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3 years ago

146536722753+63723623416

Mathematics

2 answers:

ch4aika [34]3 years ago

7 0

Answer:

what is this man.....

......what

OverLord2011 [107]3 years ago

6 0

Answer:

210,260,346,169

Step-by-step explanation:

I just used the calculator.

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What is the perimeter of this quadrilateral? (5,5) (2, 4) (4, 1) (6, 1)

lbvjy [14]

$~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{5}~,~\stackrel{y_1}{5})\qquad B(\stackrel{x_2}{2}~,~\stackrel{y_2}{4}) ~\hfill AB=\sqrt{[ 2- 5]^2 + [ 4- 5]^2} \\\\\\ AB=\sqrt{(-3)^2+(-1)^2}\implies \boxed{AB=\sqrt{10}} \\\\[-0.35em] ~\dotfill\\\\ B(\stackrel{x_1}{2}~,~\stackrel{y_1}{4})\qquad C(\stackrel{x_2}{4}~,~\stackrel{y_2}{1}) ~\hfill BC=\sqrt{[ 4- 2]^2 + [ 1- 4]^2}$

$BC=\sqrt{2^2+(-3)^2}\implies \boxed{BC=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ C(\stackrel{x_1}{4}~,~\stackrel{y_1}{1})\qquad D(\stackrel{x_2}{6}~,~\stackrel{y_2}{1}) ~\hfill CD=\sqrt{[ 6- 4]^2 + [ 1- 1]^2} \\\\\\ CD=\sqrt{2^2+0^2}\implies \boxed{CD=2} \\\\[-0.35em] ~\dotfill\\\\ D(\stackrel{x_1}{6}~,~\stackrel{y_1}{1})\qquad A(\stackrel{x_2}{5}~,~\stackrel{y_2}{5}) ~\hfill DA=\sqrt{[ 5- 6]^2 + [ 5- 1]^2}$

$DA=\sqrt{(-1)^2+4^2}\implies \boxed{DA=\sqrt{17}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Perimeter}}{\sqrt{10}~~ + ~~\sqrt{13}~~ + ~~2~~ + ~~\sqrt{17}}~~ \approx ~~ 12.89$

8 0

3 years ago

The length of time a full length movie runs from opening to credits is normally distributed with a mean of 1.9 hours and standar

Margarita [4]

Answer:

0.26
0.91
1.43

Step-by-step explanation:

given data

mean = 1.9 hours

standard deviation = 0.3 hours

solution

we get here first random movie between 1.8 and 2.0 hours

so here

P(1.8 < z < 2 )

z = (1.8 - 1.9) ÷ 0.3

z = -0.33

and

z = (2.0 - 1.9) ÷ 0.3

z = 0.33

z = 0.6293

P(-0.333 < z < 0.333 )

= 0.26

so random movie is between 1.8 and 2.0 hours long is 0.26

and

A movie is longer than 2.3 hours.

P(x > 2.3)

P( $\frac{x-\mu }{\sigma}$ > $\frac{2.3-\mu }{\sigma}$ )

P (z > $\frac{2.3-1.9 }{0.3}$ )

P (z > 1.333 )

= 0.091

so chance a movie is longer than 2.3 hours is 0.091

and

length of movie that is shorter than 94% of the movies is

P(x > a ) = 0.94

P(x < a ) = 0.06

P( $\frac{x-\mu }{\sigma }$ < $\frac{a-\mu }{\sigma }$ )

$\frac{a-1.9 }{0.3 } = -1.55$

a = 1.43

so length of the movie that is shorter than 94% of the movies about 1.4 hours.

3 0

3 years ago

Use the formula A=bh , where A is the area, b is the base length, and h is the height of the parallelogram, to solve this proble

Anettt [7]

A=bh
1056=32×b
÷32 ÷32
33=h
The height is 33 inches.

8 0

3 years ago

Read 2 more answers

Using the functions f(x) = 2x + 7, g(x) = x2 - 4, and h(x) = 5x. Find all of the following. Make sure your answer is in simplest

Stolb23 [73]

Answer:

Given

f(x) = 2x+7

g(x) = x^2-4

h(x) = 5x

a. 4h(x)

= 4 * 5x

= 20x

b. f(x) - g(x)

f(x) - g(x) = 2x + 7 - (x^2 - 4)

= 2x+7-x^2+4

=-x^2+2x+7+4

=-x^2+2x+11

c. f(g(x)) = 2(g(x))+7

=2(x^2-4) +7

=2x^2-8+7

=2x^2-1

d. g(x)h(x) = (x^2-4)(5x)

= 5x^3 - 20x

e. g(x) / f(x) = x2 - 4/ 2x + 7

7 0

3 years ago

Is 1.353 and 0.84closer to 0,1,2, or 3?

pochemuha

Both are closer to 1

3 <5

8>5

7 0

3 years ago

Read 2 more answers