All categories

1 year ago

8) 3/

Mathematics

1 answer:

Vaselesa [24]1 year ago

7 0

Answer:

0.0085× 100= 0.85℅

Step-by-step explanation:

okay so you want 3/8 from 44 as percentage but how can you put the 3/8 on top of the 44 to multilply it by 100 to get the percentage, ofcourse we will have to change the 3/8 into a whole number. To do that we are going to do the following.

= 3 ÷ 8 = 0.375, so here you got a number that you can put on the 44.

Now to get the percentage we are going to do this:

0.375/44 × 100

= 0.0085 × 100 ( I solved the 0.375/44 as you know that division always comes first in maths)

= 0.85℅ ( I solved the 0.0085 × 100)

°•°•° I hope this helped sorry for taking long °•°•° May I get brainliest if you liked my ans im trying to lvl up ty and its okay if you don't want to

You might be interested in

Use the polynomial 3x^2−4x^11+4x^4−2x^10−5+^8x^8 to answer the question.

Anit [1.1K]

Answer: 11

<u>Step-by-step explanation:</u>

3x² - 4x¹¹ + 4x⁴ - 2x¹⁰ - 5 + 8x⁸

To find the degree, look for the largest exponent. <em>11</em>

If you put this polynomial in standard form (from largest to smallest exponent), the first term will give you the degree of the polynomial and the leading coefficient.

- 4x¹¹ - 2x¹⁰ + 8x⁸ + 4x⁴ + 3x² - 5

4 0

2 years ago

Read 2 more answers

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 $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .