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SOVA2 [1]
3 years ago
12

Prove that √3cosec20° - sec20° = 4​

Mathematics
1 answer:
podryga [215]3 years ago
4 0

Answer:

LHS,

  • √3cosec20°-sec20°

hope it helps.

<h2>stay safe healthy and happy.</h2>

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20 points!!!!! The table represents a linear function
Darya [45]
I believe the answer is 5. I multiplied 5 to -4 and that’s -20 and I’m guessing since slope-intercept form is y=Mx+b I’m thinking b=4 and the slope being 5 matches the output. -16=5(-4)+4
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3 years ago
In May Mr. Lee's earings were 60 percent of the Lee family's total income. In June Mr. Lee earned 20 percent more than in May. I
murzikaleks [220]

Answer: 64.29%

Step-by-step explanation:

Let the total earnings of the Lee family in May be x

Amount earned by Mr Lee in May = 60% of x = 0.6x

Amount earned by other family members in May = x - 0.6x (total earnings - Mr Lee's earnings) = 0.4x

Amount earned by Mr Lee in June = 20% more than what Mr Lee earned in May.

That is, amount earned by Mr Lee in June = 1.2 × 0.6x = 0.72x

Amount earned by other members of the family in June remains unchanged, that is = 0.4x

Total amount earned by the Lee family in June = 0.72x + 0.4x = 1.12x

Percentage of total earnings in June earned by Mr. Lee = (0.72x/1.12x) × 100% = 64.29%.

QED!

6 0
3 years ago
Square of (1\4A+1\4B)^2
miskamm [114]

Answer:4a = 2⋅(2a)⋅1 4 a = 2 ⋅ ( 2 a) ⋅ 1

Step-by-step explanation:

Factor 4a^2-4a+1. 4a2 − 4a + 1 4 a 2 - 4 a + 1. Rewrite 4a2 4 a 2 as (2a)2 ( 2 a) 2. (2a)2 − 4a+1 ( 2 a) 2 - 4 a + 1. Rewrite 1 1 as 12 1 2. (2a)2 − 4a+12 ( 2 a) 2 - 4 a + 1 2. Check that the middle term is two times the product of the numbers being squared in the first term and third term. 4a = 2⋅(2a)⋅1 4 a = 2 ⋅ ( 2 a) ⋅ 1.

7 0
2 years ago
A calculator is required to obtain the final answer on this question. A solid metal sphere at room temperature 20oC is dropped i
yKpoI14uk [10]

Answer:

T = 47.1875°C

Step-by-step explanation:

Given:

Surrounding temp, Ts = 100°C

Initial Temperature ,T0= 20°C

Increase in temperature = 15°C

Final temperature, T = 20 + 15 = 35°C

Time, t = 9 seconds

Let's take Newton's law of cooling:

T - Ts = (T_0 - Ts)e^k^t

We'll solve for k

35 - 100 = (20 - 100)e^9^k

-65 = (-80)e^9^k

Divide both sides by -5

13 = (16)e^9^k

9k = ln (\frac{13}{16})

k = \frac{1}{9} ln (\frac{13}{16})

k = -0.02307104053

Let's now find the temperature of the ball after 18 seconds in boiling water.

Use the Newton's equation again:

T - Ts = (T_0 - Ts)e^k^t

T - 100 = (20 - 100)e^-0.0^2^3^0^7^1^0^4^0^5^3^*^1^8

T - 100 = (-80)e^-0.0^2^3^0^7^1^0^4^0^5^3^*^1^8

T - 100= -52.8125

T = 100 - 52.8125

T = 47.1875°C

Temperature of the ball after 18 seconds in boiling water is 47.1875°C

5 0
3 years ago
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