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vaieri [72.5K]
3 years ago
8

By what number must you multiply a quantity in order to increase it by 25% in a single step?

Mathematics
1 answer:
Brrunno [24]3 years ago
3 0

Answer:

5/4

Step-by-step explanation:

Lets suppose a quantity A that has a value of x

A = x

25% of A is = x/4

now adding this 25% to get 25% increase in the original value

A' = x+ x/4

    A' = 5x/4

    A' = 5*A/4

This means that , in order to increase the value by 25%  in a single step we must multiply the the original value by 5/4.

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Find the slope of the line that is perpendicular to the joining line P (2,-3) and Q (-3,-2)
azamat

m =(y2-y1)/(x2-x1)

m =(-2 + 3 )/(-3 -2)

m = 1/-5

m = -1/5

slope of the line that is perpendicular  = 5


8 0
3 years ago
A large container holds 5 gallons of water. It begins leaking at a constant rate. After 10 minutes, the container has 3 gallons
Anna007 [38]
5 - 10 x = 3
10 x = 5 - 3
10 x = 2
x = 10 : 5
x = 0.2
The rate is : 0.2 gallons / minute
5 - t * 0.2 = 0
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t = 5 : 0.2
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Answer: After 25 minutes the container will be empty.
5 0
3 years ago
The whole sale cost of bicycle is $75. There is a mark up of 55%. What is the new sale price of the bike
blsea [12.9K]

Answer:

$116.25

Step-by-step explanation:

10% = $7.50

5% = $3.75

7.50 x 5 = 37.50

37.50 + 3.75 = $41.25

$75 + $41.25 = $116.25

Hope this helps !

8 0
3 years ago
Lines l and m are parallel.<br> Use the diagram to determine the measure of ∠3.
Pavlova-9 [17]

Answer:

∠3 is 70°

Step-by-step explanation:

∠1 is 50° since it is across from another 50° angle

and then you use the equation x+25+2x+50=180 to find x

x then equals 35°

you then use the alternate interior angle theorem to say that ∠3 is 2x, or 70°

5 0
3 years ago
P(x)= 3x^3-5x^2-14x-4
nexus9112 [7]
   
\displaystyle\\&#10;P(x)=3x^3-5x^2-14x-4\\\\&#10;D_{-4}=\{-4;~-2;~\underline{\bf -1};~1;~2;~4\}\\\\&#10;\text{We observe that } \frac{-1}{3} \text{ is a solution of the equation:}\\&#10;3x^3-5x^2-14x-4=0\\\\&#10;

\displaystyle\\&#10;\text{Verification}\\\\&#10;3x^3-5x^2-14x-4=\\\\&#10;=3\times\Big(-\frac{1}{3}\Big)^3-5\times\Big(-\frac{1}{3}\Big)^2-14\times\Big(-\frac{1}{3}\Big)-4=\\\\&#10;=-\frac{1}{9}-\frac{5}{9}+\frac{14}{3}-4=\\\\&#10;=-\frac{6}{9}+\frac{14}{3}-4=\\\\&#10;=-\frac{6}{9}+\frac{42}{9}- \frac{4\times 9}{9}=\\\\&#10; =-\frac{6}{9}+\frac{42}{9}- \frac{36}{9}= \frac{42-6-36}{9}=\frac{42-42}{9}=\frac{0}{9}=0\\\\&#10;\Longrightarrow~~~P(x)~\vdots~\Big(x+ \frac{1}{3}\Big)\\\\&#10;\Longrightarrow~~~P(x)~\vdots~(3x+1)


\displaystyle\\&#10;3x^3-5x^2-14x-4=0\\&#10;~~~~~-5x^2 = x^2 - 6x^2\\&#10;~~~~~-14x =-2x-12x \\&#10;3x^3+x^2 - 6x^2-2x-12x-4=0\\&#10;x^2(3x+1)-2x(3x+1) -4(3x+1)=0\\&#10;(3x+1)(x^2-2x -4)=0\\\\&#10;\text{Solve: } x^2-2x -4=0\\\\&#10;x_{12}= \frac{-b\pm  \sqrt{b^2-4ac}}{2a}=\\\\=\frac{2\pm  \sqrt{4+16}}{2}=\frac{2\pm  \sqrt{20}}{2}=\frac{2\pm  2\sqrt{5}}{2}=1\pm\sqrt{5}\\\\&#10;x_1 =1+\sqrt{5}\\&#10;x_2 =1-\sqrt{5}\\&#10;\Longrightarrow P(x)= 3x^3-5x^2-14x-4 =\boxed{(3x+1)(x-1-\sqrt{5})(x-1+\sqrt{5})}



7 0
3 years ago
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