All categories

3 years ago

Is 144 is 30% of 480

Mathematics

2 answers:

Usimov [2.4K]3 years ago

6 0

Answer:

Yes it is 144

Step-by-step explanation:

Finding out 30% of 480

We can write 30% of 480 as

30/100×480

So this becomes after cancelling zeroes

=3×48

=144

Hence 30% of 480 is 144 so the answer is yes

svlad2 [7]3 years ago

5 0

Answer:

Yes it is true.

Step-by-step explanation:

144 / 480 x 100% = 30%

<em>Hope this helps </em>

<em>-Amelia The Unknown</em>

You might be interested in

How many inches are in 240 feet?

Tema [17]

2880 ...................................

7 0

3 years ago

Read 2 more answers

F(x) = 5x - 3 for x = -7

erastovalidia [21]

Answer:

Step-by-step explanation:

f(-7) = 5(-7) - 3 = -35 - 3 = -38

6 0

3 years ago

If doughnuts are usually 75 cents each, but there is a sale on Friday advertising them as 1 ½ dozen for $10.80, what is the new

Rzqust [24]

This deal is cheap! Well, the answer is 60 cents. To get these problems, divide 10.80 by 18, which is equal to 1.5 a dozen.

3 0

3 years ago

Read 2 more answers

The area of a rectangular lawn is 255 metre square if its length is 15 M find its perimeter

geniusboy [140]

Answer:

Step-by-step explanation:

width is 255/15=17

perimeter = 17*2+15*2=64

7 0

2 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

4 0

3 years ago