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MatroZZZ [7]
2 years ago
15

How many solutions does this system of equations have?

Mathematics
1 answer:
Alja [10]2 years ago
5 0

Answer:

Infinitely many

Step-by-step explanation:

The second equation is basically multiplying the first equation by 2, so this means that they will have the same rate of change and initial value. Therefore, there are infinitely many solutions.

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Which function has a greater y-value if x=2<br> Blue<br> Red<br> S
-Dominant- [34]
Red has a greater value.
Blue = -2
Red = 3
8 0
3 years ago
What is the opposite of 3.75 on a number line?​
vodomira [7]

Answer:

-3.75

Step-by-step explanation:

the opposite changes the sign from positive to negative, so the opposite of 3.75 would be -3.75

5 0
3 years ago
Read 2 more answers
Please help me out :P
Veseljchak [2.6K]

cos θ = \frac{-4\sqrt{65} }{65}, sin θ = \frac{-7\sqrt{65} }{65}, cot  θ  = 4/7, sec  θ = \frac{-\sqrt{65} }{4}, cosec  θ  = \frac{-\sqrt{65} }{7}

<h3>What are trigonometric ratios?</h3>

Trigonometric Ratios are values of all the trigonometric functions based on the value of the ratio of sides in a right-angled triangle.

Sin θ: Opposite Side to θ/Hypotenuse

Tan θ: Opposite Side/Adjacent Side & Sin θ/Cos

Cos θ: Adjacent Side to θ/Hypotenuse

Sec θ: Hypotenuse/Adjacent Side & 1/cos θ

Analysis:

tan θ = opposite/adjacent = 7/4

opposite = 7, adjacent = 4.

we now look for the hypotenuse of the right angled triangle

hypotenuse = \sqrt{7^{2} + 4^{2} } = \sqrt{49+16} = \sqrt{65}

sin θ = opposite/ hyp = \frac{7}{\sqrt{65} }

Rationalize, \frac{7}{\sqrt{65} } x \frac{\sqrt{65} }{\sqrt{65} } = \frac{7\sqrt{65} }{65}

But θ is in the third quadrant(180 - 270) and in the third quadrant only tan and cot are positive others are negative.

Therefore, sin θ = - \frac{7\sqrt{65} }{65}

cos   θ  = adj/hyp = \frac{4}{\sqrt{65} }

By rationalizing and knowing that cos  θ  is negative, cos θ  = -\frac{-4\sqrt{65} }{65}

cot θ  = 1/tan θ  = 1/7/4 = 4/7

sec θ  = 1/cos θ  = 1/\frac{4}{\sqrt{65} } = -\frac{-\sqrt{65} }{4}

cosec θ  = 1/sin θ  = 1/\frac{\sqrt{65} }{7} = \frac{-\sqrt{65} }{7}

Learn more about trigonometric ratios: brainly.com/question/24349828

#SPJ1

5 0
1 year ago
Can you please explain this?​
Leni [432]

Answer:

The question is giving you pairs of points in space which can be used to define lines. It is then asking you to determine if the lines defined by those points are parallel, perpendicular, or neither.

Step-by-step explanation:

Two key things you need to know to solve this is that the lines will be parallel if their slopes are the same, and perpendicular if one slope is the negative reciprocal of the other (i.e. s_{1} = -s_{2}^{-1})

Let's start with question 11. You are given two pairs of points, each of which describes a distinct line:

(3,5)-(1,1) and (0, 2)-(5, 12)

To find the slope of each pair, take the vertex with the lesser x co-ordinate, and subtract it from the vertex with the greater x co-ordinate.  That will give us a valid Δx and Δy to get the slope.

In the first pair, 3 > 1, so we'll subtract the second point from the first:

s = \Delta y / \Delta x\\s = \frac{5 - 1}{3 - 1}\\s = 4/2\\s = 2

So the first pair of vectors describe a line with a slope of 2.  Let's look at the other pair:

s = \Delta y / \Delta x\\s = (12 - 2) / (5 - 0)\\s = 10 / 5\\s = 2

That also gives us a slope of 2, meaning that the two lines are parallel.

This same process will need to be done for the other three questions.  We can't answer questions 12 or 14 here, as the last point is cut off on the edge of the image.  For question 3 though, one line has a slope of 7/3, and the other 3/7. That puts them in the "neither" category, as one is not the negative reciprocal of the other, but instead the positive reciprocal.

6 0
3 years ago
Can someone help me find the annulus of the shaded area? much appreciated ​
Dimas [21]

Answer:

34.83

Step-by-step explanation:

7 0
2 years ago
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