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AlladinOne [14]

2 years ago

13

The volume vv of a pyramid is given by the formula v=13bhv=13bh, where bb is the area of the base and hh is the height.

1 answer:

Setler [38]2 years ago

4 0

Volume of pyramid(V) = $\frac{1}{3}bh$

where 'b' is the base area and 'h' is the height.

We have to determine the formula for height 'h'.

$V = \frac{1}{3}bh$

Multiplying by '3' on both the sides of the equation.

$3V = bh$

$h = \frac{3V}{b}$

So, the formula for height 'h' of pyramid is $h = \frac{3V}{b}$ .

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The ages of the people who attend a weekly yoga class at the local yoga studio are shown in the table below.

Igoryamba

Answer:

b

Step-by-step explanation:

4 0

2 years ago

<img src="https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7Bx%2By%3D1%7D%20%5Catop%20%7Bx-2y%3D4%7D%7D%20%5Cright.%20%5C%5C%5Clef

brilliants [131]

Answer:

<em>(a) x=2, y=-1</em>

<em>(b) x=2, y=2</em>

<em>(c)</em> $\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

<em>(d) x=-2, y=-7</em>

Step-by-step explanation:

<u>Cramer's Rule</u>

It's a predetermined sequence of steps to solve a system of equations. It's a preferred technique to be implemented in automatic digital solutions because it's easy to structure and generalize.

It uses the concept of determinants, as explained below. Suppose we have a 2x2 system of equations like:

$\displaystyle \left \{ {{ax+by=p} \atop {cx+dy=q}} \right.$

We call the determinant of the system

$\Delta=\begin{vmatrix}a &b \\c &d \end{vmatrix}$

We also define:

$\Delta_x=\begin{vmatrix}p &b \\q &d \end{vmatrix}$

And

$\Delta_y=\begin{vmatrix}a &p \\c &q \end{vmatrix}$

The solution for x and y is

$\displaystyle x=\frac{\Delta_x}{\Delta}$

$\displaystyle y=\frac{\Delta_y}{\Delta}$

(a) The system to solve is

$\displaystyle \left \{ {{x+y=1} \atop {x-2y=4}} \right.$

Calculating:

$\Delta=\begin{vmatrix}1 &1 \\1 &-2 \end{vmatrix}=-2-1=-3$

$\Delta_x=\begin{vmatrix}1 &1 \\4 &-2 \end{vmatrix}=-2-4=-6$

$\Delta_y=\begin{vmatrix}1 &1 \\1 &4 \end{vmatrix}=4-3=3$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{3}{-3}=-1$

The solution is x=2, y=-1

(b) The system to solve is

$\displaystyle \left \{ {{4x-y=6} \atop {x-y=0}} \right.$

Calculating:

$\Delta=\begin{vmatrix}4 &-1 \\1 &-1 \end{vmatrix}=-4+1=-3$

$\Delta_x=\begin{vmatrix}6 &-1 \\0 &-1 \end{vmatrix}=-6-0=-6$

$\Delta_y=\begin{vmatrix}4 &6 \\1 &0 \end{vmatrix}=0-6=-6$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-6}{-3}=2$

The solution is x=2, y=2

(c) The system to solve is

$\displaystyle \left \{ {{-x+2y=0} \atop {x+2y=5}} \right.$

Calculating:

$\Delta=\begin{vmatrix}-1 &2 \\1 &2 \end{vmatrix}=-2-2=-4$

$\Delta_x=\begin{vmatrix}0 &2 \\5 &2 \end{vmatrix}=0-10=-10$

$\Delta_y=\begin{vmatrix}-1 &0 \\1 &5 \end{vmatrix}=-5-0=-5$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-10}{-4}=\frac{5}{2}$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-5}{-4}=\frac{5}{4}$

The solution is

$\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

(d) The system to solve is

$\displaystyle \left \{ {{6x-y=-5} \atop {4x-2y=6}} \right.$

Calculating:

$\Delta=\begin{vmatrix}6 &-1 \\4 &-2 \end{vmatrix}=-12+4=-8$

$\Delta_x=\begin{vmatrix}-5 &-1 \\6 &-2 \end{vmatrix}=10+6=16$

$\Delta_y=\begin{vmatrix}6 &-5 \\4 &6 \end{vmatrix}=36+20=56$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{16}{-8}=-2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{56}{-8}=-7$

The solution is x=-2, y=-7

4 0

3 years ago

Select the term that best describes the statement A triangle is acute provided all the angles have a measure of less than 90 deg

Aloiza [94]

I think the correct answer is conjunction...

4 0

2 years ago

Read 2 more answers

Then find the values of x and y?

attashe74 [19]

Y+8=3y
-2y=-8
y=4

104*2=208
360-208=152
152/2=76

x=76°

6 0

2 years ago

Read 2 more answers

What is the value of "c" in the following quadratic? (Make sure the equation is in

garri49 [273]

$\rule{50}{1}\large\blue\textsf{\textbf{\underline{Given question:-}}}\rule{50}{1}$

<em>What is the value of c in the quadratic </em> $\large\text{$x^2+28=-11x$}$ ?

$\rule{50}{1}\large\textsf{\textbf{\underline{Answer and how to solve:-}}}\rule{50}{1}$

Before starting to solve, you should notice something - the

quadratic is not in its standard form!

We can easily fix it by adding $\large\textit{11x}$ on both sides:-

$\large\text{$x^2+28-11x=0$}$

We can switch the order of 28 and -11x:-

$\large\text{$x^2-11x+28=0$}$

Now, the quadratic is in its standard form, so we can get down to

finding the value of "c".

Remember, the standard form of a quadratic looks like so:-

$\large\text{$ax^2+bx+c=0$}$

Now we can just write our <u>quadratic</u> here:-

$\large\text{$x^2-11x+28=0$}$

Now, can you see what the value of "c" is?

An easy way to <u>remember</u> "c" in quadratics is:-

The "c" in quadratics is the constant.

Henceforth, we conclude that the value of "c" in the given quadratic is:-

$\Large\textbf{28}\Large\checkmark$

<h3> Good luck with your studies.</h3>

$\rule{50}{1}\smile\smile\smile\smile\smile\smile\rule{50}{1}$

5 0

2 years ago