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3 years ago

Find the value of sin(θ) for an angle θ in standard position with a terminal ray that passes through the point (-5, -12)

Mathematics

2 answers:

IceJOKER [234]3 years ago

8 0

Answer:

c. -12/13

Step-by-step explanation:

it will be the y-coordinate divided by the distance from the origin, or (-12) / $\sqrt{5^2+12^2}$ = -12/13

netineya [11]3 years ago

6 0

Answer:

C. -12/13

Step-by-step explanation:

(-5,-12) is in the third quadrant

So sin would be negative

Hypotenuse = sqrt(5² + 12²)

sqrt(169)

sin(theta) = -12/13

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Amy is planning the seating arrangement for her wedding reception. Each round table can sit 12 guests. The head table can sit th

storchak [24]

Answer:

23 tables.

Step-by-step explanation:

282- 8 = 274 274 divided by 12 = 22.8 round to the nearest whole number which is 23.

7 0

3 years ago

Luisa recieved $108 for a necklace she sold on an online auction. The auction site took a fe of 10% of selling price. What was t

Vera_Pavlovna [14]

I believe it might be 107.9 after if that’s what your asking-like taking off 10% of $108.

3 0

3 years ago

Please help me out with this

Ivanshal [37]

Answer:

374.1

Step-by-step explanation:

Area of Hexagon = $\frac{3\sqrt{3} }{2} * a^{2}$

The ' a ' is the side length.

So now we just plug in the values

$\frac{3\sqrt{3} }{2} *12^{2}$ = 374.12

3 0

3 years ago

From experience, it is known that on average 10% of welds performed by a particular welder are defective. if this welder is requ

bulgar [2K]

Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)3. Probability is known and remains constant throughout the trials (p)4. All trials are random and independent of the othersThe number of successes, x, is then given by $P(x)=C(n,x)p^x(1-p)^{n-x}$ where $C(n,x)=\frac{n!}{x!(n-x)!}$
Here we're given
p=0.10 [ success = defective ]
n=3

(a) x=0
$P(x)=C(n,x)p^x(1-p)^{n-x}$
$=C(3,0)0.1^0(1-0.1)^{3-0}$
$=1(1)(0.729)$
$=0.729$

(b) x=2
$P(x)=C(n,x)p^x(1-p)^{n-x}$
$=C(3,2)0.1^2(1-0.1)^{3-2}$
$=3(0.01)(0.9)$
$=0.027$

(c) x ≥ 2
$P(x)=\sum_{x=2}^3C(n,x)p^x(1-p)^{n-x}$
$=P(2)+P(3)$
$=C(n,2)p^2(1-p)^{n-2}+C(n,3)p^3(1-p)^{n-3}$
$=C(3,2)0.1^2(1-0.1)^{3-2}+C(3,3)0.1^3(1-0.1)^{3-3}$
$=3(0.01)(0.9)+1(0.001)1$
$=0.027+0.001$
$=0.028$

8 0

3 years ago

Johnathan ran 5 days this week. The most he ran in one day was 3.5 miles. Write an inequality that shows the distance johnathan

Mekhanik [1.2K]

An inequality that shows the distance Johnathan could of ran any day this week is:

$x\leq 3.5$

Solution:

Let "x" be the distance Johnathan can run any day of this week

Given that,

Johnathan ran 5 days this week. The most he ran in one day was 3.5 miles

Therefore,

Number of days ran = 5

The most he ran in 1 day = 3.5 miles

Thus, the maximum distance he ran in a week is given as:

$distance = 5 \times 3.5 = 17.5$

The maximum distance he ran in a week is 17.5 miles

If we let x be the distance he can run any day of this week then, we get a inequality as:

$x\leq 3.5$

If we let y be the total distance he can travel in a week then, we may express it as,

$y\leq 17.5$

8 0

3 years ago