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natali 33 [55]
3 years ago
9

Describe and correct the error made in solving the system of equations by elimination

Mathematics
1 answer:
lesya692 [45]3 years ago
6 0
The mistake was when she multiplied the 2nd equation by 5, she forgot to also multiply that last number 10 by 5 as well.

correction :
5x + 3y - z = 15
-x + 2y + 3z = 10...multiply by 5
---------------------
5x + 3y - z = 15
-5x + 10y + 15z = 50 (result of multiplying by 5)
--------------------add
13y + 14z = 65

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Please help!<br><br> Thank you!
Dafna1 [17]

Answer:

(-16,17)

Step-by-step explanation:

The answer for the and point D is negative 16, 17. how you will solve this is you would take them in point and the endpoint c. to find that x of D you will do negative 4 minus 8 this equals negative 12. what you will do next is you will take the negative 12 and added to -4. this is how you get negative 16. you will then find the y of D. to do this you will take negative 7 and subtract 5. this equals negative 12. you will then take the midpoint and do 5 + 12. this equals 17. Hope this helps!

6 0
2 years ago
Which of the following equations is not equivalent to 5x-10=40 *
vredina [299]
10x=100 because tbh if you Calculate 5x-10=40 it’s give you your answer and go off of that.
7 0
3 years ago
Perform the indicated operations. Write the answer in standard form, a+bi.<br> 5-3i / -2-9i
Vsevolod [243]

\huge \boxed{\mathfrak{Answer} \downarrow}

\large \bf\frac { 5 - 3 i } { - 2 - 9 i } \\

Multiply both numerator and denominator of \sf \frac{5-3i}{-2-9i} \\ by the complex conjugate of the denominator, -2+9i.

\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{\left(-2-9i\right)\left(-2+9i\right)})  \\

Multiplication can be transformed into difference of squares using the rule: \sf\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{\left(-2\right)^{2}-9^{2}i^{2}})  \\

By definition, i² is -1. Calculate the denominator.

\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{85})  \\

Multiply complex numbers 5-3i and -2+9i in the same way as you multiply binomials.

\large \bf \: Re(\frac{5\left(-2\right)+5\times \left(9i\right)-3i\left(-2\right)-3\times 9i^{2}}{85})  \\

Do the multiplications in \sf5\left(-2\right)+5\times \left(9i\right)-3i\left(-2\right)-3\times 9\left(-1\right).

\large \bf \: Re(\frac{-10+45i+6i+27}{85})  \\

Combine the real and imaginary parts in -10+45i+6i+27.

\large \bf \: Re(\frac{-10+27+\left(45+6\right)i}{85})  \\

Do the additions in \sf-10+27+\left(45+6\right)i.

\large \bf Re(\frac{17+51i}{85})  \\

Divide 17+51i by 85 to get \sf\frac{1}{5}+\frac{3}{5}i \\.

\large \bf \: Re(\frac{1}{5}+\frac{3}{5}i)  \\

The real part of \sf \frac{1}{5}+\frac{3}{5}i \\ is \sf \frac{1}{5} \\.

\large  \boxed{\bf\frac{1}{5} = 0.2} \\

3 0
2 years ago
Select the numbers that have a factor of 5. Mark all that apply.
faust18 [17]

Answer:

a number ending in 0 or 5 has a factor of 5

Step-by-step explanation:

5 is a factor of

5, 10, 15, 20, 25...

any number that ends in 0 or 5 would qualify

5 0
3 years ago
Answer quickly thank you so much!!!!!!!!!!!!!!!!!!!!!
marin [14]
The answer is C I think
3 0
2 years ago
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