5 hundreds 4 hundredths = blank

What is 200 divided by 300000

aev [14]

200 divided by 300000 = 0.000666666666667

6 0

3 years ago

Read 2 more answers

Use trigonometry to solve the triangle shown. Round all side lengths to one decimal place and

Eddi Din [679]

Answer/Step-by-step explanation:

Recall: SOH CAH TOA

✔️Find <A:

Reference angle (θ) = A

Opposite side = 14 cm

Hypotenuse = 20 cm

Apply SOH:

Sin θ = Opp/Hyp

Substitute

Sin A = 14/20

A = $sin^{-1}(\frac{14}{20})$

m<A = 44° (nearest whole number)

✔️Find <C:

Reference angle (θ) = C

Adjacent side = 14 cm

Hypotenuse = 20 cm

Apply CAH:

Cos θ = Adj/Hyp

Substitute

Cos C = 14/20

C = $cos^{-1}(\frac{14}{20})$

m<C = 46° (nearest whole number)

✔️Find AB:

Reference angle (θ) = C = 46°

Opposite side = AB

Hypotenuse = 20 cm

Apply SOH:

Sin θ = Opp/Hyp

Sin 46° = AB/20

20*Sin 46° = AB

AB = 14.4 cm (one decimal place)

3 0

3 years ago

What happens when you add and subtract integers

Furkat [3]

When u add it stays the same but for subtraction u change the sign.

Example: 2- (-5)=7

5 0

3 years ago

Read 2 more answers

Maci and I are making a small kite. Two sides are 10". Two sides are 5". The shorter diagonal is 6". Round all your answers to t

Art [367]

Answer:

A. 4".

B. Approximately 9.54".

C. Approximately 13.54".

Step-by-step explanation:

Please find the attachment.

Let x be the distance from the peak of the kite to the intersection of the diagonals and y be the distance from the peak of the kite to the intersection of the diagonals.

We have been given that two sides of a kite are 10 inches and two sides are 5 inches. The shorter diagonal is 6 inches.

A. Since we know that the diagonals of a kite are perpendicular and one diagonal (the main diagonal) is the perpendicular bisector of the shorter diagonal.

We can see from our attachment that point O is the intersection of both diagonals. In triangle AOD the side length AD will be hypotenuse and side length DO will be one leg.

We can find the value of x using Pythagorean theorem as:

$(AO)^2=(AD)^2-(DO)^2$

$x^{2}=5^2-3^2$

$x^{2}=25-9$

$x^{2}=16$

Upon taking square root of both sides of our equation we will get,

$x=\sqrt{16}$

$x=\pm 4$

Since distance can not be negative, therefore, the distance from the peak of the kite to the intersection of the diagonals is 4 inches.

B. We can see from our attachment that point O is the intersection of both diagonals. In triangle DOC the side length DC will be hypotenuse and side length DO will be one leg.

We can find the value of y using Pythagorean theorem as:

$(OC)^2=(DC)^2-(DO)^2$

Upon substituting our given values we will get,

$y^2=10^2-3^2$

$y^2=100-9$

$y^2=91$

Upon taking square root of both sides of our equation we will get,

$y=\sqrt{91}$

$y\pm 9.539392$

$y\pm\approx 9.54$

Since distance can not be negative, therefore, the distance from intersection of the diagonals to the top of the tail is approximately 9.54 inches.

C. We can see from our diagram that the length of longer diagram will be the sum of x and y.

$\text{The length of the longer diagonal}=x+y$