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kykrilka [37]
3 years ago
6

Which statement best illustrates using the vertical line test to determine if the graph below is a function of x ?

Mathematics
2 answers:
eimsori [14]3 years ago
4 0
If something is a function, every x-value, only has one (1) y-value. If the line x = 0 intersects the graph in two points, there are to values for y (instead of one). Therefore the graph is not a function of x.

Your answer is A.
-BARSIC- [3]3 years ago
4 0

Answer:

A

Step-by-step explanation:

i did the quiz

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HELPP PLEASE!!!! I have a screenshot of the question because I cant copy and paste it!
mr_godi [17]

2³+3²-3×4-5²÷5+(7×4)

= 8+9-3×4-25÷5+7×4

=8+9-12-5+28

=28

6 0
3 years ago
Read 2 more answers
7 1/2% of 60 please help i give brainliest
Fofino [41]

Answer:

4.5

Step-by-step explanation:

7 1/2 X 6 = 45.

Then you just add the decimal point to get 4.5.

7 0
3 years ago
Multiply. give your answer in standard form. (2n^2-7n+2)*(3n+1)
noname [10]
For this case we have the following polynomials:
 (2n ^ 2-7n + 2)
 (3n + 1)
 Multiplying we have:
 (2n ^ 2-7n + 2) * (3n + 1)
 (6n ^ 3-21n ^ 2 + 6n) + (2n ^ 2-7n + 2)
 Rewriting we have:
 6n ^ 3 + n ^ 2 (-21 + 2) + n (6-7) + 2
 6n ^ 3 - 19n ^ 2 - n + 2
 Answer:
 
6n ^ 3 - 19n ^ 2 - n + 2
6 0
3 years ago
If sin theta = 2/3 and sex theta < 0 , find cos theta and tan theta
FromTheMoon [43]

Answer:

\displaystyle cos\theta=-\frac{\sqrt{5}}{3}

\displaystyle tan\theta=-\frac{2\sqrt{5}}{5}

Step-by-step explanation:

<u>Trigonometric Formulas</u>

To solve this problem, we must recall some basic relations and concepts.

The main trigonometric identity relates the sine to the cosine:

sin^2\theta+cos^2\theta=1

The tangent can be found by

\displaystyle tan\theta=\frac{sin\theta}{cos\theta}

The cosine and the secant are related by

\displaystyle cos\theta=\frac{1}{sec\theta}

They both have the same sign.

The sine is positive in the first and second quadrants, the cosine is positive in the first and fourth quadrants.

The sine is negative in the third and fourth quadrants, the cosine is negative in the second and third quadrants.

We are given

\displaystyle sin\theta=\frac{2}{3}

Find the cosine by solving

sin^2\theta+cos^2\theta=1

\displaystyle \left(\frac{2}{3}\right)^2+cos^2\theta=1

\displaystyle cos^2\theta=1-\left(\frac{2}{3}\right)^2=1-\frac{4}{9}=\frac{5}{9}

\displaystyle cos\theta=\sqrt{\frac{5}{9}}=-\frac{\sqrt{5}}{3}

\boxed{\displaystyle cos\theta=-\frac{\sqrt{5}}{3}}

We have placed the negative sign because we know the secant ('sex') is negative and they both have the same sign.

Now compute the tangent

\displaystyle tan\theta=\frac{sin\theta}{cos\theta}=\frac{\frac{2}{3}}{-\frac{\sqrt{5}}{3}}=-\frac{2}{\sqrt{5}}

Rationalizing

\displaystyle tan\theta=-\frac{2}{\sqrt{5}}=-\frac{2\sqrt{5}}{5}

\boxed{\displaystyle tan\theta=-\frac{2\sqrt{5}}{5}}

5 0
4 years ago
On what interval is the function f(x)=x^3-4x^2+5x concave upward?
taurus [48]
Calculus 1?

To find concavity you must take the second derivative.  

As you would to find your local maximums and minimums (critical points) in the first derivative by setting y' = 0, to find points of inflection you set acceleration, y" = 0.  

Now that you know where the point in which the function is neither concave up or concave down (at the points of inflection) plug x-values between them into the second derivative for x.  If y" is positive between those particular points will be concave up and if y" is negative it will be concave down between that interval.

For a better understanding you might find a good video on Youtube explaining this if you search "Points of Inflections" or "Concavity of a function".

Cheers.
3 0
4 years ago
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