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Mkey [24]
4 years ago
9

Solve the system. I need help solving these problems. but I have to show my work

Mathematics
1 answer:
Yuki888 [10]4 years ago
4 0
3x - 5y = 7
2x + 3y = 11

multiply the first equation by 2 and the second by -3:-

6x - 10y = 14
-6x - 9y = -33   Adding these 2 equations we eliminate the term in x:-
-19y =  -19
y = 1
Substtute  y = 1 in the first equation:_
3x - 5(1) = 7
3x = 12
x = 4

Answer is x=4,y=1  Answer

Second system of equation:-s

Here one equation is linear and the other is quadratic  so we substitute fro y in the quadratic:-
From second equation y = 10x - 38, so substituting:- 
10x - 38  = x^2 - 4x - 17

Gotta go right now  .  Sorry




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JulijaS [17]
Y=17 If it's just 1037/61
7 0
3 years ago
What is the answer to 4x<12
Korolek [52]
All you'd do is divide by 4 on both sides,

4/4 = 1
12/4 = 3

Your Answer:
x < 3

5 0
3 years ago
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can someone please help me answer this thank you, will give brainliest if i can just please no spam answers, thank you!
Gelneren [198K]

Answer:

11

Step-by-step explanation:

w=11

I hope this helps

(15-4) = 11

5 0
3 years ago
What is the equation of the sinusoid shown in the graph?
ololo11 [35]

Answer:

<em>Answer: C</em>

Step-by-step explanation:

<u>The Cosine Function</u>

The graph of a cosine function is a sinusoid that starts at its maximum value of 1 at x=0 and takes x=2π radians to complete a full cycle. The function of the parent cosine function is:

y=\cos x

Both the amplitude A and the angular frequency w of a cosine can be modeled by the function

y=A\cos(\omega x)

The graph of the cosine function shown in the figure has an amplitude of A=3 and it completes a full cycle at x=π/2, thus:

\frac{\pi}{2}\omega =2\pi

Thus:

\omega = 4

Therefore, the equation of the sinusoid is:

y=3\cos (4x)

Answer: C

3 0
3 years ago
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Is (6,1)(8,-3)(6,7) a function
pochemuha

Answer:

No It's not

Step-by-step explanation:

Since

x=6 produces y=1 and y=7, the relation (6,1),(8,−3),(6,7) is not a function.

The relation is not a function.

6 0
3 years ago
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