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3 years ago

21.

Mathematics

1 answer:

erica [24]3 years ago

4 0

I think its true im not sure tho

hope it helps

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if a and b are the measures of two first quadrant angles and sin a = 4/5 and sin b = 5/13, find sin a+b.

Blababa [14]

<span> given that sin a= 4/5
cos a= √(1-sin²a)
   = √(1-16/25)
   = √(9/25)
   = 3/5
sin b=5/13
cos b= √(1-sin²b)
   = √(1-25/169)
   = √(144/169)
   = 12/13

sin(a+b)= sin a· cos b + cos a · sin b
      = 4/5· 12/13 + 3/5· 5/13
      = 48/65+15/65
   = 63/65
hope it helps
</span>

8 0

2 years ago

Read 2 more answers

Please help me ASAP!!!

vlada-n [284]

Answer:

a) 6 1/4% of $3,875 is $242.18.

b) 16 2/3% of 96 lamps is 16 lamps.

c) 87 1/2% of 540.72 is 473.11

d) 33 1/3% of 750 boys is 252.52 boys

e) 66 1/2 % of 240 kg is 159.6 kg.

Step-by-step explanation:

x% of y is calculated as $\frac{x}{ 100} \times y$

Here calculating the given percentages:

a) 6 1/4% of $3,875

Now, 6 1/4 = 6+ 1/4 = 6.25

⇒6.25% of $3,875 = $\frac{6.25}{100} \times 3,875 = 242.18$

Hence, 6 1/4% of $3,875 is $242.18.

b) 16 2/3% of 96 lamps

Now, 16 2/3 = 16+ 2/3 = 16.67

⇒16.67% of 96 lamps = $\frac{16.67}{100} \times 96 = 16$

Hence, 16 2/3% of 96 lamps is 16 lamps.

c) 87 1/2% of 540.72

Now, 87 1/2 = 87+ 1/2 = 87.5

⇒87.5 % of 540.72 = $\frac{87.5}{100} \times 540.72 = 473.11$

Hence, 87 1/2% of 540.72 is 473.11

d) 33 1/3% of 750 boys

Now, 33 1/3 = 33+ 1/3 = 33.67

⇒33.67 % of 750 boys = $\frac{33.67}{100} \times 750 = 252.52$

Hence, 33 1/3% of 750 boys is 252.52 boys

e) 66 1/2 % of 240 kg

Now, 66 1/2 = 66+ 1/2 = 66.50

⇒66 1/2 % of 240 kg = $\frac{66.50}{100} \times 240 kg = 159.6$

Hence, 66 1/2 % of 240 kg is 159.6 kg.

4 0

3 years ago

A Cepheid variable star is a star whose brightness alternately increases and decreases. For a certain star, the interval between

fomenos

Answer:

0.41 (correct to 2 decimal place)

Step-by-step explanation:

If $B(t)=4.0+0.35sin\frac{2\pi t }{5.4}$

Let $u=\frac{2\pi t }{5.4}$ , then $B(u)=4.0+0.35sin u$

We want to determine the rate of increase $\frac{dB}{dt}$ after one day

$du=\frac{2\pi dt }{5.4}$ and $\frac{dB}{du} =0.35cos u$

$\frac{dB}{dt}=\frac{2\pi }{5.4}0.35cos (\frac{2\pi t }{5.4})=0.407cos (\frac{2\pi t }{5.4})$

$B^{'} (t)=0.407cos (\frac{2\pi t }{5.4})$

Rate of increase after one day, i.e. t=1

$B^{'} (1)=0.407cos (\frac{2\pi X 1 }{5.4})$ = 0.407 =0.41 (correct to 2 decimal place)

5 0

3 years ago

How do i solve for x and y <br> 4x+y=8<br> x=5-y

vlada-n [284]

Please brainless if it’s wrong please message me

8 0

3 years ago

PLEASE HELP ASAP!!!!!

erastova [34]

Answer:

B = (3, 3)

Step-by-step explanation:

Since the line segment is fifteen units long, from the looks of it, chopping ⅓ of A's side, led to the ordered pair of (3, 3).

I hope this helps you out alot, and as always, I am joyous to assist anyone at any time.

7 0

2 years ago