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3 years ago

Whats the answer to these? (These are called Proper fractions.)

Mathematics

2 answers:

AnnZ [28]3 years ago

7 0

Answer:

1: 7/10

2: 3/10

3: 4/8 (2/4 or 1/2 if you need to simplify it)

Step-by-step explanation:

1: Convert 2/5 to tenths by multiplying the denominator and numerator by two. 3/10 plus 4/10 is 7/10

2: Convert 4/5 and 1/2 to tenths.

4/5 is going to be 8/10, 1/2 is going to be 5/10.

Subtract 5/10 from 8/10 to get 3/10

3: Convert 1/4 to eighths to get 2/8. Then add 2/8 and 2/8 to get 4/8.

aleksandrvk [35]3 years ago

5 0

Jeremiah: convert the 5ths to 10ths
3/10 + 4/10 = 7/10

Daniel: convert both fractions to 10ths
8/10 - 5/10 = 3/10

Jared: reduce 8ths to 4ths
1/4 + 1/4 = 2/4 = 1/2

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Matt picks a marble from a bag, replaces it and then chooses a second

krek1111 [17]

Answer:

Step-by-step explanation:

6 0

3 years ago

PLEASE HELP IF U CAN!

12345 [234]

Answer:

$x = 8$

Step-by-step explanation:

Given

The attached triangle

Required

Determine the value of x

x represents the height of the triangle.

And the area of a triangle is calculated as:

$Area = \frac{1}{2} * Base * Height$

Where

$Base = 4$

$Area = 16$

$Height = x$

The expression becomes

$16 = \frac{1}{2} * 4* x$

$16 = 2* x$

$16 = 2x$

Divide through by 2

$8 = x$

$x = 8$

4 0

3 years ago

Somebody help me please

andrew-mc [135]

The justification for step 3 is incorrect. It should be x = 10/8.

Hope this helps =)

7 0

4 years ago

I need help in partial fraction!! With simple explaination would be nice!

WITCHER [35]

$\dfrac{x^4-7x^2+17x-10}{x(x^2-3)}$

The degree of the numerator (4) is larger than the degree of the denominator (3), so first you need to divide. (Added screenshot of long division procedure.)

$\dfrac{x^4-7x^2+17x-10}{x(x^2-3)}=x-\dfrac{4x^2-17x+10}{x(x^2-3)}$

Now the second term can be decomposed into partial fractions.

$\dfrac{4x^2-17x+10}{x(x^2-3)}=\dfrac{r_1}x+\dfrac{r_2x+r_3}{x^2-3}$
$\dfrac{4x^2-17x+10}{x(x^2-3)}=\dfrac{r_1(x^2-3)+x(r_2x+r_3)}{x(x^2-3)}$
$4x^2-17x+10=r_1(x^2-3)+x(r_2x+r_3)$
$4x^2-17x+10=(r_1+r_2)x^2+r_3x-3r_1$
$\implies\begin{cases}r_1+r_2=4\\r_3=-17\\-3r_1=10\end{cases}\implies r_1=-\dfrac{10}3,r_2=\dfrac{22}3,r_3=-17=-\dfrac{51}3$
$\implies\dfrac{4x^2-17x+10}{x(x^2-3)}=-\dfrac{10}{3x}+\dfrac{22x-51}{x^2-3}$

So

$\dfrac{x^4-7x^2+17x-10}{x(x^2-3)}=x+\dfrac{10}{3x}-\dfrac{22x-51}{x^2-3}$

3 0

3 years ago

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boyakko [2]

The answer would be Quadratic

5 0

3 years ago