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pantera1 [17]
3 years ago
14

Jakob is asking to simplify the expression -3a + 4b + 5a + (-7b).

Mathematics
2 answers:
777dan777 [17]3 years ago
6 0

Answer:B

Step-by-step explanation:

garik1379 [7]3 years ago
3 0

Answer:

B. Commutative

Step-by-step explanation:

He can do that because it's a sum

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A square with vertices (5, 0) and (0, -5) with a side length of 5 units​
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whats the question that is a statement

Step-by-step explanation:

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408 rounded to the nearest tenth
andreev551 [17]
408 round to the nearest tenth: 410
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Solve the following <br> 2(x-2)=8
Talja [164]

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2

Step-by-step explanation:

2 (x-2)=8 equal to 2x-4=8, put -4 to the other side by subtracting 4 on both sides once you do you get 2x=4 so 4 divided by 2 equals 2.

7 0
3 years ago
Read 2 more answers
Show that the line 4y = 5x-10 is perpendicular to the line 5y + 4x = 35 ​
Shkiper50 [21]

Step-by-step explanation:

<h2><em><u>concept :</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.Solution:</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.Solution:Given equations of lines are</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.Solution:Given equations of lines are4y = 5x-10</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.Solution:Given equations of lines are4y = 5x-10or, y = (5/4)x(5/2).</u></em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>(</em><em>1</em><em>)</em></h2><h2 /><h2><em><u>5y + 4x = 35</u></em></h2><h2 /><h2><em><u>5y + 4x = 35ory = (-4/5)x + 7.</u></em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>(</em><em>2</em><em>)</em></h2><h2 /><h2><em><u>Let m and n be the slope of equations i and ii, respectively.</u></em></h2><h2 /><h2><em><u>Let m and n be the slope of equations i and ii, respectively.Here, m = 5/4</u></em></h2><h2 /><h2><em><u>Let m and n be the slope of equations i and ii, respectively.Here, m = 5/4n= -4/5</u></em></h2><h2 /><h2><em><u>Let m and n be the slope of equations i and ii, respectively.Here, m = 5/4n= -4/5therefore, mx n = -1</u></em></h2><h2 /><h2><em><u>Let m and n be the slope of equations i and ii, respectively.Here, m = 5/4n= -4/5therefore, mx n = -1Hence, the lines are perpendicular.</u></em></h2>
8 0
3 years ago
Solve the system of equations by transforming a matrix representing the system of equation into reduced row echelon form.
Gekata [30.6K]

Take the augmented matrix,

\left[\begin{array}{ccc|c}2&1&-3&-20\\1&2&1&-3\\1&-1&5&19\end{array}\right]

Swap the row 1 and row 2:

\left[\begin{array}{ccc|c}1&2&1&-3\\2&1&-3&-20\\1&-1&5&19\end{array}\right]

Add -2(row 1) to row 2, and -1(row 1) to row 3:

\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&-5&-14\\0&-3&4&22\end{array}\right]

Add -1(row 2) to row 3:

\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&-5&-14\\0&0&9&36\end{array}\right]

Multiply through row 3 by 1/9:

\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&-5&-14\\0&0&1&4\end{array}\right]

Add 5(row 3) to row 2:

\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&0&6\\0&0&1&4\end{array}\right]

Multiply through row 2 by -1/3:

\left[\begin{array}{ccc|c}1&2&1&-3\\0&1&0&-2\\0&0&1&4\end{array}\right]

Add -2(row 2) and -1(row 3) to row 1:

\left[\begin{array}{ccc|c}1&0&0&-3\\0&1&0&-2\\0&0&1&4\end{array}\right]

So we have \boxed{x=-3,y=-2,z=4}.

3 0
3 years ago
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