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2 years ago

Solve the right triangle. Round decimal answers to the nearest tenth.

Mathematics

1 answer:

AnnZ [28]2 years ago

5 0

Answer:

See solution below

Step-by-step explanation:

According to pythagoras theorem

hyp^2 = opp^2 + adj^2

10^2 = 4^2 + b^2

100 = 16 + b^2

b^2 = 100-16

b^2 = 84

b = √84

b ≈ 9

Using SOH CHA TOA identity

sin B = AC/AB

Sin B = 9/10

B = arcsin0.9

B ≈ 64degrees

sinA = BC/AB

Sin A = 4/10

A = arcsin0.4

A ≈ 24degrees

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Use numerals instead of words. If necessary, use / for the fraction bar.

Step2247 [10]

Here, AB ║ CD ; EF ⊥ AB

Number of 90 degree formed by the intersections of EF and two parallel lines AB and CD is 8

Hope this helps!

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3 years ago

Question 1!!! PLEASE HELP

andreev551 [17]

Quickly using a calculator (or long division) reveals that 191/238=0.8025=80.25%, which is close to 80%=4/5
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7 0

3 years ago

Please please help me

Gekata [30.6K]

Answer:

$\large\boxed{\dfrac{\boxed{8}}{81}}$

Step-by-step explanation:

$\text{If}\ a_1,\ a_2,\ a_3,\ a_4,\ ...,\ a_n\ \text{is the geometric sequence, then}\\\\\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=\dfrac{a_4}{a_3}=...=\dfrac{a_n}{a_{n-1}}=constant=r-\text{common ration}.\\\\\text{We have}\ \dfrac{1}{2},\ \dfrac{1}{3},\ \dfrac{2}{9},\ \dfrac{4}{27},\ ...\\\\\dfrac{\frac{1}{3}}{\frac{1}{2}}=\dfrac{1}{3}\cdot\dfrac{2}{1}=\dfrac{2}{3}\\\\\dfrac{\frac{2}{9}}{\frac{1}{3}}=\dfrac{2}{9}\cdot\dfrac{3}{1}=\dfrac{2}{3}\\\\\dfrac{\frac{4}{27}}{\frac{2}{9}}=\dfrac{4}{27}\cdot\dfrac{9}{2}=\dfrac{2}{3}$

$\bold{CORRECT}\\\\\dfrac{x}{\frac{4}{27}}=\dfrac{2}{3}\qquad\text{multiply both sides by}\ \dfrac{4}{27}\\\\x=\dfrac{2}{3}\cdot\dfrac{4}{27}\\\\x=\dfrac{8}{81}$

8 0

2 years ago

Which expression and diagram represent "three times a number"?

pav-90 [236]

Answer:

3x the second one

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Joe has one book each for algebra, geometry, history, psychology, Spanish, English and Physics in his locker. How many different

Alisiya [41]

Answer:

There are 35 different sets of 3 books Joe could choose

Step-by-step explanation:

* Lets explain how to solve the problem

- Combination is a collection of the objects where the order doesn't

matter

- The formula for the number of possible combinations of r objects from

a set of n objects is nCr = n!/r!(n-r)!

- n! = n(n - 1)(n - 2)................. × 1

Lets solve the problem

- Joe has one book each for algebra, geometry, history, psychology,

Spanish, English and Physics in his locker

∴ He has seven books in the locker

- He wants to chose three of them

∵ The order is not important when he chose the books

∴ We will use the combination nCr to find how many different sets

of three books he can choose

- The total number of books is 7

∴ n = 7

∵ He chooses 3 of them

∴ r = 3

∵ 7C3 = 7!/3!(7 - 3)! = 7!/3!(4!)

∴ $7C3=\frac{(7)(6)(5)(4)(3)(2)(1)}{[(3)(2)(1)][(4)(3)(2)(1)]}=35$

∴ 7C3 = 35

* There are 35 different sets of 3 books Joe could choose

3 0

3 years ago