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Rainbow [258]
2 years ago
15

Cassie earned $40 in July. She earned $48 in August..

Mathematics
2 answers:
Dahasolnce [82]2 years ago
7 0

Answer: 20% increase

Step-by-step explanation:

miv72 [106K]2 years ago
7 0

Answer:

C. 20% increase

Step-by-step explanation:

if it was decreasing it would be less and 20% is even and seven is also even

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What is the correct decimal expansion for 2 1/11 ​?<br> A. 2.09<br> B. 2.57<br> C. 2.09<br> D. 2.11
hjlf

Your choices A and C are the same

but the decimal of 1/11 is 0.090909

so your answer would be 2.09

3 0
2 years ago
Find the solution set of the inequality 14- 3x &lt; -1.14
Brrunno [24]

Answer:

14-3x<-1.14

-3x<-1.14-14

-3x<-15.14

x<5.04

8 0
3 years ago
Read 2 more answers
Pls help me I’m stuck
Korvikt [17]
90 + 49 = 139

180 - 139 = 41

x = 41
3 0
3 years ago
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Solve for xxx. Your answer must be simplified. \dfrac x{-6}\geq-20 −6 x ​ ≥−20
Tems11 [23]

Answer:

x ≤ - 27

Step-by-step explanation:

\frac{x}{-9}\ge \:3\\\\\mathrm{Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)}\\\\\frac{x\left(-1\right)}{-9}\le \:3\left(-1\right)\\\\Simplify\\\\\frac{x}{9}\le \:-3\\\\\mathrm{Multiply\:both\:sides\:by\:}9\\\\\frac{9x}{9}\le \:9\left(-3\right)\\\\\mathrm{Simplify}\\\\x\le \:-27

7 0
3 years ago
Consider the following set of vectors. v1 = 0 0 0 1 , v2 = 0 0 3 1 , v3 = 0 4 3 1 , v4 = 8 4 3 1 Let v1, v2, v3, and v4 be (colu
Alona [7]

Answer:

We have the equation

c_1\left[\begin{array}{c}0\\0\\0\\1\end{array}\right] +c_2\left[\begin{array}{c}0\\0\\3\\1\end{array}\right] +c_3\left[\begin{array}{c}0\\4\\3\\1\end{array}\right] +c_4\left[\begin{array}{c}8\\4\\3\\1\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]

Then, the augmented matrix of the system is

\left[\begin{array}{cccc}0&0&0&8\\0&0&4&4\\0&3&3&3\\1&1&1&1\end{array}\right]

We exchange rows 1 and 4 and rows 2 and 3 and obtain the matrix:

\left[\begin{array}{cccc}1&1&1&1\\0&3&3&3\\0&0&4&4\\0&0&0&8\end{array}\right]

This matrix is in echelon form. Then, now we apply backward substitution:

1.

8c_4=0\\c_4=0

2.

4c_3+4c_4=0\\4c_3+4*0=0\\c_3=0

3.

3c_2+3c_3+3c_4=0\\3c_2+3*0+3*0=0\\c_2=0

4.

c_1+c_2+c_3+c_4=0\\c_1+0+0+0=0\\c_1=0

Then the system has unique solution that is (c_1,c_2c_3,c_4)=(0,0,0,0) and this imply that the vectors v_1,v_2,v_3,v_4 are linear independent.

4 0
3 years ago
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