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ira [324]
3 years ago
13

an image of a right triangle is shown with an angle labeled y If sin y° = 7 divided by q and tan y� = 7 divided by r, what is th

e value of cos y°?
Mathematics
1 answer:
trapecia [35]3 years ago
3 0

Answer:

cos y = r/q

Step-by-step explanation:

Sine is opposite divided by hypotenuse. Here, since sin y = 7/q, we can say that the opposite side length is 7 and the hypotenuse side length is q.

Tangent is opposite divided by adjacent. Here, since tan y = 7/r, we can say that the opposite side is 7 (just like above with sine) and the adjacent side is r.

Now, we have a triangle with its two legs labelled 7 and r and its hypotenuse labelled q.

Cosine is adjacent over hypotenuse. Because the adjacent side has a length of r and the hypotenuse has length q, we know that cos y must be r/q.

Hope this helps!

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Identify the function g(x) that is a vertical stretch by a factor of 6, a shift right 3 units, and
tia_tia [17]
Well the actual answer is 6x=6788. Na I’m just kidding juss needa stack up these points✨
8 0
3 years ago
The perimeter of a rectangle is 34 units. Its width is 6.5, point
Sliva [168]

Answer:

Length = 10.5units,Area = 68.25 unit²

Step-by-step explanation:

Perimeter =34 units

Width =6.5 units

Perimeter = l+l+w+w

Where l= length and w= width

34 = l + l + 6.5+ 6.5

34.= 2l + 13

Subtract 13 from both sides

2l = 34 - 13

2l = 21

Divide both sides by 2

L= 21/2

Length = 10.5units

If we are to find the area.

Area = length x width

Area = 10.5 × 6.5

Area = 68.25 unit²

I hope this was helpful, please mark as brainliest

6 0
3 years ago
In sunlight, a vertical yardstick casts a 1 ft shadow at the same time that a nearby tree casts a 15 ft shadow. How tall is the
kati45 [8]

Answer: A) 45 ft

Step-by-step

The yardstick is 3ft tall since 1 yard = 3ft. It is 3 times as tall as its shadow. The tree has a 15ft long shadow. The tree should also be 3 times as tall as its shadow. The tree is 15*3ft tall, so it is 45ft tall. I attached an image of the diagram I made.

4 0
3 years ago
Pls show full working out
sweet-ann [11.9K]

Answer:

  • Question 1a. i) x=1.8m

  • Question 1a. ii)   Volume=27.6m^3

  • Question 1b) Volume=65.9m^3

Explanation:

<u><em>Question 1 a. i) Find the value of x.</em></u>

         tan(\theta )=\dfrac{opposite\text{ }leg}{adjacent\text{ }leg}

For the smalll triangle you can write:

        tan(\theta )=\dfrac{x}{1m}

For tthe big triangle:

      tan(\theta )=\dfrac{x+2.7m}{2.5m}

Substitute:

        \dfrac{x}{1m}=\dfrac{x+2.7m}{2.5m}

Solve for x:

        2.5x=x+2.7m\\\\2.5x-x=2.7m\\\\1.5x=2.7m\\\\x=2.7m/1.5\\\\x=1.8m

<u><em>Question 1a ii) Find the volume of the frustrum</em></u>

  • Find the volume of a cone with height = 2.7m + 1.8m = 4.5m, and radius = 2.5m

Formula:

         Volume=(1/3)\pi \times radius^2\times height

Substitute:

         Volume=(1/3)\pi \times (2.5m)^2\times 4.5m=9.375\pi m^3

  • Find the volume of a cone with heigth = 1.8m and radius = 1m

        Volume=(1/3)\pi \times (1m)^2\times 1.8m=0.6\pi m^3

  • Subtract the volume of the small cone from the volume of the big cone

        Volume\text{ }of\text{ }frustrum=9.375\pi m^3-0.6\pi m^3=8.775\pi m^3\approx 27.6m^3

<u><em>Question 1b. Calculate the volume of the bin</em></u>

<u>i) Upper frustrum</u>

This is the same frustrum from the equation of above, thus ist volume is 27.6m³.

<u>ii) Lower frustrum</u>

            \dfrac{x}{2.0m}=\dfrac{x+2.4m}{2.5m}

           2.5x=2(x+2.4m)\\\\2.5x=2x+4.8m\\\\0.5x=4.8m\\\\x=9.6m

        Volume=(1/3)\pi \times (2.5m)^2\times (9.6m+2.4m)-(2.0m)^2\times (9.6m)

       Volume=38.3m^3

<u>iii) Add the volume of the two frustrums</u>

  • Volume=27.6m^3+38.3m^3=65.9m^3

6 0
3 years ago
Point R is on line segment QS. Given QS= 5x-2, QR= 3x-6, and RS=4x-2, determine the numerical length of RS.
pashok25 [27]

Answer:

The numerical length of RS is 10 units

Step-by-step explanation:

* <em>Lets explain How to solve the problem</em>

- Point R is on line segment QS

- The length of QS is (5x - 2)

- The length of QR is (3x - 6)

- The length of RS is (4x - 2)

- <em>The length of QS is the sum of the lengths of QR and RS</em>

∵ QS = QR + RS

∴ (5x - 2) = (3x - 6) + (4x - 2)

- Simplify the right hand side by adding like terms

∴ 5x - 2 = (3x + 4x) + (-6 + -2)

∴ 5x - 2 = 7x + -8 ⇒ (+)(-) = (-)

∴ 5x - 2 = 7x - 8

- Add 8 for both sides

∴ 5x + 6 = 7x

- Subtract 5x from both sides

∴ 6 = 2x

- Divide both sides by 2

∴ x = 3

- <em>To find the length of RS substitute the value of x in its expression</em>

∵ RS = 4x - 2

∵ x = 3

∴ RS = 4(3) - 2 = 12 - 2 = 10

∴ The numerical length of RS is 10 units

8 0
4 years ago
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