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2 years ago

The product of two rational numbers is 6.If one of them is 8,find the other number.

Mathematics

1 answer:

frez [133]2 years ago

6 0

Answer:

0.75

Step-by-step explanation:

The "product" is a result of multiplication, so we know we are multiplying eight with something to get six. 0.75 is a rational number, and when multiplied by eight, gives six.

So, 0.75 * 8 = 6

The other number is 0.75

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Kayla and two of her friends form an investment club. Each contributes $250 to an account that earns 5% interest, compounded ann

vaieri [72.5K]

The multiplier of each account's value is 1.05³ = 1.157625.

The $750 in the club's account will grow to
$750×1.157625 = 868.22
and each member will get 1/3 of that amount, $289.

The $250 in Sheri's individual account will grow to
$250×1.157625 = $289.

Sheri, Kayla, and Kayla's two friends will all have the same amount as a result of their investments.

6 0

3 years ago

Read 2 more answers

The sum of two numbers is 27 and their product is 50. Find the numbers.?<br><br>

san4es73 [151]

Answer:

a + b = 27

=> a = 27 - b

ab = 50

So, (27 - b) x b = 50

27b - b^2 = 50

b^2 - 27b + 50 = 0

Solving, b = 25, 2

Hence,

Case 1: b = 25, a = 2

Case 2, b = 2, a = 25

5 0

3 years ago

Read 2 more answers

V= 512^2/3* 19683^1/3*256^1/4

irina1246 [14]

Answer:

V=512^2/3* 19683^1/3*256^1/4 = 6912

Step-by-step explanation:

V= $(512^{\frac{2}{3} } ) \times (19683^{\frac{1}{3} } )\times (256^{\frac{1}{4} } )\\$

$2^{9} = 512\\2^{8} = 256\\19683^{\frac{1}{3} } = 27\\$

putting values in equation

V= $2^{9(^{\frac{2}{3} } )} \times 27 \times 2^{8^({\frac{1}{4}) } } \\$ let it be equation 1

$2^9^{\frac{2}{3} } = 2^6 according\,to\,exponential\,rule\\2^8^\frac{1}{4} = 2^2 according\,to\,exponential\,rule\\\\$

putting values in equation 1

$V= 2^6\times 27\times 2^2$

According to exponential rule $2^6 \times 2^2 = 2^8$

V= $2^8 \times 27\\$

$V=256\times 27\\V=6912$

V= 6912

Keywords: Algebra

Learn more about algebra at:

#learnwithBrainly

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cint%5C%20%28x%5E3-6x%5E2%2B9x%2B3%29%283x%5E2-12x%2B9%29%20dx" id="TexFormula1" title="\int

AlexFokin [52]

If the integral is simply

$\displaystyle\int(x^3-6x^2+9x+3)(3x^2-12x+9)\,\mathrm dx$

then notice that

$\mathrm d(x^3-6x^2+9x+3)=(3x^2-12x+9)\,\mathrm dx$

which means you can compute the integral easily with a substitution

$u=x^3-6x^2+9x+3\implies\mathrm du=(3x^2-12x+9)\,\mathrm dx$

Under this transformation, the integral is

$\displaystyle\int u\,\mathrm du=\frac{u^2}2+C=\boxed{\frac{(x^3-6x^2+9x+3)^2}2+C}$

On the other hand, in case you're missing a symbol and the integral is actually

$\displaystyle\int\frac{x^3-6x^2+9x+3}{3x^2-12x+9}\,\mathrm dx$

then first carry out the division:

$\dfrac{x^3-6x^2+9x+3}{3x^2-12x+9}=\dfrac x3-\dfrac23-\dfrac{2x-9}{3x^2-12x+9}$

Now, $3x^2-12x+9=3(x-3)(x-1)$ , so to integrate the remainder term you can decompose it into partial fractions:

$-\dfrac{2x-9}{3(x-3)(x-1)}=\dfrac a{x-3}+\dfrac b{x-1}$

$9-2x=a(x-1)+b(x-3)$

$x=1\implies7=-2b\implies b=-\dfrac72$

$x=3\implies3=2a\implies a=\dfrac32$

$\implies-\dfrac{2x-9}{3(x-3)(x-1)}=\dfrac 3{2(x-3)}-\dfrac 7{2(x-1)}$

Then the integral would be

$\displaystyle\int\frac{x^3-6x^2+9x+3}{3x^2-12x+9}\,\mathrm dx=\boxed{\frac{x^2}6-\frac{2x}3+\frac32\ln|x-3|-\frac72\ln|x-1|+C}$

which can be rewritten in several ways, such as

$\dfrac{x^2-4x}6+\dfrac12ln\left|\dfrac{(x-3)^3}{(x-1)^7}\right|+C$

6 0

4 years ago

What is 9 plus 10 equals ?

nexus9112 [7]

Answer:

Step-by-step explanation:

9+10=19

3 0

3 years ago