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3 years ago

Find the solutions of the following equations.

Mathematics

1 answer:

Katyanochek1 [597]3 years ago

5 0

Answer:

a.) x = -1.5, 1.5

b.) x = 2

c.) x = 1.5

d.) x = -1, 1

e.) x = -0.943, 0.943

f.) x = -0.447, 0.447

g) x = -0.75

h.) x = -5

You can plug all these equations on Desmos

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It x, y, and z are positive intergers and 3x = 4y = 7z, then the least possible value of x + y + z is???

Dmitry [639]

$3x = 4y = 7z \\ \Rightarrow \begin{cases}3x - 4y = 0 \\4y - 7z = 0 \\ 7z - 3x = 0\end{cases} \\ \Leftrightarrow \begin{cases}x = \frac{4}{3} y\\4y - 7z = 0 \\ 7z - 3( \frac{4}{3}y) = 0\end{cases} \\ \Leftrightarrow \begin{cases}x = \frac{4}{3} y\\4y - 7z = 0 \\ 7z - 4y= 0\end{cases} \\ \Leftrightarrow \begin{cases}x = \frac{4}{3} y\\ y \\ z = \frac{4}{7}y \end{cases} \\ x,y,z\: are\: intergers\Rightarrow y=LCM(3,7)=21\Rightarrow x = 28,z = 12 \\ \Rightarrow x + y + z = 61$

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2 years ago

Max's grade on the first math exam was a 94. His grade on his seond math exam was an 89. What was the percent of change in Max's

Tom [10]

We are given that

Max's grade on the first math exam was a 94

so, first exam grade =94

His grade on his seond math exam was an 89

so, second exam grade =89

change in grade=(first exam grade)-(second exam grade)

change in grade=94-89

change in grade=5

Percentage = ( change in grade)/( first exam grade) *100

so, percentage is

$=\frac{5}{94}\times 100$

so,

the percent of change in Max's grade to the nearest whole percentage was 5%..........Answer

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3 years ago

Read 2 more answers

On a number line, the distance from zero to -7 is 7 units.

aivan3 [116]

Answer: Option A. |-7| = 7

Solution:

The distance is a positive value, then we must not take in account the sign, and using:

|a| =-a if a<0

with a=-7

|-7| = -(-7)

|-7| = 7

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3 years ago

_____ a sequence in which a fixed amount is added on to get the next term

Mariana [72]

Answer: Arithmetic

Explanation:
An Arithmetic Sequence is when a fixed amount is getting added on to the next term, which basically means it is adding a number at a constant rate.

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3 years ago

Find the vertical and horizontal asympotote for (×-2) (×+3)/ (2×+2)

nekit [7.7K]

$\bf \cfrac{(x-2)(x+3)}{2x+2}\implies \cfrac{x^2+x-6}{2x+2}~~ \begin{array}{llll} \leftarrow \textit{2nd degree polynomial}\\ \leftarrow \textit{1st degree polynomial} \end{array} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{vertical asymptote}}{2x+2=0}\implies 2x=-2\implies x=-\cfrac{2}{2}\implies x=-1$

when the degree of the numerator is greater than the denominator's, then it has no horizontal asymptotes.

quick note:

when the degree of the numerator is 1 higher than the degree of the denominator, then it has an slant-asymptote, so this one has a slant-asymptote.

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2 years ago