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3 years ago

60 out of 300 CDs are classical music CDs. What percent of the CDs are classical music CDs?

Mathematics

2 answers:

AnnyKZ [126]3 years ago

7 0

300 divided by 60 is 5 so that mean 60 goes into 300 5 times so each time is 20 %

valkas [14]3 years ago

6 0

60/300=m/100
300 divided by 3=100
60 divided by 3= 20
20/100
20%

sometimes anything over 100 can be used as a percent

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What is the value of h in the figure below? In this diagram BAD~CBD

liberstina [14]

Answer:

The correct option is E. 8

The value of h is 8 unit.

Step-by-step explanation:

Given:

Δ BAD ~ Δ CBD

AC = 20

DC = 4

∴ $AD = AC - DC=20-4=16$

To Find:

h = ?

Solution:

Δ BAD ~ Δ CBD ................Given

If two triangles are similar then their sides are in proportion.

$\frac{BD}{CD} =\frac{AD}{BD} =\frac{BA}{CB}\ \textrm{corresponding sides of similar triangles are in proportion}\\$

On substituting the given values we get

$\dfrac{BD}{CD} =\dfrac{AD}{BD}$

$\dfrac{h}{4} =\dfrac{16}{h}\\\therefore h^{2}=64\\\therefore h=8\ unit$

The value of h is 8 unit.

8 0

3 years ago

Read 2 more answers

Starting at the point (-3,2) go left 9 and down 8. What ordered pair gives the location of this new point

tigry1 [53]

The new a=plot would be:

(-3-9,2-8)=

5 0

2 years ago

Add.<br> (4x2 - 2x) + (5x-7)

Sloan [31]

Answer:

4x²+3x-7 is the answer

Step-by-step explanation:

=(4x²-2x)+(5x-7)

opening brackets

=4x²-2x+5x-7

=4x²+3x-7

i hope this will help you :)

7 0

3 years ago

There are 34 beads on Brenda's necklace. 24 of the beads are pink and the rest are purple.

djyliett [7]

the correct answer is 24:10

6 0

2 years ago

Find two unit vectors orthogonal to both given vectors. i j k, 4i k

Maurinko [17]

The cross product of two vectors gives a third vector $\mathbf v$ that is orthogonal to the first two.

$\mathbf v=(\vec i+\vec j+\vec k)\times(4\,\vec i+\vec k)=\begin{vmatrix}\vec i&\vec j&\vec k\\1&1&1\\4&0&1\end{vmatrix}=\vec i+3\,\vec j-4\,\vec k$

Normalize this vector by dividing it by its norm:

$\dfrac{\mathbf v}{\|\mathbf v\|}=\dfrac{\vec i+3\,\vec j-4\,\vec k}{\sqrt{1^2+3^2+(-4)^2}}=\dfrac1{\sqrt{26}}(\vec i+3\,\vec j-4\vec k)$

To get another vector orthogonal to the first two, you can just change the sign and use $-\mathbf v$ .

6 0

3 years ago