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3 years ago

Prove: lim x^3 = 8. x approaches 2

1 answer:

5 0

When we compute the value of a limit, all that we do is to change the value of the parameter
<span>lim x^3 = (2)^3= 2x2x2=8
</span><span>x approaches 2</span>

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Area of a triangle with points at (-9,5), (6,10), and (2,-10)

Ann [662]

First we are going to draw the triangle using the given coordinates.
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Distance from point A to point B:
$d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}$
$d_{AB}= \sqrt{(6+9)^2+(10-5)^2}$
$d_{AB}= \sqrt{(15)^2+(5)^2}$
$d_{AB}= \sqrt{225+25}$
$d_{AB}= \sqrt{250}$
$d_{AB}=15.81$

Distance from point A to point C:
$d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}$
$d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}$
$d_{AC}= \sqrt{11^2+(-15)^2}$
$d_{AC}= \sqrt{121+225}$
$d_{AC}= \sqrt{346}$
$d_{AC}= 18.60$

Distance from point B from point C
$d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}$
$d_{BC}= \sqrt{(-4)^2+(-20)^2}$
$d_{BC}= \sqrt{16+400}$
$d_{BC}= \sqrt{416}$
$d_{BC}=20.40$

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
$s= \frac{AB+AC+BC}{2}$
$s= \frac{15.81+18.60+20.40}{2}$
$s= \frac{54.81}{2}$
$s=27.41$

Finally, to find the area of our triangle, we are going to use Heron's formula:
$A= \sqrt{s(s-AB)(s-AC)(s-BC)}$
$A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}$
$A= \sqrt{27.41(11.6)(8.81)(7.01)}$
$A=140.13$

We can conclude that the perimeter of our triangle is 140.13 square units.

3 0

3 years ago

tell whether the system has one solution, infinitely many solutions, or no solution. x = -7y + 24 x + 7y = 32

mylen [45]

This system has no solution.
x = -7y + 24 can be rewritten as x = 24 - 7y
to simplify x + 7y = 32, we have to subtract 7y from both sides
x = 32 - 7y
Now, our system is as follows:
x = 24 - 7y
x = 32 - 7y

7 0

3 years ago

Please help with #35 I don’t get it and I’m really stressed

skad [1K]

Don't let it get to ya. If they didn't think you're capable of answering this question, they wouldn't ask you. So they know your abilities better than you do.

We don't know the unknown number, so in order to discuss it, we have to give it some kind of a name. We could call it 'x', 'y', 'Ana, 'Q', or 'Ralph'. Let's call it 't'.

The unknown number . . . . . t

Five times the number . . . . . 5t

Nine less than five times the number . . . . . 5t - 9 . Hold on to this quantity.

Twice the number . . . . . 2t .

Three more than twice the number . . . . . 2t + 3. Hold on to this quantity too.

The question says that these two quantities are equal. So the equation is

5t - 9 = 2t + 3 . This is the equation in both choices 'c' and 'd' . I guess we'll have to solve it to find out which choice is the correct one.

The equation is . . . . . 5t - 9 = 2t + 3

Add 9 to each side . . . . . 5t = 2t + 12

Subtract 2t from each side . . . 3t = 12

Divide each side by 3 . . . . . t = 4 .

It looks like choice-d is the winner.

3 0

3 years ago

Read 2 more answers

Geometry math question no Guessing and Please show work thank you

vesna_86 [32]

First let us find the length of JN

JN =JK +KN

JN = 82+105 = 187

JN=187..............(1)

UC= JN -( JH +HU+CN)

We are given :

JH= 22, HU = 96 and CN = 51

plugging all the values we get

UC = 187-( 22+96+51)

UC =187 -169

UC = 18

Answer is UC =18 ( option B)

8 0

3 years ago

Look at problem 4. Will three soccer balls weigh more than 1 kilogram Explain

alex41 [277]

Answer:Yes

Step-by-step explanation:

So thats say each soccer ball weigh 450 grams and a kilogram is 1000 grams so we will do 450 times 3 which is 1350 and 1350 is more than 1000 grams which more than a killogram so the answer is yes and its by 350 more.

8 0

3 years ago