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irina [24]
3 years ago
10

Help???????? Please??

Mathematics
1 answer:
densk [106]3 years ago
7 0
Hey there!
In order to see if it's a function or not, it can't have the same x value twice, or it just has to pass the vertical line test.
This means that you could put a vertical line anywhere and it would not pass between two points.
If we look, we could indeed try the vertical line test, and and there would be no intersection between two points.
Additionally, we also do not have any repeating x values on this quadratic function's parabola.
Hope this helps!
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A point's coordinates are changed from (–7, –2) to (–7, 2). What effect does this have on the location of the point
Inessa [10]
It will move it from one section to another and make the x axis 4 units more to the right
7 0
3 years ago
Solve the Law of Cosine: c^2 = a^2+ b^2 - 2abcosC for cos C.
Andrew [12]

Answer:

The Law of Cosine :  cos C = \frac{a^{2}+ b^{2}-c^{2}}{2ab}

Step-by-step explanation:

See the figure to understand the proof :

Let A Triangle ABC with sides a,b,c,

Draw a perpendicular on base AC of height H meet at point D

Divide base length b as AD = x -b   and    CD = x

By Pythagoras Theorem

In Triangle BDC             And     In Triangle BDA

a² = h² + x²     (  1  )                        c² = h² + (x-b)²

                                                      c² = h² + x² + b² - 2xb   ...(. 2)

From above eq 1 and 2

c² = (a² - x²) + x² + b² - 2xb

or, c² = a² + b² - 2xb                    .....(3)

Again in ΔBDC

cos C = \frac{BD}{BC}

Or, cos C = \frac{x}{a}

∴ x= a cos C

Now put ht value of x in eq 3

I.e, c² = a² + b² - 2ab cos C

Hence , cos C = \frac{a^{2}+ b^{2}-c^{2}}{2ab}      Proved   Answer

6 0
3 years ago
Can anyone tell me what the answer please
Tasya [4]

What you're looking at in this image are two alternate exterior angles.

Alternate exterior angles, are, essentially, exterior angles of a transversal that runs through two parallel lines. These angles are on alternating sides. These angles, assuming the lines are parallel, must be equal.

There's a proof behind why these angles are equal, but I won't bore you with the specifics as the question doesn't require it.

So - we know that these two angles must be equal to prove that the lines A and B are parallel. Knowing this, we can write an equation:

5x + 20 = 3x + 60

<u>Subtract 20 from both sides:</u>

5x + 20 - 20 = 3x + 60 - 20

5x = 3x + 40

<u>Subtract 3x from both sides:</u>

5x - 3x = 3x - 3x + 40

2x = 40

x = 20

If you have any questions on how I got to the answer, just ask!

- breezyツ

3 0
3 years ago
Show that the triangles are similar by comparing the ratios of the corresponding sides. Simplify your answer completely in order
kirill [66]

Answer/Step-by-step explanation:

AC = 1.2

AB = 4

BC = 2.6

DF = 3

DE = 10

EF = 6.5

Thus:

\frac{DE}{AB} = \frac{10}{4} = \frac{5}{2}

\frac{DF}{AC} = \frac{3}{1.2} = \frac{3*10}{1.2*10} = \frac{30}{12} = \frac{5}{2}

\frac{EF}{BC} = \frac{6.5}{2.6} = \frac{6.5*10}{2.6*10} = \frac{65}{26} = \frac{5}{2}

The ratio of their corresponding sides are all equal to ⁵/2. Therefore, both triangles are similar.

4 0
3 years ago
Please help what is the answer
Vika [28.1K]
I give a good guess of 42
6 0
3 years ago
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