Which triangle is similar to triangle PQR using the pieces of Right Triangles Similarity Theorem?

In rectangle ABCD the diagonals intersect each other at point O and m∠ABD = 30°. Find BC if AC = 16 in

pashok25 [27]

16(sin 30degrees) = BC

BC = 8

4 0

3 years ago

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White raven [17]

The ratio of the volume of pipe a to the volume of pipe b is 27:1 .

<h3>What is cylinder?</h3>

A cylinder is a three-dimensional shape consisting of two parallel circular bases, joined by a curved surface. The center of the circular bases overlaps each other to form a right cylinder.

Given that,

Radius of pipe a is 6 cm, and the radius of pipe b is 2 cm.

We know that the volume of cylinder is :- $\pi r^{2}h$

where r is radius and h is height of the cylinder.

If two figures are similar then ratio of volume is equal to the cube of any dimension.

The ratio of the volume of Pipe a to the volume of Pipe b :-

$\frac{V(a)}{V(b)}=\frac{6^{3} }{2^{3} }$

= $\frac{27}{1}$

Hence, The ratio of the volume of pipe a to the volume of pipe b is 27:1 .

To learn more about cylinder from the given link:

brainly.com/question/1392572

#SPJ4

7 0

1 year ago

In Exploration 5.4.2 Question 2, what conclusion can you make about the value of the derivative at

givi [52]

The value of the derivative at the maximum or minimum for a continuous function must be zero.

<h3>What happens with the derivative at the maximum of minimum?</h3>

So, remember that the derivative at a given value gives the slope of a tangent line to the curve at that point.

Now, also remember that maximums or minimums are points where the behavior of the curve changes (it stops going up and starts going down or things like that).

If you draw the tangent line to these points, you will see that you end with horizontal lines. And the slope of a horizontal line is zero.

So we conclude that the value of the derivative at the maximum or minimum for a continuous function must be zero.

If you want to learn more about maximums and minimums, you can read:

brainly.com/question/24701109

4 0

2 years ago

What is the answer to this solution. -7r−4≥ 4r+2

tia_tia [17]

Answer:

The solution is:

$-7r-4\ge \:\:4r+2\quad \::\quad \:\begin{bmatrix}\mathrm{Solution:}\:&\:r\le \:\:-\frac{6}{11}\:\\ \:\:\mathrm{Decimal:}&\:r\le \:\:-0.54545\dots \:\\ \:\:\mathrm{Interval\:Notation:}&\:(-\infty \:\:,\:-\frac{6}{11}]\end{bmatrix}$

Please check the attached line graph below.

Step-by-step explanation:

Given the expression

$-7r-4\ge \:4r+2$

Add 4 to both sides

$-7r-4+4\ge \:4r+2+4$

Simplify

$-7r\ge \:4r+6$

Subtract 4r from both sides

$-7r-4r\ge \:4r+6-4r$

Simplify

$-11r\ge \:6$

Multiply both sides by -1 (reverses the inequality)

$\left(-11r\right)\left(-1\right)\le \:6\left(-1\right)$

Simplify

$11r\le \:-6$

Divide both sides by 11

$\frac{11r}{11}\le \frac{-6}{11}$

Simplify

$r\le \:-\frac{6}{11}$

Therefore, the solution is:

$-7r-4\ge \:\:4r+2\quad \::\quad \:\begin{bmatrix}\mathrm{Solution:}\:&\:r\le \:\:-\frac{6}{11}\:\\ \:\:\mathrm{Decimal:}&\:r\le \:\:-0.54545\dots \:\\ \:\:\mathrm{Interval\:Notation:}&\:(-\infty \:\:,\:-\frac{6}{11}]\end{bmatrix}$

Please check the attached line graph below.

7 0

2 years ago

Find the equation of the line specified. The slope is -4, and it passes through ( 5, 8).

ollegr [7]

Answer:

B. y= -4x + 28

Step-by-step explanation:

To find the equation of this line first put it into slope-point form, $y-y_{1} =m(x-x_{1})$ where $y_{1}$ and $x_{1}$ are the y and x coordinates and m is the slope. So plug in the information to get the equation, y-8=-4(x-5). Then solve for y,

First, distribute the -4

y-8=-4x+20

Next, add 8 to both sides

y=-4x+28

4 0

3 years ago

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