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raketka [301]
3 years ago
12

5x-2y=-13 2x+y=11 Fine the solution using the substitution method.

Mathematics
1 answer:
dexar [7]3 years ago
3 0
Y= 11-2x
Substitute in:
5x-2(11-2x)=-13
5x-22+4x=-13(combine like-terms)
9x-22=-13
9x=-13+22
9x=9
x=1

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Solve the equation 5.4g + 4 = 1.4g +20.
katovenus [111]
A. g=4
b. first you're going to subtract 1.4g from both sides;
5.4g + 4 - 1.4g = 1.4g + 20 - 1.4g
then simplify it;
4g+4=20
then, you'll subtract 4 from both sides;
4g + 4 - 4 = 20 - 4
then simplify again;
4g=16
lastly, you will divide 4 from both sides;
4g/4 = 16/4
since 16/4=4, then g=4
5 0
3 years ago
What is the value of |-6|-|6| -(-6)
bixtya [17]

Answer:

6

Step-by-step explanation:

Solve the absolutes: |-6| becomes 6

and |6| doesn't change and still is 6

Substitute: 6-6-(-6)

- * - = +

6-6+6 = 6

7 0
3 years ago
Mr. Neiland bought 12 tickets to a chili supper and spent a total of $60. He bought a combination of adult tickets for $10 each
Drupady [299]

Answer:

434.60

Step-by-step explanation:

7 0
3 years ago
What is the slope of a line that is perpendicular to the graph of y=-3x?
lutik1710 [3]
<span>y=-3x has slope = -3
</span><span>
perpendicular lines, slope is opposite and reciprocal
so slope of </span>perpendicular  = 1/3

answer
1/3
5 0
3 years ago
Evaluate the line integral, where C is the given curve. (x + 6y) dx + x2 dy, C C consists of line segments from (0, 0) to (6, 1)
Dima020 [189]

Split C into two component segments, C_1 and C_2, parameterized by

\mathbf r_1(t)=(1-t)(0,0)+t(6,1)=(6t,t)

\mathbf r_2(t)=(1-t)(6,1)+t(7,0)=(6+t,1-t)

respectively, with 0\le t\le1, where \mathbf r_i(t)=(x(t),y(t)).

We have

\mathrm d\mathbf r_1=(6,1)\,\mathrm dt

\mathrm d\mathbf r_2=(1,-1)\,\mathrm dt

where \mathrm d\mathbf r_i=\left(\dfrac{\mathrm dx}{\mathrm dt},\dfrac{\mathrm dy}{\mathrm dt}\right)\,\mathrm dt

so the line integral becomes

\displaystyle\int_C(x+6y)\,\mathrm dx+x^2\,\mathrm dy=\left\{\int_{C_1}+\int_{C_2}\right\}(x+6y,x^2)\cdot(\mathrm dx,\mathrm dy)

=\displaystyle\int_0^1(6t+6t,(6t)^2)\cdot(6,1)\,\mathrm dt+\int_0^1((6+t)+6(1-t),(6+t)^2)\cdot(1,-1)\,\mathrm dt

=\displaystyle\int_0^1(35t^2+55t-24)\,\mathrm dt=\frac{91}6

6 0
3 years ago
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