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2 years ago

What is the value of y in the figure if x = 30°? y =

Mathematics

2 answers:

QveST [7]2 years ago

7 0

Answer:

y is 150

Step-by-step explanation:

a straight angle always measures 180 degrees. if x is 30, the simply minus 30 from 180. that should give u 150 which is your answer.

hope this helped!

kow [346]2 years ago

5 0

The answer is 150 because a straight angle always measures a 180 then you subtract the 30 from it then you get 150 for the Y.!

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PLEASE HELP !! ILL GIVE BRAINLIEST *EXTRA 40 POINTS* DONT SKIP :(( .!

Charra [1.4K]

Answer: no solution

Step-by-step explanation:

They have the same slope which means they are parallel lines that will never cross or intersect. This mean that there is no solution

7 0

3 years ago

Jerry has a miniature model of a boat. He knows that the boat is 3 3/4 inches wide and 5 1/2 inches long. What is the actual len

FromTheMoon [43]

Width of the miniature model = $3\frac{3}{4}$ inches

Length of the miniature model = $5\frac{1}{2}$ inches

The actual width of the boat = 15 feet

Let us find the actual length of the boat.

Write it in a proportion form.

width of model : length of model : : width of actual : length of actual

$$\Rightarrow 3\frac{3}{4}:5\frac{1}{2}::15:x$

$$\Rightarrow\frac{15}{4}:\frac{11}{2}::15:x$

$$\Rightarrow\frac{\frac{15}{4}}{\frac{11}{2}} =\frac{15}{x}$

Do cross multiplication.

$$\Rightarrow{\frac{15}{4}x =15\times\frac{11}{2}$

$$\Rightarrow{\frac{15}{4}x =\frac{165}{2}$

$$\Rightarrow x =\frac{165}{2}\times{\frac{4}{15}$

⇒ x = 22

Hence the actual length of the boat is 22 feet.

6 0

3 years ago

Messed up and didnt put the attachment on the last post

Margarita [4]

Answer :-

9(3+√3) feet

Step by step explanation :-

A triangle is given to us. In which one angle is 30° and length of one side is 18ft ( hypontenuse) .So here we can use trignometric Ratios to find values of rest sides. Let's lable the figure as ∆ABC .

Now here the other angle will be = (90°-30°)=60° .

In ∆ABC ,

=> sin 30 ° = AB / AC

=> 1/2 = AB / 18ft

=> AB = 18ft/2

=> AB = 9ft .

Again In ∆ ABC ,

=> cos 30° = BC / AC

=> √3/2 = BC / 18ft

=> BC = 18 * √3/2 ft

=> BC = 9√3 ft .

Hence the perimeter will be equal to the sum of all sides = ( 18 + 9 + 9√3 ) ft = 27 + 9√3 ft = 9(3+√3) ft .

<h3>Hence the perimeter of the triangular pathway shown is 9 ( 3 + √3 ) ft .</h3>

3 0

2 years ago

60 EXTRA POINTS!!!! I need help with writing an inequality for a graph. It is below. :)

ch4aika [34]

Answer:

The answer to your question is: y ≥ - x + 6 and y ≥ x - 4

Step-by-step explanation:

Look of a pairs of points for each graph

Line 1 Line 2

A(0, 6) B (5, 1) C (7,3) D(5, 1)

m = (1- 6) / (5- 0) m = (1 - 3) / (5 - 7)

m = -5/5 = -1 m = -2 / -2 = 1

Find the line equation for each slope (y - y1) = m (x - x1)

(y - 1) = -1 (x - 5) (y - 1) = 1 (x - 5)

y - 1 = -x + 5 y -1 = x - 5

y = -x + 5 + 1 y = x - 5 + 1

y = -x + 6 y = x - 4

Inequality

y ≥ - x + 6 y ≥ x - 4

8 0

3 years ago

This is a "water tank" calculus problem that I've been working on and I would really appreciate it if someone could look at my w

Sedaia [141]

Part A

Everything looks good but line 4. You need to put all of the "2h" in parenthesis so the teacher will know you are squaring all of 2h. As you have it right now, you are saying "only square the h, not the 2". Be careful as silly mistakes like this will often cost you points.

============================================================

Part B

It looks like you have the right answer. Though you'll need to use parenthesis to ensure that all of "75t/(2pi)" is under the cube root. I'm assuming you made a typo or forgot to put the parenthesis.

dh/dt = (25)/(2pi*h^2)
2pi*h^2*dh = 25*dt
int[ 2pi*h^2*dh ] = int[ 25*dt ] ... applying integral to both sides
(2/3)pi*h^3 = 25t + C
2pi*h^3 = 3(25t + C)
h^3 = (3(25t + C))/(2pi)
h^3 = (75t + 3C)/(2pi)
h^3 = (75t + C)/(2pi)
h = [ (75t + C)/(2pi) ]^(1/3)

Plug in the initial conditions. If the volume is V = 0 then the height is h = 0 at time t = 0
0 = [ (75(0) + C)/(2pi) ]^(1/3)
0 = [ (0 + C)/(2pi) ]^(1/3)
0 = [ (C)/(2pi) ]^(1/3)
0^3 = (C)/(2pi)
0 = C/(2pi)
C/(2pi) = 0
C = 0*2pi
C = 0

Therefore the h(t) function is...
h(t) = [ (75t + C)/(2pi) ]^(1/3)
h(t) = [ (75t + 0)/(2pi) ]^(1/3)
h(t) = [ (75t)/(2pi) ]^(1/3)

Answer:
h(t) = [ (75t)/(2pi) ]^(1/3)

============================================================

Part C

Your answer is correct.
Below is an alternative way to find the same answer

--------------------------------------

Plug in the given height; solve for t
h(t) = [ (75t)/(2pi) ]^(1/3)
8 = [ (75t)/(2pi) ]^(1/3)
8^3 = (75t)/(2pi)
512 = (75t)/(2pi)
(75t)/(2pi) = 512
75t = 512*2pi
75t = 1024pi
t = 1024pi/75
At this time value, the height of the water is 8 feet

Set up the radius r(t) function
r = 2*h
r = 2*h(t)
r = 2*[ (75t)/(2pi) ]^(1/3) .... using the answer from part B

Differentiate that r(t) function with respect to t
r = 2*[ (75t)/(2pi) ]^(1/3)
dr/dt = 2*(1/3)*[ (75t)/(2pi) ]^(1/3-1)*d/dt[(75t)/(2pi)]
dr/dt = (2/3)*[ (75t)/(2pi) ]^(-2/3)*(75/(2pi))
dr/dt = (2/3)*(75/(2pi))*[ (75t)/(2pi) ]^(-2/3)
dr/dt = (25/pi)*[ (75t)/(2pi) ]^(-2/3)

Plug in t = 1024pi/75 found earlier above
dr/dt = (25/pi)*[ (75t)/(2pi) ]^(-2/3)
dr/dt = (25/pi)*[ (75(1024pi/75))/(2pi) ]^(-2/3)
dr/dt = (25/pi)*[ (1024pi)/(2pi) ]^(-2/3)
dr/dt = (25/pi)*(1/64)
dr/dt = 25/(64pi)
getting the same answer as before

----------------------------

Thinking back as I finish up, your method is definitely shorter and more efficient. So I prefer your method, which is effectively this:
r = 2h, dr/dh = 2
dh/dt = (25)/(2pi*h^2) ... from part A
dr/dt = dr/dh*dh/dt ... chain rule
dr/dt = 2*((25)/(2pi*h^2))
dr/dt = ((25)/(pi*h^2))
dr/dt = ((25)/(pi*8^2)) ... plugging in h = 8
dr/dt = (25)/(64pi)
which is what you stated in your screenshot (though I added on the line dr/dt = dr/dh*dh/dt to show the chain rule in action)

8 0

3 years ago