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Veronika [31]
3 years ago
15

What is the lcd of: 8/9x 2/3x=1/9x 45?

Mathematics
1 answer:
spayn [35]3 years ago
5 0
LCD would be 18

8/9 = 16/18
2/3 = 12/18
1/9 = 2/18
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What is the result when 3/8x + 2 1/5 is subtracted from 3 1/2x - 4 1/10?<br><br> Don’t put a link.
Umnica [9.8K]

Answer:

Option 3 3 1/8x - 6 3/10

Step-by-step explanation:

(3 1/2x - 4 1/10) - (3/8x + 2 1/5)

Divide it into parts:

(3 1/2x - 3/8x) + (-4 1/10 - 2 1/5)

3 1/2x - 3/8x = 3 4/8x - 3/8x = 3 1/8x

-4 1/10 - 2 1/5 = -4 1/10 - 2 2/10 = -6 3/10

So we get:

2 1/4x - 6 3/10

6 0
3 years ago
Read 2 more answers
A physical education teacher plans to divide the seventh graders at Wilson Middle School into teams of equal size of year-ending
Crank

Step-by-step explanation:

Consider the provided information.

The total number of students are: 24+26+21=71

He wants each team will have the same number of students between 5 and 9.

The number between 5 to 9 are 6, 7 and 8.

The number 71 is a prime number so we need to divide the 71 students in group so that minimum students left without team.

Dividing 6 students in each group.

Number of teams: \frac{71}{6} = 11 teams and 5 students.

Dividing 7 students in each group.

Number of teams: \frac{71}{7} = 10 teams and 1 students.

Dividing 8 students in each group.

Number of teams: \frac{71}{8} = 8 teams and 7 students.

Thus, we can not divide 71 students into equal teams, but 70 divided by 7 gives the least students left out i.e 10 teams each with 7 students and only 1 student will remain at last.

6 0
3 years ago
How to determine if a graph is symmetrical
Lisa [10]

Answer:

If you fold it in half.

Step-by-step explanation:

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Evaluate Dx / ^ 9-8x - x2^
Solnce55 [7]
It depends on what you mean by the delimiting carats "^"...

Since you use parentheses appropriately in the answer choices, I'm going to go out on a limb here and assume something like "^x^" stands for \sqrt x.

In that case, you want to find the antiderivative,

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}

Complete the square in the denominator:

9-8x-x^2=25-(16+8x+x^2)=5^2-(x+4)^2

Now substitute x+4=5\sin y, so that \mathrm dx=5\cos y\,\mathrm dy. Then

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\int\frac{5\cos y}{\sqrt{5^2-(5\sin y)^2}}\,\mathrm dy

which simplifies to

\displaystyle\int\frac{5\cos &#10;y}{5\sqrt{1-\sin^2y}}\,\mathrm dy=\int\frac{\cos y}{\sqrt{\cos^2y}}\,\mathrm dy

Now, recall that \sqrt{x^2}=|x|. But we want the substitution we made to be reversible, so that

x+4=5\sin y\iff y=\sin^{-1}\left(\dfrac{x+4}5\right)

which implies that -\dfrac\pi2\le y\le\dfrac\pi2. (This is the range of the inverse sine function.)

Under these conditions, we have \cos y\ge0, which lets us reduce \sqrt{\cos^2y}=|\cos y|=\cos y. Finally,

\displaystyle\int\frac{\cos y}{\cos y}\,\mathrm dy=\int\mathrm dy=y+C

and back-substituting to get this in terms of x yields

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\sin^{-1}\left(\frac{x+4}5\right)+C
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Answer:

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