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2 years ago

What is the measure of 4?

Mathematics

1 answer:

zlopas [31]2 years ago

8 0

Answer:

∠ 4 = 80°

Step-by-step explanation:

∠ 5 and 100° are vertical angles and are congruent , so

∠ 5 = 100°

∠ 4 and ∠ 5 are same- side interior angles and sum to 180° , that is

∠ 4 + ∠ 5 = 180°

∠ 4 + 100° = 180° ( subtract 100° from both sides )

∠ 4 = 80°

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Express in simplest radical form.<br> V100

Elanso [62]

Answer:

I think it is 10.

Step-by-step explanation:

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2 years ago

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Help me this is due this period please please ill mark brainiest with a bunch of points.

Verdich [7]

Answer:

(-4,1)

Step-by-step explanation:

The answer is just where the two lines intersect

I'm rather tired so double check me on this but I'm getting about (-4,1) and it's asking for an approximation so think (-4,1) is fine

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2 years ago

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Three friends fill bags with trash at a neighborhood cleanup. Randall's bag weighs

BabaBlast [244]

Answer:

Randall picks up 68 pounds

Seth picks up 40 pounds

Joanna picks up 52 pounds

Step-by-step explanation:

Let us make an equation in x to find it

∵ Randall's bag weighs 3x - 7 pounds

∵ Joanna's bag weighs 2x + 2 pounds

∵ Seth's bag weighs 2x - 10 pounds

∵ Randall's and Joanna's bags weigh 3 times as much as Seth's bag

- That means add Randall's bag and Joanna's bag and equate

the sum by 3 times Seth's bag

∵ Randall's bag + Joanna's bag = 3(Seth's bag)

∴ (3x - 7) + (2x + 2) = 3(2x - 10)

- Simplify the two sides

∴ (3x + 2x) + (-7 + 2) = 3(2x) - 3(10)

∴ 5x + (-5) = 6x - 30

∴ 5x - 5 = 6x - 30

- Add 5 to both sides

∴ 5x = 6x - 25

- Subtract 6x from both sides

∴ -x = -25

- Divide both sides by -1

∴ x = 25

Substitute x by 25 in the weight of each bag

∵ Randall's bag weighs = 3(25) - 7 = 75 - 7

∴ Randall's bag weighs = 68 pounds

∵ Seth's bag weighs = 2(25) - 10 = 50 - 10

∴ Seth's bag weighs = 40 pounds

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7 0

3 years ago

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Please calculate this limit <br>please help me

Tasya [4]

Answer:

We want to find:

$\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n}$

Here we can use Stirling's approximation, which says that for large values of n, we get:

$n! = \sqrt{2*\pi*n} *(\frac{n}{e} )^n$

Because here we are taking the limit when n tends to infinity, we can use this approximation.

Then we get.

$\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n} = \lim_{n \to \infty} \frac{\sqrt[n]{\sqrt{2*\pi*n} *(\frac{n}{e} )^n} }{n} = \lim_{n \to \infty} \frac{n}{e*n} *\sqrt[2*n]{2*\pi*n}$

Now we can just simplify this, so we get:

$\lim_{n \to \infty} \frac{1}{e} *\sqrt[2*n]{2*\pi*n} \\$

And we can rewrite it as:

$\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n}$

The important part here is the exponent, as n tends to infinite, the exponent tends to zero.

Thus:

$\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n} = \frac{1}{e}*1 = \frac{1}{e}$

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3 years ago

A dog kennel charges $40 to board a dog for one night and $35 per night each night after that. Henry

Lostsunrise [7]

Answer: 6 nights

Step-by-step explanation:Step-by-step

Minus 40 from 215 then divide it by 35

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2 years ago