All categories

3 years ago

A price decreases from $254 to $213.36 . What is the percent of change?

Mathematics

1 answer:

sattari [20]3 years ago

8 0

16% discount, this is because 213.36-254 over the original price (254) times 100% equals -16%

You might be interested in

Use order of operations to make this true 7 11 4 2 = 10

Keith_Richards [23]

Answer:

$7 + 11 - 4 \times 2 = 10$

Step-by-step explanation:

Using order of operations, the first thing to be calculated would be $4 \times 2$ .

That would turn our equation into this:

$7 + 11 - 8 = 10$

You can then calculate left-to-right as usual to get the answer 10.

5 0

3 years ago

How do I find:<br> If 6^x=1/216,find the value of x with and without a Casio calculator?

san4es73 [151]

Answer:

The value of x is -3.

Step-by-step explanation:

To find x, we have to put everything in base 6 and continue to simplify. Since we know that 216 is 6^3, we can replace that.

6^x = 1/216

6^x = 1/(6^3)

Now, when there is a power in the denominator, we know we can move it to the numerator by making it negative. We can do that to get the bases to match.

6^x = 1/(6^3)

6^x = 6^-3

Now that they are both put in simple base 6 systems, we can just eliminate the bases and assume that the powers are equal. This lets us know that x = -3.

8 0

3 years ago

A decorative window is made up of a rectangle with semicircles at either end. The ratio of AD to AB is 3:2 and AB is 30 inches.

scZoUnD [109]

Answer: The required ratio will be

$84:1034$

Step-by-step explanation:

Since we have given that

Ratio of AD to AB is 3:2

Length of AB = 30 inches

So, it becomes

$2x=30\\\\x=\frac{30}{2}=15\ inches$

So, Length of AD becomes

$3x=3\times 15=45\ inches$

Now, at either end , there is a semicircle.

Radius of semicircle along AB is given by

$\frac{30}{2}=15\ inches$

So, Area of semicircle along AB and CD is given by

$2\times \frac{\pi r^2}{2}\\\\=\frac{22}{7}\times 15\times 15\\\\=\frac{4950}{7}\ in^2$

Radius of semicircle along AD is given by

$\frac{45}{2}=22.5\ inches$

Area of semicircle along AD and BC is given by

$2\times \frac{1}{2}\pi r^2\\\\=\frac{22}{7}\times \frac{45}{2}\times \frac{45}{2}\\\\=\frac{445500}{28}\ in^2$

And the combined area of the semicircles is given by

$\frac{4950}{7}+\frac{445500}{28}\\\\=\frac{465300}{28}\ in^2$

Area of rectangle is given by

$Length\times width\\\\=AD\times AB\\\\=45\times 30\\\\=1350\ in^2$

Hence, Ratio of the area of the rectangle to the combined area of the semicircles is given by

$1350:\frac{465300}{28}\\\\=1350\times 28:465300\\\\=37800:465300\\\\=84:1034$

Hence, the required ratio will be

$84:1034$

8 0

3 years ago

HELPPPPPPPPPP PLEASE

Vlad1618 [11]

Answer:

a) 27. b) 8/3

This is your answer ☺️☺️☺️

3 0

3 years ago

How do I do 8b(ii) ? Please help me thank you!

Mila [183]

Step One
======
Find the length of FO (see below)

All of the triangles are equilateral triangles. Label the center as O
FO = FE = sqrt(5) + sqrt(2)

Step Two
======
Drop a perpendicular bisector from O to the midpoint of FE. Label the midpoint as J. Find OJ

Sure the Pythagorean Theorem. Remember that OJ is a perpendicular bisector.

FO^2 = FJ^2 + OJ^2
FO = sqrt(5) + sqrt(2)
FJ = 1/2 [(sqrt(5) + sqrt(2)] \
OJ = ??

[Sqrt(5) + sqrt(2)]^2 = [1/2(sqrt(5) + sqrt(2) ] ^2 + OJ^2
5 + 2 + 2*sqrt(10) = [1/4 (5 + 2 + 2*sqrt(10) + OJ^2
7 + 2sqrt(10) = 1/4 (7 + 2sqrt(10)) + OJ^2 Multiply through by 4
28 + 8* sqrt(10) = 7 + 2sqrt(10) + 4 OJ^2 Subtract 7 + 2sqrt From both sides
21 + 6 sqrt(10) = 4OJ^2 Divide both sides by 4
21/4 + 6/4* sqrt(10) = OJ^2
21/4 + 3/2 * sqrt(10) = OJ^2 Take the square root of both sides.
sqrt OJ^2 = sqrt(21/4 + 3/2 sqrt(10) )
OJ = sqrt(21/4 + 3/2 sqrt(10) )

Step three
find h
h = 2 * OJ
h = 2* sqrt(21/4 + 3/2 sqrt(10) ) <<<<<< answer.

7 0

3 years ago