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Verdich [7]
3 years ago
8

The hypotenuse of a right triangle is 28 units and the length of one of the legs is 14 units. What is the length of the other le

g in the simplest radical form?
Mathematics
1 answer:
timurjin [86]3 years ago
5 0

Answer:

The length of the other leg in the simplest radical form is 14\sqrt{3}

Step-by-step explanation:

Let's assume

hypotenuse of a right triangle is 'c'

one of the leg is 'a'

other leg is 'b'

we are given

The hypotenuse of a right triangle is 28 units

so, c=28

the length of one of the legs is 14 units

so,a=14

now, we can use Pythagoras theorem

c^2=a^2+b^2

now, we can plug values

28^2=14^2+b^2

now, we can solve for b

b^2=588

b=\sqrt{588}

b=14\sqrt{3}

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kozerog [31]

Answer:

It's 11.72

Step-by-step explanation:

Area=)1/2×15/4×25/4

Answer》11.72

Hope it helps...

Have a great day

3 0
3 years ago
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42 points!
kupik [55]

Answer:

C

Step-by-step explanation:

30x20=600

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3 years ago
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If u help me your dreams will come true ya know
wariber [46]

Answer:

Area of larger circle = \pi *r^2=3.14(11^2)=379.94

Area of smaller circle = \pi *r^2=3.14(3^2)=28.26

Probability that it lands in the shaded area (smaller circle):

28.26 ÷ 379.94 = 0.07438... = 0.07 (nearest hundredth)

8 0
3 years ago
Given: ∆PQR, m∠R = 90° m∠PQR = 75° M ∈ PR , MP = 18 m∠MQR = 60° Find: RQ
tensa zangetsu [6.8K]

Answer:    RQ= 8.99 ( approx)

Step-by-step explanation:

Let MR= x

Since, In triangle, PRQ, tan 75°= \frac{18+x}{RQ}

⇒ RQ=  \frac{18+x}{tan 75^{\circ}}

Now, In triangle MRQ,

tan 60°= \frac{18+x}{RQ}

⇒ RQ=  \frac{x}{tan 60^{\circ}}

On equating both values of RQ,

\frac{18+x}{tan 75^{\circ}}=\frac{x}{tan 60^{\circ}}

⇒\frac{18+x}{x}=\frac{tan 75^{\circ}}{tan 60^{\circ}}

⇒\frac{18+x}{x}=\frac{tan 75^{\circ}}{tan 60^{\circ}}

⇒\frac{18+x}{x}=2.15470053838

⇒18=2.15470053838x-x

⇒x=15.5884572681≈15.60

Thus RQ=8.99999999999≈8.99


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3 years ago
Please helppp Probability Maths
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Answer:

16/40 = 0.4

Step-by-step explanation:

Find total marbles then divide the number of green by total.

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