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3 years ago

I need help! Please someone i’ll give brainliest!

Mathematics

2 answers:

Vinvika [58]3 years ago

7 0

You need to use SohCahToa
I recommend you use Toa
Tan117=y/21
x21 x21
21tan117=y
Put that in your calculator and you should get 20.0023654637
And round it however your teacher asks you to

Zielflug [23.3K]3 years ago

5 0

I’m literally stuck on the same kind of stuff. i don’t get it

You might be interested in

How do we represent a percent in a equation

Anastasy [175]

Answer: heres an example

Step-by-step explanation:Convert the problem to an equation using the percentage formula: Y/X = P%

X is 60, Y is 12, so the equation is 12/60 = P%

Do the math: 12/60 = 0.20

Important! The result will always be in decimal form, not percentage form. You need to multiply the result by 100 to get the percentage.

Converting 0.20 to a percent: 0.20 * 100 = 20%

7 0

2 years ago

Find the missing lengths of the sides.

uranmaximum [27]

Answer:

b = 8√3 and c = 16

Step-by-step explanation:

<u>Points to remember</u>

If the angles of a right angled triangle with angles are 30°, 60, and 90 the their sides are in the ratio 1 : √3 : 2

<u>To find the unknown side lengths</u>

From the given figure we can see a right angled triangle

Angles are 30°, 60° and 90°

sides are in the ratio 1 : √3 : 2

1 : √3 : 2 = 8 : b : c

b = 8√3 and c = 8 * 2 = 16

Therefore b = 8√3 and c = 16

6 0

3 years ago

WILL AWARD BRAINLIEST ANSWER

Tpy6a [65]

Answer:

$(x - 5)^2 + (y - 7)^2 = 13^2$

$(x - 5)^2 + (y - 7)^2 = 169$

Step-by-step explanation:

Well, the center origin of the circle is given (h,k) = (5,7).

We have to find our radius as they gave us a point. from origin to the edge of the circle.

Using the formula: (x - h)^2 + (y - k)^2 = r^2

Plug in our (h,k) = (5,7) and (x,y) = (10,19) to solve for radius.

(x - h)^2 + (y - k)^2 = r^2

(10 - (5))^2 + (19 - (7)^2 = r^2

(5)^2 + (12)^2 = r^2

25 + 144 = r^2

r^2 = 169

r = 13

5 0

2 years ago

What values of x make the two expression below equal?

nasty-shy [4]

The denominator can't equal 0.
$11(x-7) \not= 0 \\ x-7 \not=0 \\ x \not= 7 \\ \\ \\ \frac{(2x+1)(x-7)}{11(x-7)}=\frac{2x+1}{11} \\ \\
(2x+1)(x-7) \cdot 11= 11(x-7) \cdot (2x+1) \\
11(2x+1)(x-7)=11(2x+1)(x-7) \\ \\
x \in R \setminus \{7 \}$

The answer is D.

7 0

3 years ago

Read 2 more answers

Suppose there are 4 defective batteries in a drawer with 10 batteries in it. A sample of 3 is taken at random without replacemen

SSSSS [86.1K]

Answer:

a.) 0.5

b.) 0.66

c.) 0.83

Step-by-step explanation:

As given,

Total Number of Batteries in the drawer = 10

Total Number of defective Batteries in the drawer = 4

⇒Total Number of non - defective Batteries in the drawer = 10 - 4 = 6

Now,

As, a sample of 3 is taken at random without replacement.

a.)

Getting exactly one defective battery means -

1 - from defective battery

2 - from non-defective battery

So,

Getting exactly 1 defective battery = ⁴C₁ × ⁶C₂ = $\frac{4!}{1! (4 - 1 )!}$ × $\frac{6!}{2! (6 - 2 )!}$

= $\frac{4!}{(3)!}$ × $\frac{6!}{2! (4)!}$

= $\frac{4.3!}{(3)!}$ × $\frac{6.5.4!}{2! (4)!}$

= $4$ × $\frac{6.5}{2.1! }$

= $4$ × $15$ = 60

Total Number of possibility = ¹⁰C₃ = $\frac{10!}{3! (10-3)!}$

= $\frac{10!}{3! (7)!}$

= $\frac{10.9.8.7!}{3! (7)!}$

= $\frac{10.9.8}{3.2.1!}$

= 120

So, probability = $\frac{60}{120} = \frac{1}{2} = 0.5$

b.)

at most one defective battery :

⇒either the defective battery is 1 or 0

If the defective battery is 1 , then 2 non defective

Possibility = ⁴C₁ × ⁶C₂ = 60

If the defective battery is 0 , then 3 non defective

Possibility = ⁴C₀ × ⁶C₃

= $\frac{4!}{0! (4 - 0)!}$ × $\frac{6!}{3! (6 - 3)!}$

= $\frac{4!}{(4)!}$ × $\frac{6!}{3! (3)!}$

= 1 × $\frac{6.5.4.3!}{3.2.1! (3)!}$

= 1× $\frac{6.5.4}{3.2.1! }$

= 1 × 20 = 20

getting at most 1 defective battery = 60 + 20 = 80

Probability = $\frac{80}{120} = \frac{8}{12} = 0.66$

c.)

at least one defective battery :

⇒either the defective battery is 1 or 2 or 3

If the defective battery is 1 , then 2 non defective

Possibility = ⁴C₁ × ⁶C₂ = 60

If the defective battery is 2 , then 1 non defective

Possibility = ⁴C₂ × ⁶C₁

= $\frac{4!}{2! (4 - 2)!}$ × $\frac{6!}{1! (6 - 1)!}$

= $\frac{4!}{2! (2)!}$ × $\frac{6!}{1! (5)!}$

= $\frac{4.3.2!}{2! (2)!}$ × $\frac{6.5!}{1! (5)!}$

= $\frac{4.3}{2.1!}$ × $\frac{6}{1}$

= 6 × 6 = 36

If the defective battery is 3 , then 0 non defective

Possibility = ⁴C₃ × ⁶C₀

= $\frac{4!}{3! (4 - 3)!}$ × $\frac{6!}{0! (6 - 0)!}$

= $\frac{4!}{3! (1)!}$ × $\frac{6!}{(6)!}$

= $\frac{4.3!}{3!}$ × 1

= 4×1 = 4

getting at most 1 defective battery = 60 + 36 + 4 = 100

Probability = $\frac{100}{120} = \frac{10}{12} = 0.83$

3 0

2 years ago