All categories

3 years ago

Find the value of y in the diagram below.

Mathematics

1 answer:

Alja [10]3 years ago

5 0

Answer:

The answer to your question is: y = 80°

Step-by-step explanation:

The sum of the internal angles in a polygon is = 180°(n - 2)

= 180° (6 - 2)

= 180(4)

= 720°

y° + y° + 2(2y -20) + 2(2y -20) = 720

2y° + 4(2y -20) = 720

2y° + 8y -80 = 720

10y = 720 + 80

10 y = 800

y = 800 / 10

y = 80°

2y - 20 = 2(80) -20 = 160 -20 = 140

80° + 80° * 140 + 140 + 140 + 140 = 720°

You might be interested in

I keep getting D wrong

blondinia [14]

Yeah d is wrong

:((( don’t know what it is supposed to be.

3 0

2 years ago

The points plotted below satisfy a polynomial. In what ranges of x-values must there be a root of the graph? Check all that appl

lisov135 [29]

Answer:

not sure,but I think the answer is 3 to 4

6 0

3 years ago

Can you explain this question please

Rasek [7]

Answer:

Step-by-step explanation:

θ and B are corresponding angles

Each is in the same position ( upper right hand corner ) in its group of four angles.

3 0

2 years ago

Read 2 more answers

An object is dropped from a height of 1000 cm above the ground after t seconds is given by the formula below-

borishaifa [10]

Hello...im new lol whats up dude

8 0

3 years ago

At 2:00 PM a car's speedometer reads 30 mi/h. At 2:20 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:20 the acce

Bad White [126]

Answer:

Let v(t) be the velocity of the car t hours after 2:00 PM. Then $\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }$ . By the Mean Value Theorem, there is a number c such that $0 < c$ with $v'(c)=60 \:{\frac{mi}{h^2}}$ . Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly $60 \:{\frac{mi}{h^2}}$ .

Step-by-step explanation:

The Mean Value Theorem says,

Let be a function that satisfies the following hypotheses:

f is continuous on the closed interval [a, b].
f is differentiable on the open interval (a, b).

Then there is a number c in (a, b) such that

$f'(c)=\frac{f(b)-f(a)}{b-a}$

Note that the Mean Value Theorem doesn’t tell us what c is. It only tells us that there is at least one number c that will satisfy the conclusion of the theorem.

By assumption, the car’s speed is continuous and differentiable everywhere. This means we can apply the Mean Value Theorem.

Let v(t) be the velocity of the car t hours after 2:00 PM. Then $v(0 \:h) = 30 \:{\frac{mi}{h} }$ and $v( \frac{1}{3} \:h) = 50 \:{\frac{mi}{h} }$ (note that 20 minutes is $20/60=1/3$ of an hour), so the average rate of change of v on the interval $[0 \:h, \frac{1}{3} \:h]$ is

$\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }$

We know that acceleration is the derivative of speed. So, by the Mean Value Theorem, there is a time c in $(0 \:h, \frac{1}{3} \:h)$ at which $v'(c)=60 \:{\frac{mi}{h^2}}$ .

c is a time time between 2:00 and 2:20 at which the acceleration is $60 \:{\frac{mi}{h^2}}$ .

4 0

3 years ago