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sammy [17]
2 years ago
9

URGENT GEOMETRY!!

Mathematics
1 answer:
Darina [25.2K]2 years ago
6 0

Answer: 140 degrees

A straight line is 180 degrees. Since we know one of the angles in the line is 40, we just subtract 40 from 180. So, the measure of angle 1 plus the measure of angle 2 is 140 degrees.

:)

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What is 5x + 3x equal to
Gekata [30.6K]
The answer is 8x.
You are combining like terms as 5 and 3 both have x so you are simply adding 5+3= 8
3 0
3 years ago
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If a cone has a volume of 338 cm^3 and a radius of 6 cm, how tall is it?
Anna35 [415]

The formula for the volume of a cone is V = πr²h ÷3

You are solving for the height, h.

V = 338 cm³ and r = 6 cm

338 = π(6)²h÷3

338 = π36h÷3

338 = π12h

28.16 = πh

8.965 = h

The height of the cone is around 9 cm.

6 0
3 years ago
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6 0
2 years ago
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Please answer quickly<br>expand and simplify: 4(5x-3y) (x-4y) -(3x-4y)(2x+3y)​
Gekata [30.6K]

Answer: 14x^2-93xy+60y^2 Hope that helps!

Step-by-step explanation:

1. Expand by distributing terms

(20x-12y)(x-4y)-(3x-4y)(2x+3y)

2. Use the Foil method:(a+b)(c+d)= ac+ad+bc+bd

20x^2-80xy-12yx+48y^2-(3x-4y)(2x+3y)

3. Use the Foil method : (a+b)(c+d)= ac+ad+bc+bd

20x^2-80xy-12yx+48y^2-(6x^2+9xy-8yx-12y^2)

4. Remove parentheses 20x^2-80xy-12yx+48y^2-6x^2-9xy+ 8yx+12y^2

5. Collect like terms (20x^2-6x^2)+(-80xy-12xy-9xy+8xy)+(48y^2+12y^2)

6. Simplify.

And your answer would be 14x^2-93xy+60y^2

5 0
3 years ago
The expected number of typographical errors on a page of a certain magazine is .2. What is the probability that an article of 10
Pavel [41]

Answer:

a) The probability that an article of 10 pages contains 0 typographical errors is 0.8187.

b) The probability that an article of 10 pages contains 2 or more typographical errors is 0.0175.

Step-by-step explanation:

Given : The expected number of typographical errors on a page of a certain magazine is 0.2.

To find : What is the probability that an article of 10 pages contains

(a) 0 and (b) 2 or more typographical errors?

Solution :

Applying Poisson distribution,

N\sim Pois(0.2)

P(N=r)=\frac{e^{-np}(np)^r}{r!}

where, n is the number of words in a page

and p is the probability of every word with typographical errors.

Here, n=10 and E(N)=np=0.2

a) The probability that an article of 10 pages contains 0 typographical errors.

Substitute r=0 in formula,

P(N=0)=\frac{e^{-0.2}(0.2)^0}{0!}

P(N=0)=\frac{e^{-0.2}}{1}

P(N=0)=e^{-0.2}

P(N=0)=0.8187

The probability that an article of 10 pages contains 0 typographical errors is 0.8187.

b) The probability that an article of 10 pages contains 2 or more typographical errors.

Substitute r\geq 2 in formula,

P(N\geq 2)=1-P(N

P(N\geq 2)=1-[P(N=0)+P(N=1)]

P(N\geq 2)=1-[\frac{e^{-0.2}(0.2)^0}{0!}+\frac{e^{-0.2}(0.2)^1}{1!}]

P(N\geq 2)=1-[e^{-0.2}+e^{-0.2}(0.2)]

P(N\geq 2)=1-[0.8187+0.1637]

P(N\geq 2)=1-0.9825

P(N\geq 2)=0.0175

The probability that an article of 10 pages contains 2 or more typographical errors is 0.0175.

6 0
3 years ago
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