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3 years ago

Help me out Wishing. What are the trigonometric ratios. Write all six.

Mathematics

2 answers:

ivolga24 [154]3 years ago

4 0

Answer:

$trigonemetric \: ratios \\ sin \: theta = \frac{opposite}{hypotenuse} \\ cos \: theta = \frac{adjacent}{hypotenuse} \\ tan \: theta = \frac{opposite}{adjacent} \\ cosec \: theta = \frac{1}{sin \: theta} \\ sec \: theta = \frac{1}{cos \: theta} \\ cot \: theta = \frac{1}{tan \: theta}$

hope this helps.........

MAVERICK [17]3 years ago

3 0

Answer:

Sine - opposite : hypotenuse

Cosine - adjacent : hypotenuse

Tangent - opposite : adjacent

Cosecant - hypotenuse : opposite

Secant - hypotenuse : adjacent

Cotangent - adjacent : opposite

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A kite four equal sides. If 5 inches is added to each side, the new perimeter will be 56 inches. Find the side length of the ori

GREYUIT [131]

51 inches because I subtracted it 56-5=51

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3 years ago

Derive the equation of the parabola with a focus at (_5, 5) and a directrix of y = -1.

mars1129 [50]

Answer: Option D is correct.

Step-by-step explanation:

Since we have given that

focus = (-5,5)

and a directrix y= -1

Since, equation of parabola in this case will be

$(x-h)^2=4.a(y-k)$

Now, here

$y=k-a=-1\\\\\text { focus =(h,k+a)}\\\\\text{So,} k+a=5\\\\\text{ by solving these two equation , we get }\\\\a=3\text{ and } k=2$

So equation will be

$(x+5)^2=4\times 3(y-2)\\\\(x+5)^2=12(y-3)\\\\y=\frac{1}{12}(x+5)^2+2$

So, option D is correct .

4 0

3 years ago

Read 2 more answers

SAT verbal scores are normally distributed with a mean of 433 and a standard deviation of 90. Use the Empirical Rule to determin

laila [671]

34% of the scores lie between 433 and 523.

Solution:

Given data:

Mean (μ) = 433

Standard deviation (σ) = 90

Empirical rule to determine the percent:

(1) About 68% of all the values lie within 1 standard deviation of the mean.

(2) About 95% of all the values lie within 2 standard deviations of the mean.

(3) About 99.7% of all the values lie within 3 standard deviations of the mean.

$$Z(X)=\frac{x-\mu}{\sigma}$

$$Z(433)=\frac{433-\ 433}{90}=0$

$$Z(523)=\frac{523-\ 433}{90}=1$

Z lies between o and 1.

P(433 < x < 523) = P(0 < Z < 1)

μ = 433 and μ + σ = 433 + 90 = 523

Using empirical rule, about 68% of all the values lie within 1 standard deviation of the mean.

i. e. $((\mu-\sigma) \ \text{to} \ (\mu+\sigma))=68\%$

Here μ to μ + σ = $\frac{68\%}{2} =34\%$

Hence 34% of the scores lie between 433 and 523.

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3 years ago

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3 years ago

Can someone help me with number 8 please? Thank you.

Naddik [55]

Answer: $-4 \le x \le 5$
Note: to write the domain in interval notation, you'd write [-4,5]
if you need the domain in set-builder notation, then you'd write $\left\{x|x\in\mathbb{R}, \ -4\le x\le 5\right\}$

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Explanation:

The domain is the set of possible x input values. Look at the left most point (-4,-1). The x coordinate here is x = -4. This is the smallest x value allowed. The largest x value allowed is x = 5 for similar reasons, but on the other side of the graph.

So that's how I got $-4 \le x \le 5$ (x is between -4 and 5; inclusive of both endpoints)

Writing [-4,5] for interval notation tells us that we have an interval from -4 to 5 and we include both endpoints. The square brackets mean "include endpoint"

Writing $\left\{x|x\in\mathbb{R}, \ -4\le x\le 5\right\}$ is the set-builder notation way of expressing the domain. The $x\in\mathbb{R}$ portion means "x is a real number"

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3 years ago