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3 years ago

Evaluate the limit using the appropriate properties of limits. (If an answer does not exist, enter DNE.)

Mathematics

1 answer:

Leviafan [203]3 years ago

8 0

Answer:

+infinity, or if there is no such choice, DNE.

Step-by-step explanation:

4x^2 − 3x^2 + x − 3

simplifies to

x^2+x-3

As x-> infinity,

both x^2 and x approach inifinity, therefore

the limit approaches +infinity, or does not exist (DNE)

Generally, it is more indicative to say +infinity or -infinitite, while reserving DNE (does not exist) for incoherent one-sided limits.

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Inessa [10]

Answer:

0.05, 0.5, 0.507, 0.57, 0.577

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Identify the polygon with vertices A(5,0), B(2,4), C(−2,1), and D(1,−3), and then find the perimeter and area of the polygon. HE

inn [45]

Answer:

Part 1) The polygon is a square

Part 2) The perimeter is equal to $20\ units$

Part 3) The area is equal to $25\ units^{2}$

Step-by-step explanation:

we have

$A(5,0), B(2,4), C(-2,1),D(1,-3)$

Plot the points

see the attached figure

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Find the distance AB

$A(5,0),B(2,4)$

substitute in the formula

$d=\sqrt{(4-0)^{2}+(2-5)^{2}}$

$d=\sqrt{(4)^{2}+(-3)^{2}}$

$d=\sqrt{25}$

$AB=5\ units$

Find the distance BC

$B(2,4), C(-2,1)$

substitute in the formula

$d=\sqrt{(1-4)^{2}+(-2-2)^{2}}$

$d=\sqrt{(-3)^{2}+(-4)^{2}}$

$d=\sqrt{25}$

$BC=5\ units$

Find the distance CD

$C(-2,1),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-1)^{2}+(1+2)^{2}}$

$d=\sqrt{(-4)^{2}+(3)^{2}}$

$d=\sqrt{25}$

$CD=5\ units$

Find the distance AD

$A(5,0),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-0)^{2}+(1-5)^{2}}$

$d=\sqrt{(-3)^{2}+(-4)^{2}}$

$d=\sqrt{25}$

$AD=5\ units$

we have that

AB=BC=CD=AD

Find the distance BD (diagonal)

$B(2,4),D(1,-3)$

substitute in the formula

$d=\sqrt{(-3-4)^{2}+(1-2)^{2}}$

$d=\sqrt{(-7)^{2}+(-1)^{2}}$

$BD=\sqrt{50}\ units$

Verify if the polygon is a square

If the triangle BDA is a right triangle, then the polygon is a square

Applying the Pythagoras theorem

$BD^{2}=AD^{2}+AB^{2}$

substitute

$(\sqrt{50})^{2}=5^{2}+5^{2}$

$50=50$ -----> is true

The triangle BDA is a right triangle

therefore

The polygon is a square

Find the Area of the polygon

The area of a square is equal to

$A=b^{2}$

we have

$b=5\ units$

$A=5^{2}=25\ units^{2}$

Find the perimeter of the polygon

The perimeter of a square is equal to

$P=4b$

we have

$b=5\ units$

$P=4(5)=20\ units$

6 0

3 years ago

A composite shape is shown.

kodGreya [7K]

Answer:

you lost me at composite

Step-by-step explanation:

Have a good day

3 0

3 years ago

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Please help me if you think you're hot beautiful skinny funny kind please help me

ehidna [41]

Answer:

Step-by-step explanation:

y= -5x/2+1

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3 years ago

Brandon picked 14 of a pound of cherries. That afternoon, Brandon's brother ate 12 of the cherries. How many pounds of cherries

stellarik [79]

Answer:

Step-by-step explanation:

Brandon picked 14 pounds of cherries but later in the afternoon Brandon's brother ate 12 so you would subtracts

14-12 = 2

8 0

3 years ago