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Effectus [21]
3 years ago
15

Kendra has a math test every 12 school days and a vocabulary test every 9 school days. She had both a math test and a vocabulary

test today. The next time she will have both tests on the same day will be in school days.
Mathematics
2 answers:
vredina [299]3 years ago
7 0

Answer: So the answer is 36

Step-by-step explanation:

<h2><u><em>In order to find the next time Kendra will have both tests on the same day, find the least common multiple of each increment of days. </em></u></h2><h2><u><em> </em></u></h2><h2><u><em>Write some multiples of each number, and then find the smallest multiple that both numbers have in common. </em></u></h2><h2><u><em> </em></u></h2><h2><u><em> </em></u></h2><h2><u><em> </em></u></h2><h2><u><em>The smallest multiple that both numbers have in common is 36. </em></u></h2><h2><u><em> </em></u></h2><h2><u><em>Therefore, Kendra will have both tests on the same day again in 36 school days. </em></u></h2><h2><u><em> </em></u></h2>

faust18 [17]3 years ago
4 0
36 school days. You need to find the LMC of both 12 and 9 which would be 36.
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Step-by-step explanation:

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There is 3/4 of a box of cereal remaining. You eat 2/5 of the remaining cereal. What fraction of the box do you eat? Hint: What
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3 years ago
Edge- a line segment where two __ meet
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Answer:\boxed{\text{verticles}}

Edge - a line segment where two <u>verticals</u> meet.

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3 0
3 years ago
How you divide 288 by 16
Ratling [72]

Answer:

18

Step-by-step explanation:

16 |‾‾‾288‾‾‾

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8 0
3 years ago
Read 2 more answers
F(x) = x^2+x-2/x^2-3x-4
Andrej [43]

i. Domain and Range

The given function is

f(x)=\frac{x^2+x-2}{x^2-3x-4}


The domain of this function is,

x^2-3x-4\ne 0

(x-4)(x+1)\ne 0

x\ne4,xne -1


The range refers to the y-values for which x is defined. x  is defined for all values of y.

The range is all real numbers. See graph

ii. x-and-y-intercept

For x- intercept intercept we put f(x)=0

This implies that;

\frac{x^2+x-2}{x^2-3x-4}=0


This will give us

x^2+x-2=0

\Rightarrow x^2+x-2=0


\Rightarrow x^2+2x--x-2=0

\Rightarrow x(x+2)-1(x+2)=0

\Rightarrow (x+2)(x-1)=0


\Rightarrow (x+2)=0,(x-1)=0

\Rightarrow x=-2,x=1

The x-intercepts are (-2,0),(1,0)


For y-intercept, we put

x=0 to obtain;

f(0)=\frac{0^2+0-2}{0^2-3(0)-4}

f(0)=\frac{1}{2}

The y-intercept is

(0,\frac{1}{2})

iii. Horizontal asyptote

Since degree of the numerator and the denominator are the same, there is a horizontal asymptote

To find the horizontal asymptote.


We divide the leading coefficient of the numerator by the leading  coefficient of the denominator.


The horizontal asymptote is y=\frac{1}{1}=1

iv. Vertical asymptote

To find the vertical asymptote, we equate the denominator to zero to get;

x^2-3x-4=0


This implies that;

x^2+x-4x-4=0

Split the middle term

x(x+1)-4(x+1)=0

Factor

(x+1)(x-4)=0

Factor further

(x+1)=0,(x-4)=0

x=-1,x=4


The vertical asymptotes are x=-1,x=4




8 0
3 years ago
Read 2 more answers
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