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3 years ago

I need help with this question

Mathematics

1 answer:

zhuklara [117]3 years ago

7 0

Answer: 5x - 1 = -6

Step-by-Step Explanation:

We have this diagram here showing numbers and symbols. Just a remainder, "X" is a variable which can be found through and equation of Algebra. Alright, so we have 5 boxes with the symbol "X". Alright, that's a start, so we have 5x as a start of the equation. We have -1 on the same side as well.

Alright, so on the other side- we have Six symbols of "-1".

That concludes to... -6. Alright, notice how all of the symbols are on a Scale, so we can heavily imply that the "=" symbol will take place between the two sets of symbols. Alright, combining them we get.. 5x - 1 = -6.

If you need further instructions/help, don't hesitate to give me a shout out. Bye, and I hope this helps.

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A solid right pyramid has a square base. The length of the base edge is4 cm and the height of the pyramid is 3 cm period what is

Illusion [34]

Answer:

The volume of this pyramid is 16 cm³.

Step-by-step explanation:

The volume $V$ of a solid pyramid can be given as:

$\displaystyle V = \frac{1}{3} \cdot b \cdot h$ ,

where

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Here's how to solve this problem with calculus without using the previous formula.

Imaging cutting the square-base pyramid in half, horizontally. Each horizontal cross-section will be a square. The lengths of these squares' sides range from 0 cm to 3 cm. This length will be also be proportional to the vertical distance from the vertice of the pyramid.

Refer to the sketch attached. Let the vertical distance from the vertice be $x$ cm.

At the vertice of this pyramid, $x = 0$ and the length of a side of the square is also $0$ .
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As a result, the length of a side of the square will be

$\displaystyle \frac{x}{3}\times 4 = \frac{4}{3}x$ .

The area of the square will be

$\displaystyle \left(\frac{4}{3}x\right)^{2} = \frac{16}{9}x^{2}$ .

Integrate the area of the horizontal cross-section with respect to $x$

from the top of the pyramid, where $x = 0$ ,
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$\displaystyle \begin{aligned}\int_{0}^{3}{\frac{16}{9}x^{2}\cdot dx} &= \frac{16}{9}\int_{0}^{3}{x^{2}\cdot dx}\\ &= \frac{16}{9}\cdot \left(\frac{1}{3}\int_{0}^{3}{3x^{2}\cdot dx}\right) & \text{Set up the integrand for power rule}\\ &= \left.\frac{16}{9}\times \frac{1}{3}\cdot x^{3}\right|^{3}_{0}\\ &= \frac{16}{27}\times 3^{3} \\ &= 16\end{aligned}$ .

In other words, the volume of this pyramid is 16 cubic centimeters.

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Step-by-step explanation:

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