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1 year ago

Use Special Right Triangles to solve for the value of y. Leave your answer in SIMPLIFIED radical form.

Mathematics

1 answer:

Scorpion4ik [409]1 year ago

7 0

We can use the sine function and tangent function in order to solve for x and y:

$\begin{gathered} sin(\theta)=\frac{opposite}{hypotenuse} \\ sin(60)=\frac{15}{x} \\ x=\frac{15}{sin(60)} \\ x=10\sqrt{3} \end{gathered}$ $\begin{gathered} tan(\theta)=\frac{opposite}{adjacent} \\ tan(60)=\frac{15}{y} \\ y=\frac{15}{tan(60)} \\ y=5\sqrt{3} \end{gathered}$

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Choose an appropriate metric unit of a typical house cat

valentina_108 [34]

Your answer is gram hope this helps that is what unit to use :)

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3 years ago

6/30=3/15 true or false

Dennis_Churaev [7]

Answer: True

Step-by-step explanation: This is because they are equivalent fractions

Hope this helps and have a nice day

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3 years ago

Read 2 more answers

Terry needs to carry a pole that

Natasha2012 [34]

Answer:

Yes, The pole will fit through the door because the diagonal width of the door is 10.8 feet, which is longer than the length of the pole.

Step-by-step explanation:

Using the Pythagorean Theorem, ( $a^2+b^2=c^2$ ) we can measure the hypotenuse of a right triangle. Since the doorway is a rectangle, and a rectangle cut diagonally is a right triangle, we can use Pythagorean Theorem to measure the diagonal width of the doorway.

Plug in the values of the length and width of the door for a and b. The c value will represent the diagonal width of the doorway:

$6^2+9^2=c^2$

$36+81=c^2$

$117=c^2$

Since 117 is equal to the value of c multiplied by c, we must find the square root of 117 to find the value of c.

$\sqrt{117} =10.8$

$10.8=c$

Yes, The pole will fit through the door because the diagonal width of the door is 10.8 feet, which is longer than the length of the pole, measuring 10 feet.

7 0

3 years ago

Graph the inequality.<br> 1

Mrrafil [7]

Answer:

hope it helps

Step-by-step explanation:

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3 years ago

nikita ran a 5-kilometer race in 39 minutes (0.65 of an hour) without training beforehand. in the first part of the race, her av

Nataliya [291]

Total distance 5 km; at 5km / 0.65 h =

Second part distance: x; at 6 km/h, during t2

First part distance: 5 - x; at 8.75 km/h, during t1

V = d/t ⇒ t = d/V

t2 = x/6
t1=[5-x]/8.75

t2 + t1 = 0.65

x/6 + [5-x]/8.75 = 0.65

x/6 + 5/8.75 - x/8.75 = 13/20

x/6 - x/8.75 = 13/20 - 5/8.75

x/6 - 4x/35 =13/20 - 20/35

35x - 24x = (35*6)(35*13 - 20*20)/(20*35)

11 x = 16.5
x = 16.5/11 = 1.5 km

5 0

3 years ago