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3 years ago

How do you write 8/50 as a percentage?

Mathematics

2 answers:

Phoenix [80]3 years ago

8 0

Answer:

16%

convert into a decimal and move the decimal point over and add the percent sign.

8/50= .16

.16

16%

Step-by-step explanation:

nikdorinn [45]3 years ago

3 0

Answer: . nem .8%

Step-by-step explanation:

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Rewrite the function f(x)=2x^2-16x-15 in vertex form

Dmitry_Shevchenko [17]

Answer: y=2 (x-4)^2 -47

and the vertex is (4,-47)

Step-by-step explanation:

your welcome :)

8 0

3 years ago

The Lewis family is building a circular swimming pool if the radius of the pool is 9 feet what is it’s circumference use TT =3.1

zzz [600]

Answer:

C =56.52 ft

Step-by-step explanation:

The circumference is found by

C = 2*pi*r where r is the radius

C = 2*3.14 * 9

C =56.52 ft

5 0

3 years ago

Angles A and B are supplementary. Angle A has a measure of 80 degrees. What is the measure of angle B?

Aleonysh [2.5K]

Supplementary angles, when added, equal 180 degrees

so < A + < B = 180...and if < A = 80
80 + < B = 180
< B = 180 - 80
< B = 100 <==

4 0

3 years ago

Read 2 more answers

Which inequality is represented by the graph? y≥35x−1.5 y≤35x−1.5 y<35x−1.5 y>35x−1.5

Setler79 [48]

Answer:

y > 0.6x - 1.5

Step-by-step Explanation:

We need two points, to get to the equation of the graph.

Since we've got the following equation for two points (x1, y1), (x2, y2):-

$\boxed{ \mathsf{ \red{y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1}) }}}$

okay soo

I found two points that lie on this graph, not on the shaded region but yeah the dotted line which defines the graph.

one point is <u>(0, -1.5)</u> which lies on the y axis(the point where the dotted line touches the y axis)

other point is <u>(2.5, 0)</u> and this lies on the x axis

placing these points in the place of (x1, y1) and (x2, y2) in the above mentioned equation

$\mathsf{\implies y - ( - 1.5) = \frac{0 -( - 1.5)}{2.5 -0 } (x -0 )}$

you can take any one as (x1, y1) or (x2, y2).

so upon solving the above equation we get

$\mathsf{\implies (y + 1.5) = \frac{0 + 1.5}{2.5 } (x )}$

$\mathsf{\implies y + 1.5 = \frac{ 1.5}{2.5 } x }$

$\mathsf{\implies y + 1.5 = \frac{ \cancel{1.5}\:\:{}^3}{\cancel{2.5}\:\:{}^5 } x }$

$\mathsf{\implies y + 1.5 = \frac{ 3}{5 } x }$

multiplying both sides by 5

$\mathsf{5y + 7.5 = 3x}$

okay so this is the required equation of the dotted line

now we'll find the inequality

for this check whether the origin (0,0) lies under the shaded region or not

in this case it does

replacing x and y with 0

$\mathsf{\implies5(0) + 7.5 = 3(0)}$

$\mathsf{\implies0 + 7.5 = 0}$

this is absurd, 7.5 is not equal to 0 so we're gonna replace that equals sign with that of inequality

7.5 is greater than 0! so,

$\mathsf{\implies7.5 > 0}$

this goes for the whole equation, since we didnt swap any thing from left to right side of the equation or vice versa we can use this sign, to obtain the required inequality

$\mathsf{5y + 7.5 > 3x}$

dividing this inequality by 5, since there's no co-efficient in front of y in the given answers

we get

y + 1.5 > 0.6x

taking 1.5 to the RHS

that is the last option

5 0

3 years ago

Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator

SpyIntel [72]

$\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}$
$\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)$

When $x=3$ , you're left with

$147=49a_1\implies a_1=\dfrac{147}{49}=3$

When $x=\sqrt2$ or $x=-\sqrt2$ , you're left with

$\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}$

Adding the two equations together gives $-10=2a_5$ , or $a_5=-5$ . Subtracting them gives $2\sqrt2=2\sqrt2a_4$ , $a_4=1$ .

Now, you have

$5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)$
$5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)$
$2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)$

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that $a_2x^4=2x^4$ and $a_3(-3)(-2)=6a_3=-6$ . These alone tell you that you must have $a_2=2$ and $a_3=-1$ .

So the partial fraction decomposition is

$\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}$

7 0

3 years ago