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Alecsey [184]
3 years ago
9

Which ratio is NOT equivalent to the other choices?

Mathematics
1 answer:
Virty [35]3 years ago
3 0

Answer:

It's not possible to answer the question with the information provided, you will need to post the ratios in the problem.

Step-by-step explanation:

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the sum of 35 and one-fifth part of itself is added to the sum of one-seventh part of itselfcand three​
Alex777 [14]

Answer:

50

Step-by-step explanation:

35/5 = 7

35/7 = 5

35 + 7 + 5 + 3 = 50

5 0
4 years ago
Y=x-7<br> Y=-4x-2<br> Solve by substitution help please
wlad13 [49]

Answer:

(1, - 6 )

Step-by-step explanation:

Given the 2 equations

y = x - 7 → (1)

y = - 4x - 2 → (2)

Substitute y = x - 7 into (2)

x - 7 = - 4x - 2 ( add 4x to both sides )

5x - 7 = - 2 ( add 7 to both sides )

5x = 5 ( divide both sides by 5 )

x = 1

Substitute x = 1 into either of the 2 equations for corresponding value of y

Substituting x = 1 into (1)

y = 1 - 7 = - 6

Solution is (1, - 6 )

6 0
3 years ago
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Zanzabum
Wednesday and bottom one on the left
8 0
3 years ago
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For the matrices below. Find all (real) eigenvalues. <br><br> A= [7 9]<br> [0 9]
Jet001 [13]

Answer:

The answer is "9 and 7".

Step-by-step explanation:

Given:

A=\left[\begin{array}{cc}7&9\\0&9\end{array}\right]

Using formula:

|A-\lambda \cdot I|= 0\\\\

\to |\left[\begin{array}{cc}7&9\\0&9\end{array}\right]-\lambda \left[\begin{array}{cc}1&0\\0&1\end{array}\right] |=0\\\\\\ \to |\left[\begin{array}{cc}7&9\\0&9\end{array}\right]- \left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right] |=0\\\\\\\to|\left[\begin{array}{cc}7-\lambda &9\\0&9-\lambda\end{array}\right]|=0\\\\\\\to|(7-\lambda)(9-\lambda)|=0\\\\\to (7-\lambda)(9-\lambda)=0\\\\\to 7-\lambda=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 9-\lambda=0\\\\

\to \lambda=7 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda=9\\\\

8 0
3 years ago
I need help now please ​
enyata [817]

Answer:

the 3rd one

Step-by-step explanation:

5 0
3 years ago
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