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2 years ago

How many fifths are the in 60/100

1 answer:

8 0

In order to calculate that, we need to divide it by 1/5 as follows:
60/100 / 1/5

We can re-write it as: 60/100 * 5 = 6/10 * 5 = 30/10 = 3

So, there are "3" 1/5ths in 60/100

Hope this helps!

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Plz help... must show steps....will pick brainiest Solve, finding all solutions in [0,2pi)

Triss [41]

A graphing calculator shows solutions to be
.. x = {π/4, 2.034, 5π/4, 5.176}

___
Asked and answered elsewhere.
brainly.com/question/8691400

3 0

3 years ago

Solve only if you know the solution and show work.

SashulF [63]

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=\int\mathrm dx+2\int\frac{\sin x+3}{\cos x+\sin x+1}\,\mathrm dx$

For the remaining integral, let $t=\tan\dfrac x2$ . Then

$\sin x=\sin\left(2\times\dfrac x2\right)=2\sin\dfrac x2\cos\dfrac x2=\dfrac{2t}{1+t^2}$
$\cos x=\cos\left(2\times\dfrac x2\right)=\cos^2\dfrac x2-\sin^2\dfrac x2=\dfrac{1-t^2}{1+t^2}$

and

$\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx\implies \mathrm dx=2\cos^2\dfrac x2\,\mathrm dt=\dfrac2{1+t^2}\,\mathrm dt$

Now the integral is

$\displaystyle\int\mathrm dx+2\int\frac{\dfrac{2t}{1+t^2}+3}{\dfrac{1-t^2}{1+t^2}+\dfrac{2t}{1+t^2}+1}\times\frac2{1+t^2}\,\mathrm dt$

The first integral is trivial, so we'll focus on the latter one. You have

$\displaystyle2\int\frac{2t+3(1+t^2)}{(1-t^2+2t+1+t^2)(1+t^2)}\,\mathrm dt=2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt$

Decompose the integrand into partial fractions:

$\dfrac{3t^2+2t+3}{(1+t)(1+t^2)}=\dfrac2{1+t}+\dfrac{1+t}{1+t^2}$

so you have

$\displaystyle2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt=4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$

which are all standard integrals. You end up with

$\displaystyle\int\mathrm dx+4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$
$=x+4\ln|1+t|+2\arctan t+\ln(1+t^2)+C$
$=x+4\ln\left|1+\tan\dfrac x2\right|+2\arctan\left(\arctan\dfrac x2\right)+\ln\left(1+\tan^2\dfrac x2\right)+C$
$=2x+4\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)+C$

To try to get the terms to match up with the available answers, let's add and subtract $\ln\left|1+\tan\dfrac x2\right|$ to get

$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)-\ln\left|1+\tan\dfrac x2\right|+C$
$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|+C$

which suggests A may be the answer. To make sure this is the case, show that

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\sin x+\cos x+1$

You have

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\cos^2\dfrac x2+\sin\dfrac x2\cos\dfrac x2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\dfrac{1+\cos x}2+\dfrac{\sin x}2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac2{\cos x+\sin x+1}$

So in the corresponding term of the antiderivative, you get

$\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|=\ln\left|\dfrac2{\cos x+\sin x+1}\right|$
$=\ln2-\ln|\cos x+\sin x+1|$

The $\ln2$ term gets absorbed into the general constant, and so the antiderivative is indeed given by A,

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=2x+5\ln\left|1+\tan\dfrac x2\right|-\ln|\cos x+\sin x+1|+C$

5 0

2 years ago

Solve the inequality -8r<-12.

Kobotan [32]

Divide -8 from both sides to get the answer

6 0

3 years ago

Read 2 more answers

I can’t figure out which formula goes with which sequence. Help……. Please???

Radda [10]

For a geometric sequence

a, ar, ar ², ar ³, …

the n-th term in the sequence is ar ⁿ ⁻ ¹.

The first sequence is

1, 3, 9, 27, …

so it's clear that a = 1 and r = 3, and so the n-th term is 3ⁿ ⁻ ¹.

The second sequence is

400, 200, 100, 50, …

so of course a = 400, and you can easily solve for r :

200 = 400r ==> r = 200/400 = 1/2

Then the n-th term is 400 (1/2)ⁿ ⁻ ¹.

Similarly, the other sequences are given by

3rd: … 4 × 2ⁿ ⁻ ¹

4th: … 400 (1/4)ⁿ ⁻ ¹

5th: … 5ⁿ ⁻ ¹

6th: … 1000 (1/2)ⁿ ⁻ ¹

7th: … 2 × 5ⁿ ⁻ ¹

7 0

2 years ago

mark and heather starr are celebrating their 25th anniversary by having a reception at local reception hall. they have budget 45

Lera25 [3.4K]

Answer:

Step-by-step explanation:

The $50 cleanup fee is a "fixed cost."

The "$30 per-person fee is a "variable cost." That is, it varies with the number of people.

"within" means "less than or equal to"

Cost = $50 + $30x [where x = number of people]

Cost ≤ $4500 ["stay within their budget"]

$50 + $30x ≤ $4500 [substitute for cost]

$30x ≤ $4450 [subtract $50 from both sides retains sense of inequality]

x ≤ 148.3333 [divide by +$30 retains sense of inequality]

Mark and Heather may invite up to 148 guests.

[note: to be "≤," always truncate, never round up]

Check:

148 people for $30 each + $50 is $4440+$50 = $4490.

One more person makes it $4520 -- that is too much.

There we go, my bad

4 0

2 years ago

Read 2 more answers