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3 years ago

The value of 8 in 5.18 is. of the value of 8 in 5.81

Mathematics

2 answers:

Rudiy273 years ago

7 0

The value of 8 in 5.18 is 0.08. The value of the 8 in 5.81 is 0.8.

PilotLPTM [1.2K]3 years ago

6 0

Answer:

0.8

Step-by-step explanation:

5 is in the ones place, 8 is in the tenths place, and 1 is in the hundredths place.

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Whats 2+2? please help i literally dont know

UkoKoshka [18]

Answer:

2 + 2 = 4

Step-by-step explanation:

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3 years ago

What is the volume of this cone?

olchik [2.2K]

Answer:

The volume of the cone is approximately 453.0 cm³

Step-by-step explanation:

The volume of a cone is one third that of a cylinder with the same height and radius. That gives us 1/3 πr²h, where r is radius and h is height.

However, we are not given the height of the cone, but the side length. We can work out the height using the Pythagorean theorem, as we have a right triangle with the height, base radius, and length. You may recall that the Pythagorean theorem states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of it's other two sides:

$a^2 = b^2 + c^2$

So we can find the height of the cone with that:

$10.8^2 = 7.4^2 + h^2\\h^2 = 10.8^2 - 7.4^2\\h^2 = 116.64 - 54.76\\h^2 = 61.88\\h = \sqrt{61.88}\\h \approx 7.9$

Now that we have the cone's height, we can solve for its volume:

$v = \frac{1}{3} \pi r^2 h\\v = \frac{1}{3} \pi \times 7.4^2 \times 7.9\\v = \frac{1}{3} \pi \times 54.76 \times 7.9\\v = \frac{1}{3} \pi \times 432.6\\v = \pi \times 144.2\\v \approx 453.0 cm^3$

5 0

3 years ago

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Which of the following statements is TRUE about the midline of a trapezoid?

Y_Kistochka [10]

Answer:

C will be the rightful answer

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2 years ago

100 POINTS TO RIGHT AWNSER!!!!

makvit [3.9K]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3000586

——————————

The answer is option D) r < 5 or r > – 1.

I'm going to graph each inequality below on a number line.

A) r > 5 or r > – 1.

$\large\begin{array}{cl} \mathsf{r>5}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\overset{*****~~~}{\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\ \mathsf{r>-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\\\ \mathsf{r>5~~or~~r>-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}} \end{array}$

The result is found just by joining those two intervals together. Actually that compound inequality only implies

r > – 1

which does not represent all real numbers.

—————

B) r > 5 or r < – 1.

$\large\begin{array}{cl} \mathsf{r>5}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\overset{*****~~~}{\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\ \mathsf{r5~~or~~r$

Numbers between – 1 and 5 (including them) are not included in the union, so you don't have all real numbers represented there either.

—————

C) r < 5 or r < – 1.

$\large\begin{array}{cl} \mathsf{r$

Numbers that are greater or equal to 5 are not in the union. So it does not represent all real numbers.

—————

D) r < 5 or r > – 1.

$\large\begin{array}{cl} \mathsf{r-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\\\ \mathsf{r-1}&\qquad\mathsf{\overset{~~~**************************~~~}{\textsf{|||}\!\!\!\underset{-1}{\bullet}\!\!\!\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}} \end{array}$

Now all real numbers are included in the union. So this is the right choice.

Answer: option D) r < 5 or r > – 1.

I hope this helps. =)

6 0

3 years ago

Using only words solve 8x+3=27 step by step

erik [133]

Answer:

X=3

Step-by-step explanation:

Subtract 3 from both sides

Simplify 27-3 to 24

Divide both sides by 8

Simplify 24/8 to 3

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3 years ago

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