Whats equivalent to 24+44

GaryK [48]

Answer:
cot θ = √15 / 7

Explanation:
csc θ = 1 / sin θ
In a right angled triangle, sin θ = opposite / hypotenuse
This means that:
csc θ = hypotenuse / opposite = 8 / 7
So, we have:
opposite = 7 units and hypotenuse = 8 units
We will use the Pythagorean theorem to get the adjacent side as follows:
(hypotenuse)^2 = (opposite)^2 + (adjacent)^2
(8)^2 = (7)^2 + (adjacent)^2
(adjacent)^2 = 15
adjacent = √15 units

Now, we want to get the value of cot θ.
We know that:
cot θ = 1 / tan θ
Since tan θ = opposite / adjacent
Therefore:
cot θ = adjacent / opposite = √15 / 7

Hope this helps :)

4 0

2 years ago

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LAST QUESTION...ILL GIVE BRAINLESS FOR BEST EXPLANATION

Dennis_Churaev [7]

Answer:

175

Step-by-step explanation:

The circle has to = 360 degrees. 360-10=350

350/2=175

The x is halfway that's why you divide by 2.

4 0

2 years ago

0 A two-inch-long grasshopper can jump a horizontal distance of 40 inches. An athlete, who is _ five feet nine, wants to cover a

Mkey [24]

Answer:

a) nearest jump is JL = 1380 inches = 115ft

b) number of jumps in 1 mile N= 46 jumps

Step-by-step explanation:

Given that the jump length is proportional to the body length.

If 2 inch grasshopper can jump 40 inches.

JL = k(BL)

k = JL/BL

where JL = jump length = 40 inches

BL = Body length = 2 inches.

k = 40/2 = 20

The constant of proportionality is 20.

For the athlete :

BL = 5ft 9 inches = 5(12)+9 = 69 inches.

The jump length of the athlete is:

JL = k(BL) = 20(69)

JL = 1380 inches. = 115ft

The number of jumps in 1 mile is

1 mile = 63360 inches

N = 63360/1380

N = 45.9 = 46

N= 46

Therefore, 46 jumps would be needed.

7 0

2 years ago

Find the Laplace transformation of each of the following functions. In each case, specify the values of s for which the integral

MAVERICK [17]

Answer:

a. $\frac {2} {s-1}$ converges to s> 1.

b. $\frac{3}{e^3 \left(s-5 \right)}$ converges to s> 5.

c. $- \frac {2}{s + 3}$ converges to s> - 3.

d. $\frac {s}{s^2 + 25}$ converges to s> 0.

e. $\frac {10} {s^2 + 1}$ converges even s> 0.

f. $\frac {12}{s^2 + 4}$ converges to s> 0.

g. $-\frac {5\left(\cos\left (1\right) s-2 \sin\left(1\right)\right)}{s^2 + 4}$ converges to s> 0.

h. $\frac {1} {s ^ 2 + 4}$ converges to s> 0.

Step-by-step explanation:

a. $L \left\{2e^t \right\} = 2L \left\{e^t \right\} = 2 \cdot \frac {1} {s-1} = \frac {2} {s-1}$ converges to s> 1.

b. $L \left\{3e^{5t-3} \right\} = 3e^{-3} L \left\{e^{5t} \right\} = 3e^{-3} L \left\{e^{5t} \right\} = \frac{3}{e^3 \left(s-5 \right)}$ converges to s> 5.

c. $L \left\{-2e^{-3t} \right\} = -2L \left\{e^{-3t} \right\} = - \frac {2}{s + 3}$ converges to s> - 3.

d. $L \left\{\cos\left (5t \right)\right\} = \frac {s}{s^2 + 25}$ converges to s> 0.

e. $L \left\{10 \sin\left(t\right)\right\} = 10L\left\{\sin\left(t\right)\right\} = \frac {10} {s^2 + 1}$ converges even s> 0.

f. $L \left\{6\sin \left(2t \right) \right\} = 6L\left\{\sin\left (2t\right)\right\} = \frac {12}{s^2 + 4}$ converges to s> 0.

g. $L \left\{-5\cos\left(2t + 1\right) \right\} = -5L\left\{\cos\left(2t + 1 \right)\right\} = -\frac {5\left(\cos\left (1\right) s-2 \sin\left(1\right)\right)}{s^2 + 4}$ converges to s> 0.

h. $L\left\{\sin \left(t\right)\cos \left(t\right)\right\} = L\left\{\sin\left(2t\right)\frac{1}{2}\right\} =\frac{1}{2}\cdot \frac{2}{s^2+4} = \frac {1} {s ^ 2 + 4}$ converges to s> 0.

7 0

3 years ago

Witch is bigger -5 or -55

Georgia [21]

The answer to which is bigger is -5 because it is closer to zero

3 0

2 years ago

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