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3 years ago

what is the volume of a triangular prism that is 75 centimeters long and that has a base with an area of 30 square centimeters

Mathematics

1 answer:

pychu [463]3 years ago

3 0

75 * 30= 2250 that's how u get the answer

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Which of the following is a rational number?

zubka84 [21]

Answer:

Step-by-step explanation:

Rational numbers have an ending decimal expansion. If you put the other numbers into a calculator you will se that there is just more numbers after the decimal point

8 0

3 years ago

Y = x – 6

zhenek [66]

-4-6= -10
-10 = Y
-4 = X
Therefore the answer is (-4,-10) which isn’t on the options

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3 years ago

A pet store has three aquariums for fish. Aquarium A contains 24 fish, Aquarium B contains 51 fish, and Aquarium C contains 36 f

bazaltina [42]

The graph that can represent the data most accurately is (a) The y-axis of a bar graph starts at zero fish. One bar is 24 units, another bar is 51 units, and the third bar is 36 units

The given parameters are:

Aquarium A: 24 fishes
Aquarium B: 51 fishes
Aquarium C: 36 fishes

The above dataset can be represented on a bar graph, where the lengths of the bars represent the number of fishes in each aquarium

Hence, the graph that can represent the data most accurately is (a)

Read more about graphs at:

brainly.com/question/25677468

5 0

3 years ago

This is mainly for catra0920<br> But anyone can help :)

garri49 [273]

Answer:

$-\frac{4}{9}$

$-\frac{8}{27}$

Step-by-step explanation:

4 0

3 years ago

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Use the definition of continuity and the properties of limit to show that the function f(x)=x sqrtx/ (x-6)^2 is continuous at x=

jasenka [17]

Answer:

The function $\\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}$ is continuous at x = 36.

Step-by-step explanation:

We need to follow the following steps:

The function is:

$\\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}$

The function is continuous at point x=36 if:

The function $\\ f(x)$ exists at x=36.
The limit on both sides of 36 exists.
The value of the function at x=36 is the same as the value of the limit of the function at x = 36.

Therefore:

The value of the function at x = 36 is:

$\\ f(36) = \frac{36*\sqrt{36}}{(36-6)^{2}}$

$\\ f(36) = \frac{36*6}{900} = \frac{6}{25}$

The limit of the $\\ f(x)$ is the same at both sides of x=36, that is, the evaluation of the limit for values coming below x = 36, or 33, 34, 35.5, 35.9, 35.99999 is the same that the limit for values coming above x = 36, or 38, 37, 36.5, 36.1, 36.01, 36.001, 36.0001, etc.

For this case:

$\\ lim_{x \to 36} f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}$

$\\ \lim_{x \to 36} f(x) = \frac{6}{25}$

Since

$\\ f(36) = \frac{6}{25}$

And

$\\ \lim_{x \to 36} f(x) = \frac{6}{25}$

Then, the function $\\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}$ is continuous at x = 36.

8 0

3 years ago