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Volgvan
3 years ago
6

Which of the following equations have no solution? Check all that apply.

Mathematics
2 answers:
stealth61 [152]3 years ago
8 0
6x+1 = 3(2x-4)
-4(x+3) = 4(x-1)
Karolina [17]3 years ago
8 0
6x+1=3(2x-4) is no solution.
Reason:
6x+1=6x-12
-6x -6x
1≠-12

The answer is just the first option. Every equation has a solution.
The third option is an infinite solution. The first option is the only one with no solution.
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The graph shows a line and two similar triangles
ss7ja [257]
Your rise is 5 and your run is 3, your y-intercept is 0, therefore y=(5/3)x
6 0
3 years ago
G(x) = x to the power of 2 - 6 over 3x + 10 & g(4) = what over what
dalvyx [7]

Answer:

We have that:

g(x) = \frac{x^2 - 6}{3*x + 10}

And we want to find the value of g(4)

Then we are evaluating the function g(x) in x = 4, this means that we need to locate all the "x" in g(x), and replace them by 4.

If we do that, we get:

g(4) = \frac{4^2 - 6}{3*4 + 10}

Now we can just solve this to get:

g(4) = \frac{4^2 - 6}{3*4 + 10} = \frac{16-6}{12-10} = \frac{10}{2}  = 5

8 0
3 years ago
PLEASEEE HELPPPPPPP 15 POINTS
anygoal [31]
The equivalent to (1/3)^3 is

(1/3) x (1/3) x (1/3) = .<span>037
</span>
The answer is 1/27 or 0.37.
3 0
2 years ago
Adam wanted to play his video games. His mother told him that he needed to finish his homework, clean out the dishwasher, and cl
11Alexandr11 [23.1K]

Answer:

2 hours and 28 mins

Step-by-step explanation:

75-15=60=1 hour     remain 15 mins

45+14=63-3=60=1 hour      remain 3 mins

3+25=28 mins        remain 0 mins

total= 2hours and 28 mins

plz mark me brainly .

3 0
3 years ago
Q.6. The equation of the ellipse whose centre is at the origin and the x-axis, the major axis, which passes
azamat

<h3>Answer:</h3>

Equation of the ellipse = 3x² + 5y² = 32

<h3>Step-by-step explanation:</h3>

<h2>Given:</h2>

  • The centre of the ellipse is at the origin and the X axis is the major axis

  • It passes through the points (-3, 1) and (2, -2)

<h2>To Find:</h2>

  • The equation of the ellipse

<h2>Solution:</h2>

The equation of an ellipse is given by,

\sf \dfrac{x^2}{a^2} +\dfrac{y^2}{b^2} =1

Given that the ellipse passes through the point (-3, 1)

Hence,

\sf \dfrac{(-3)^2}{a^2} +\dfrac{1^2}{b^2} =1

Cross multiplying we get,

  • 9b² + a² = 1 ²× a²b²
  • a²b² = 9b² + a²

Multiply by 4 on both sides,

  • 4a²b² = 36b² + 4a²------(1)

Also by given the ellipse passes through the point (2, -2)

Substituting this,

\sf \dfrac{2^2}{a^2} +\dfrac{(-2)^2}{b^2} =1

Cross multiply,

  • 4b² + 4a² = 1 × a²b²
  • a²b² = 4b² + 4a²-------(2)

Subtracting equations 2 and 1,

  • 3a²b² = 32b²
  • 3a² = 32
  • a² = 32/3----(3)

Substituting in 2,

  • 32/3 × b² = 4b² + 4 × 32/3
  • 32/3 b² = 4b² + 128/3
  • 32/3 b² = (12b² + 128)/3
  • 32b² = 12b² + 128
  • 20b² = 128
  • b² = 128/20 = 32/5

Substituting the values in the equation for ellipse,

\sf \dfrac{x^2}{32/3} +\dfrac{y^2}{32/5} =1

\sf \dfrac{3x^2}{32} +\dfrac{5y^2}{32} =1

Multiplying whole equation by 32 we get,

3x² + 5y² = 32

<h3>Hence equation of the ellipse is 3x² + 5y² = 32</h3>
8 0
3 years ago
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