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3 years ago

How many of possible outcomes could be a number biggger than 6 on either cube

Mathematics

1 answer:

Naily [24]3 years ago

3 0

There is no number bigger than 6 on dice cubes.

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Lee ran a mile in 7 1/3 minutes. His friend Sam ran the same mile in 8 5/9 minutes. How many minutes faster did Lee run? (First

Yakvenalex [24]

He ran the mile 1 and 3/9 minutes faster

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3 years ago

A pack of 5 smiley stickers cost $6.50. What is the unit price?

nika2105 [10]

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$1.30

Step-by-step explanation:

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2 years ago

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3 years ago

The curves y = √x and y=(2-x) and the Cartesian axes form two distinct regions in the first quadrant. Find the volumes of rotati

makkiz [27]

Answer:

Step-by-step explanation:

If you graph there would be two different regions. The first one would be

$y = \sqrt{x} \,\,\,\,, 0\leq x \leq 1 \\$

And the second one would be

$y = 2-x \,\,\,\,\,, 1 \leq x \leq 2$ .

If you rotate the first region around the "y" axis you get that

$A_1 = 2\pi \int\limits_{0}^{1} x\sqrt{x} dx = \frac{4\pi}{5} = 2.51$

And if you rotate the second region around the "y" axis you get that

$A_2 = 2\pi \int\limits_{1}^{2} x(2-x) dx = \frac{4\pi}{3} = 4.188$

And the sum would be 2.51+4.188 = 6.698

If you revolve just the outer curve you get

If you rotate the first region around the x axis you get that

$A_1 =\pi \int\limits_{0}^{1} ( \sqrt{x})^2 dx = \frac{\pi}{2} = 1.5708$

And if you rotate the second region around the x axis you get that

$A_2 = \pi \int\limits_{1}^{2} (2-x)^2 dx = \frac{\pi}{3} = 1.0472$

And the sum would be 1.5708+1.0472 = 2.618

7 0

3 years ago

If H is in the interior of ∠EFG, m∠EFH = 75°, and m∠HFG = (10x)°, and m∠EFG = (20x − 5)°, then x = ? and m∠HFG = ?

Diano4ka-milaya [45]

Answer:

$x=8$

$\m\angle HFG= 80^\circ$

Step-by-step explanation:

Given that:

Point $H$ is interior of $\angle EFG$ .

$m\angle EFH = 75^\circ\\m\angle HFG = (10x)^\circ\\m\angle EFG = (20x-5)^\circ$

To find:

$x = ?\\m\angle HFG = ?$

Solution:

First of all, let us represent the given values in the form of a diagram.

Kindly refer to the attached image for the given points and values of angles.

We can clearly see that:

$m\angle EFG = m\angle EFH + m\angle HFG$

Putting all the values given in above equation, we get:

$(20x-5)^\circ = 75^\circ + 10x^\circ\\\Rightarrow 20x-10x=75+5\\\Rightarrow 10x =80\\\Rightarrow \bold{x =8}$

$m\angle HFG =10x^\circ\\\Rightarrow m\angle HFG =10\times 8^\circ\\\Rightarrow m\angle HFG = \bold{80^\circ}$

8 0

3 years ago