All categories

2 years ago

HELP!!! Is this a function or not a function Why or why not

Mathematics

1 answer:

defon2 years ago

6 0

Answer:

Yes, it is a function

Step-by-step explanation:

anytime you want to test to see if the graph is a function, perform a vertical line test.

Basically if you drew a vertical line and the vertical line crosses the graph at only one point that means the graph is a function.

If it crosses on more than one point the graph is not a function.

You might be interested in

Mia discovered that if she takes half her age and adds 3, it produces the same results as taking one-fourth of her age and addin

Vikentia [17]

See https://web2.0calc.com/questions/help_29833.

5 0

3 years ago

ALONDRA Help me please check my answer and help with the others that I haven't done

nadya68 [22]

1) is correct.
2) is correct.
3) is correct.
4) is correct.
5) is not there.
6) is your first choice.
7) is incorrect is your second choice.
8) is your first choice.
13) is your last choice.
14) is your last choice.
15) is your third choice.

5 0

4 years ago

Have students write an explanation of how to find the slope of a line given the points A(1, 5) and B(−7, 8), also give me the an

Grace [21]

Answer:

3/-8

Step-by-step explanation:

Slope= y2-y2/x2-x1

substitute in: 8-5/-7-1= 3/-8

4 0

3 years ago

Integrate the following

enot [183]

I suppose you mean to have the entire numerator under the square root?

$\displaystyle\int_2^4\frac{\sqrt{x^2-4}}{x^2}\,\mathrm dx$

We can use a trigonometric substitution to start:

$x=2\sec t\implies\mathrm dx=2\sec t\tan t\,\mathrm dt$

Then for $x=2$ , $t=\sec^{-1}1=0$ ; for $x=4$ , $t=\sec^{-1}2=\frac\pi3$ . So the integral is equivalent to

$\displaystyle\int_0^{\pi/3}\frac{\sqrt{(2\sec t)^2-4}}{(2\sec t)^2}(2\sec t\tan t)\,\mathrm dt=\int_0^{\pi/3}\frac{\tan^2t}{\sec t}\,\mathrm dt$

We can write

$\dfrac{\tan^2t}{\sec t}=\dfrac{\frac{\sin^2t}{\cos^2t}}{\frac1{\cos t}}=\dfrac{\sin^2t}{\cos t}=\dfrac{1-\cos^2t}{\cos t}=\sec t-\cos t$

so the integral becomes

$\displaystyle\int_0^{\pi/3}(\sec t-\cos t)\,\mathrm dt=\boxed{\ln(2+\sqrt3)-\frac{\sqrt3}2}$

7 0

3 years ago

Fifteen years ago, you deposited $12,500 into an investment fund. Five years ago, you added an

True [87]

Answer:

a) 8.16%

b) $65,762.50

c) $39,701.07

Step-by-step explanation:

Given:

15 years ago, Initial investment = $12500

5 years ago, Investment = $20000

Nominal interest = 8% semi annually for first 10 years

Interest2= 6.5% compunded annually for last five years

a) for the effective annual interest rate (EAR) in the first 10 years, let's use the formula:

[1+(nominal interest rate/number of compounding periods)]^ number of compounding periods-1

$EAR = [1 + (\frac{0.08}{2})]^2 - 1$

$= (1 + 0.04)^2 - 1$

$= (1.04)^2 - 1$

= 1.0816 - 1

= 0.0816

= 8.16%

The effective annual interesting rate, EAR = 8.16 %

b) To find the amount in my account today.

Let's first find the amount for $12500 for 10 years compounded semi annually

= 12,500 +( 12,500 * 8.160% * 10)

= $ 22,700

Let's also find the amount for $32,500($12,500+$20,000) for 5 Years compoundeed annually

$32,500 + ($32,500 * 6.5% *5)

= $ 43,062.50

Money in account today will be:

$22,700 + $43,062.50

= $65,762.50

c) Let's the amount I should have invested to be X

For first 10 years at 8.160%, we have:

Interest Amount = ( X * 8.160% * 10 ) = 0.8160 X

For next 5 years at 6.5%, we have:

Interest Amount = (X * 6.5% * 5) = 0.325 X

Therefore the total money at the end of 15 Years = 85000

0.8160X + 0.3250X + X = $85,000

= 2.141X = $85,000

X = 85,000/2.141

X = 39,701.074 ≈ $39,700

If I wish to have $85,000 now, I should have invested $39,700 15 years ago

7 0

3 years ago