*NEED HELP A.S.A.P*

A child is flying a kite 145ft above the ground. The string makes an angle of 57 degrees from the child's hands. How much has be

Arturiano [62]

Answer: the length of string that been let out to fly the kite this high is 172.89 ft

Step-by-step explanation:

The length of string attached to the kite, the vertical height of the kite above the ground and the ground distance forms a right angle triangle.

With an angle of 57 degrees, the length of the string that is attached to the kite represents the hypotenuse of the right angle triangle.

The height of the kite above the ground represents the opposite side of the triangle

To determine h, the length of the string that has been let out to fly the kite this high, we would apply the

Sine trigonometric ratio which is expressed as

Sine θ = opposite side/hypotenuse

Sin 57 = 145/h

h = 145/Sin57 = 145/0.8387

h = 172.89

7 0

3 years ago

How many grams of NH3 are required to produce 4.65 g of HF

nignag [31]

1.326 grams of NH3 are required to produce 4.65 g of HF.

Step-by-step explanation:

Balanced chemical reaction is written first to know the number of moles taking part in original reaction.

NH3 +3 F2 ⇒3 HF + NF3

Given:

mass of HF = 4.65

First the number of moles of HF in 4.65 grams is calculated by using the formula:

number of moles (n) = $\frac{mass}{atomic mass of 1 mole}$

atomic mass of HF = 20 grams/mole

putting the values in the above equation number of moles can be found.

n = $\frac{4.65}{20}$

= 0.235 moles of HF are given.

From the equation it can be said that:

1 mole of NH3 reacts to form 3 moles of HF

so, x moles of NH3 would react to form 0.235 moles of HF

$\frac{3}{1}$ = $\frac{0.235}{x}$

3x = 0.235

x = $\frac{0.235}{3}$

x = 0.078 moles of NH3 is required.

The moles are converted to mass by applying the formula:

mass = atomic mass X number of moles (atomic mass of NH3 = 17 grams/mole)

putting the values in the formula

mass = 17 X 0.078

mass = 1.326 grams

8 0

3 years ago

What is 78 in scientific notation

4vir4ik [10]

Well first you take 78 and turn it to 7.8
then multiply by 10000
therefore 7.8*10000= ?

3 0

3 years ago

Mason walked on Monday, Wensday, and Friday. Use these clues to find how far he walked each day. These distances were six-eights

krok68 [10]

Answer:

Monday he walked $\frac{1}{4}miles$

Wednesday he walked $\frac{6}{8}miles$

Friday he walked $\frac{1}{6}miles$

Step-by-step explanation:

Given Mason walked on Monday, Wednesday and Friday. These distances were six-eights mile, one-fourth mile, and one-sixth mile. He did not walk the farthest on Monday. He walked less on Friday than Monday. we have to find how far he walked each day.

distances are $\frac{6}{8}, \frac{1}{4}, \frac{1}{6}$ that are 0.75, 0.25 and 0.17 respectively.

Now, he didn't walk farthest on Monday and also walked less on Friday than Monday.

∴ Less distance travelled is $\frac{1}{6}$ which is on friday and then $\frac{1}{4}$ on monday.

Rest distance which is $\frac{6}{8}$ on wednesday.

Hence, Monday he walked $\frac{1}{4}miles$

Wednesday he walked $\frac{6}{8}miles$

Friday he walked $\frac{1}{6}miles$

4 0

3 years ago

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Calculus A hard problem help please

dexar [7]

∑x = 1 + 2 + 3 + 4 + 5 + 6 = 21
∑y = 8 + 3 + 0 + 1 + 2 + 1 = 15
∑x^2 = 1 + 4 + 9 + 16 + 25 + 36 = 91
∑y^2 = 64 + 9 + 0 + 1 + 4 + 1 = 79
∑xy = 8 + 6 + 0 + 4 + 10 + 6 = 34

r = (n∑xy - ∑x∑y)/(sqrt(n∑x^2 - (∑x)^2)*sqrt(n∑y^2 - (∑y)^2)) = (6(34) - 21(15))/(sqrt(6(91) - (21)^2)*sqrt(6(79) - (15)^2)) = (204 - 315)/(sqrt(546 - 441)*sqrt(474 - 225)) = -111/(sqrt(105)*sqrt(249)) = -111/(10.25*15.78) = -111/161.7 = -0.68

3 0

3 years ago

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