

And finally divide both sides by $66t$ :

$\frac{11tc+5g}{66t}=\frac{66nt}{66t}\\\\\frac{11tc+5g}{66t}=n$

3 0

3 years ago

A trampolinist steps off from 15 feet above ground to a trampoline 13 feet below. The function h (t) = -16 t 2 + 15, where t rep

kondor19780726 [428]

Answer:

Trampolinist will land on the trampoline after 0.9 seconds.

Step-by-step explanation:

The function h(t) = -16t² + 15 represents the relation between height 'h' above the ground and the time 't' of the trampolinist.

We have to find the time when trampolinist lands on the ground.

That means we have to find the value of 't' when h(t) = 15 - 13 = 2

[Since trampoline is 2 feet above the ground]

When we plug in the value h(t) = 2

2 = -16t² + 15

2 + 16t² = -16t² + 16t² + 15

16t² + 2 = 15

16t² + 2 - 2 = 15 - 2

16t² = 13

$\frac{16t^{2}}{16}=\frac{13}{16}$

$t^{2}=\frac{13}{16}$

t = $\sqrt{\frac{13}{16}}$

t ≈ 0.9 seconds

Therefore, trampolinist will land on the trampoline at 0.9 seconds.

5 0

3 years ago