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2 years ago

When a system contains the equations of two parabolas, there can only be one solution.

Mathematics

2 answers:

Phoenix [80]2 years ago

6 0

It should be false it would come out with more then one solution

Dvinal [7]2 years ago

4 0

False, there can be 0, 1, or 2 solutions depending on how many points they intersect

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Anyone know the answer i don’t need the steps or the work just the answer. ty!

NeX [460]

Answer:

= x^4 y^5 / 3

Step-by-step explanation:

I got this answer = x^4 y^5 / 3

But my answer is not found directly

And I think they deliberately set up this arrangement to make us mess up

, but in your Options I think it's ( B )

7 0

2 years ago

In Woodinville the sales tax is 10%. What is the total cost including tax on a $28 purchase?

Brut [27]

Answer:

$30.80

Step-by-step explanation:

10% of $28 is 2.80

add 28+2.80=30.80

7 0

3 years ago

Solve for x or find x

padilas [110]

Answer:

x=20

Step-by-step explanation:

3 0

2 years ago

Evaluate the function at the given numbers( correct to six decimal places). Use the results to guess the value of the limit or e

netineya [11]

Answer:

Value of the limit is 0.5.

Step-by-step explanation:

Given,

$F(x)=\frac{e^x-1-x}{x^2}$

When,

$x=1,F(1)=frac{e^1-1-1}{1}=e-2=0.718281$

$x=0.5, F(0.5)=\frac{e^0.5-1-0.5}{(0.5)^2}=0.594885$

$x=0.1, F(0.1)=\frac{e^0.1-1-0.1}{(0.1)^2}=0.517091$

$x=0.05, F(0.05)=\frac{e^0.05-1-0.05}{(0.05)^2}=0.508438$

$x=0.01, F(0.01)=\frac{e^0.01-1-0.01}{(0.01)^2}=0.501670 \hfill (1)$

Correct upto six decimal places.

Now,

$\lim_{x\to 0}F(x)=\lim_{x\to 0}\frac{e^x-1-x}{x^2}$ $(\frac{0}{0})$ form, applying L-Hospital rule that is differentiating numerator and denominator we get,

$\lim_{x\to 0}F(x)$

$=\lim_{x\to 0}\frac{e^x-1}{2x}$ $(\frac{0}{0})$ form.

$=\lim_{x\to 0}\frac{e^x}{2}=\frac{1}{2}=0.5\hfill (2)$

Limit exist and is 0.5. That is according to (1) we can see as the value of x lesser than 1 and tending to near 0, value of the function decreases respectively. And from (2) it shows ultimately it decreases and reach at 0.5, consider as limit point of F(x).

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%20%5Cfrac%7B2%7D%7B15y%7D%20%20%2B%20%20%5Cfrac%7B3%7D%7B5y%20%7B%7D%5E%7B3%7D%20%7D" id="

Xelga [282]

Answer:

2y² + 9

---------------

15y³

Step-by-step explanation:

Start by identifying the LCD, and then change each fraction so that its denominator is the LCD.

Here the LCD is 15y³, which is evenly divisible by 15y and 5y³.

Focus now on the first fraction: 2 / (15y). Multiplying numerator and denominator of this fraction by y² results in

y²·2 2y²

--------- → ----------

y²·15y 15y³ ←This is the correct LCD

Multiplying numerator and denominator of the second fraction by 3 results in:

3·3 9

------------ → ---------

3·5y³ 15y³ ←This is the correct LCD

So now those two original terms look like:

2y² 9

--------- + --------

15y³ 15y³

and this can be written in simpler form as:

2y² + 9

---------------

15y³

3 0

3 years ago