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laiz [17]
3 years ago
5

1 2 3 4 5 6 7 8 9 10

Mathematics
1 answer:
klemol [59]3 years ago
4 0

Answer:

the first one mate

Step-by-step explanation:

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Eloise bought 3 boxes of crackers to share with her friends over the weekend. They ate 1/2 box of crackers on Saturday and 2/3 b
sveticcg [70]

Answer:

C.) 1 5/b boxes

Number of boxes left over

= 3 - 1/2 - 2/3

Convert each  fraction to one in terms of the LCM:

= 18/6 - 3/6 - 4/6

= 11/6

= 1 5/6 boxes.

4 0
3 years ago
What is the answer?<br> 2500/x+20<br> =2500/(48.59)+20
sammy [17]
The answer would be 71.45
7 0
3 years ago
You have just timed a person doing a hair cut for the first time. It took 50 minutes. What unit improvement factor learning curv
Sladkaya [172]

Answer: C. 70 percent

Step-by-step explanation:

Given, Time for the first unit = 50 minutes

Time for the second unit = 35 minutes

The unit improvement factor learning curve = (The time for the second unit) ÷ (time for the first unit) x 100.

So, The unit improvement factor learning curve  = 35÷ 50 × 100 = 70 percent.

Hence, the correct option is "C. 70 percent".

3 0
3 years ago
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
4 0
3 years ago
What is the area of this figure? I will give brainliest! :)
jenyasd209 [6]

Answer:

Step-by-step explanation:

Area = (17×35) - (6×15) = 595 - 90 = 505 ft²

3 0
3 years ago
Read 2 more answers
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