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kotykmax [81]
3 years ago
9

Solve the system by using elimination 3x+3y=-9 3x-3y=21

Mathematics
2 answers:
LekaFEV [45]3 years ago
6 0

Answer:

\large\boxed{x=2,\ y=-5\to(2,\ -5)}

Step-by-step explanation:

\underline{+\left\{\begin{array}{ccc}3x+3y=-9\\3x-3y=21\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad\qquad6x=12\qquad\text{divide both sides by 6}\\.\qquad\qquad\boxed{x=2}\\\\\text{Put the value of x to the first equation}\\\\3(2)+3y=-9\\6+3y=-9\qquad\text{subtract 6 from both sides}\\3y=-15\qquad\text{divide both sides by 3}\\\boxed{y=-5}

e-lub [12.9K]3 years ago
3 0

Answer:

y=-5

x=2

Step-by-step explanation:

3x                  +                          3y=-9

3x                  -                          3y=21

-

0+6y=-30

6y=-30

y=-5

3x-15=-9

x=2

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A skateboard ramp is 40 feet long and rises from the ground at an angle of 33 degrees. How tall is the ramp?
olga_2 [115]

Answer:

The ramp is 21.8 ft tall.

Step-by-step explanation:

If you convert this into a triangle, 40 ft is the hypotenuse, 33° is the angle next to the right angle (on the horizontal line) and we need to find how tall the ramp is, which will be x. It is very helpful to draw a picture.

We will use the sine ratio because, starting from the given degree of 33, we have the hypotenuse value and need to find the value opposite the degree:

sine=\frac{opposite}{hypotenuse}

Insert values:

sin33=\frac{x}{40}

Multiply 40 to both sides to isolate the variable:

40(sin33)=40(\frac{x}{40})\\\\40*sin33=x\\\\x=40*sin33

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Round if necessary:

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Done.

6 0
3 years ago
20 POINTS!!
lbvjy [14]

Answer:

Step-by-step explanation:

2x-10y=10

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6 0
3 years ago
Read 2 more answers
Help I’m not sure with this one!! Urgent help
Nastasia [14]

This system has exactly one solution, because the two equation are linearly independent (i.e. one is not a multiple of the other).

We can explicitly find the solutions: from the first equation we deduce

x = \dfrac{c-by}{a}

Plug this value for x in the second equation:

\dfrac{a}{3}\cdot \dfrac{c-by}{a}+\dfrac{by}{2}=\dfrac{c}{6} \iff \dfrac{c-by}{3}+\dfrac{by}{2}=\dfrac{c}{6} \iff \dfrac{2(c-by)+3by}{6}=\dfrac{c}{6}

Multiply both sides by 6 to get

2(c-by)+3by=c \iff 2c-2by+3by=c \iff by=-c \iff y=-\dfrac{c}{b}

Plug this value for y in the expression for x:

x = \dfrac{c-by}{a} = \dfrac{c-b\cdot\frac{-c}{b}}{a}=\dfrac{2c}{a}

Since both a,b are not zero, both solutions are well defined.

8 0
3 years ago
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