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2 years ago

I need help with problem 14

Mathematics

2 answers:

aleksandrvk [35]2 years ago

5 0

Answer:

x = 50

Step-by-step explanation:

The sum of the interior angles of a quadrilateral = 360°

sum the 4 angles and equate to 360

2x + 15 + 3x - 20 + x + x + 15 = 360 , that is

7x + 10 = 360 ( subtract 10 from both sides )

7x = 350 ( divide both sides by 7 )

x = 50

never [62]2 years ago

4 0

Answer:

x = 50

Step-by-step explanation:

The sum of the interior angles of a quadrilateral = 360°

Therefore, add together the expressions for all 4 angles, equal to 360, and solve for x:

(2x + 15) + (3x - 20) + x + (x + 15) = 360

Gather like terms: 2x + 3x + x + x + 15 - 20 + 15 = 360

Combine like terms: 7x + 10 = 360

Subtract 10 from both sides: 7x = 350

Divide both sides by 7: x = 50

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g A 5 foot tall man walks at 10 ft/s toward a street light that is 20 ft above the ground. What is the rate of change of the len

vredina [299]

Answer:

$-\frac{10}{3}ft/s$

Step-by-step explanation:

We are given that

Height of man=5 foot

$\frac{dy}{dt}=-10ft/s$

Height of street light=20ft

We have to find the rate of change of the length of his shadow when he is 25 ft form the street light.

ABE and CDE are similar triangle because all right triangles are similar.

$\frac{20}{5}=\frac{x+y}{x}$

$4=\frac{x+y}{x}$

$4x=x+y$

$4x-x=y$

$3x=y$

$3\frac{dx}{dt}=\frac{dy}{dt}$

$\frac{dx}{dt}=\frac{1}{3}(-10)=-\frac{10}{3}ft/s=-\frac{10}{3}ft/s$

Hence, the rate of change of the length of his shadow when he is 25 ft from the street light= $-\frac{10}{3}ft/s$

7 0

2 years ago

A) 4

Shkiper50 [21]

Answer:

i think c

Step-by-step explanation:

8 0

3 years ago

The hockey team won 63 out of the 84 games it played this year.what percent of the games did the team win

Deffense [45]

63 is 75% of 84.
Percent of a Total

3 0

3 years ago

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Dmitry_Shevchenko [17]

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2 years ago

Suppose that a simulation with a six-sided number cube provided these results during testing.

lyudmila [28]

Theoretical probability:
1 ... (16 and 2/3) %
2 ... (16 and 2/3) %
3 ... (16 and 2/3) %
4 ... (16 and 2/3) %
5 ... (16 and 2/3) %
6 ... (16 and 2/3) %

Experimental results:
1 ... 18
2 ... 16
3 ... 16
4 ... 17
5 ... 16
6 ... 17

The total number of rolls in the experiment was

(18 + 16 + 16 + 17 + 16 + 17) = 100

so the expected frequency for each outcome was 16-2/3 times,
and the SIMULATION probabilities were

1 ... 18%
2 ... 16%
3 ... 16%
4 ... 17%
5 ... 16%
6 ... 17%

To me, this looks fantastically close. The cube
could hardly be more fair than it actually is.

8 0

3 years ago