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3 years ago

Find the values ofx and y for the triange. x 60° 30° у

Mathematics

2 answers:

Scrat [10]3 years ago

8 0

Answer:

X=36.4 or 36.37306696.... and Y=42

Step-by-step explanation:

So you have a 30-60-90 triangle!

Do Not Worry this is a hard subject to understand, but I will try to do my best with helping you. Right now you need to find x and y. All you have is 21. Under your 60 angle, which is pointing downward, is the 21. And 21 is initially in the spot of x or 1. so 21 is the x(But not the x on the side of 90). I am sorry if this is a bit confusing.

You are then going to take your side which is x, the side that is directly in the path of 90, and it is going to be 21 $\sqrt{3$ . Then on the side of your y, which is in the path of your 60, it will be 2 times your x.

The general layout of this is:

x is underneath the side of 90 degrees.

then on the side where 60 degrees and 30 degrees is, is going to be x $\sqrt{3$ .

Then the last side, where 30 degrees and 90 degrees lay, will be 2(x).

So all you have to do is plug everything in.

x=36.4 in decimal

y=42(because all you had to do was plug 21 where the x is, because that is where x is in the general layout.)

egoroff_w [7]3 years ago

6 0

Answer:

x = 42, y = 36.37

Step-by-step explanation:

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Now, the triangle ABC and A'B'C' are similar triangles and we know that similar triangles angles are congruent. Thus;

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The measure of angle BCA = The measure of angle C prime A prime B prime

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2 years ago

. A biologist is comparing the growth of a population of flies per week to the number of flies an iguana will consume per week.

Veseljchak [2.6K]

We are given the equation <span>3^x = 2x + 1 in which x as the number of days of the population of flies and the number of flies consumed by iguana each week. In this case, the other equation described when x =7 population is not more than 8. The graph is a logarithmic graph</span>

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3 years ago

Read 2 more answers

Suppose a marketing company computed a 94% confidence interval for the true proportion of customers who click on ads on their sm

Ksivusya [100]

Answer:

d. There is a 98% chance that the true proportion of customers who click on ads on their smartphones is between 0.56 and 0.62.

Step-by-step explanation:

Confidence interval:

x% confidence

Of a sample

Between a and b.

Interpretation: We are x% sure(or there is a x% probability/chance) that the population mean is between a and b.

In this question:

I suppose(due to the options) there was a small typing mistake, and we have a 98% confidence interval between 0.56 and 0.62.

Interpreation: We are 98% sure, or there is a 98% chance, that the true population proportion of customers who click on ads on their smartphones is between 0.56 and 0.62. Option d.

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4 years ago

In parallelogram EFGH, the measure of angle G is (4x + 16)° and the measure of angle H is (2x − 16)°. What is the measure of ang

asambeis [7]

Answer:

m∠H = 44°

Step-by-step explanation:

Consecutive angles of a parallelogram are supplementary. So,

∠H and ∠G are supplementary

4x + 16 + 2x - 16 = 180

6x = 180

x = 30

m∠H = 2x - 16 = 2(30) - 16 = 60 - 16 = 44°

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3 years ago

Find the average value of f(x+y)=sin(x+y) over

Troyanec [42]

For each given region $D$ , the average value of $f$ over $D$ is the integral of $f$ over $D$ divided by the area of $D$ . In both cases, $D$ is a rectangle so the area is trivial.

a. The area of $D$ is $\dfrac\pi3\cdot\dfrac{5\pi}3=\dfrac{5\pi^2}9$ . The integral of $f$ over $D$ is

$\displaystyle\int_0^{5\pi/3}\int_0^{\pi/3}\sin(x+y)\,\mathrm dx\,\mathrm dy$

$\displaystyle=\int_0^{5\pi/3}\left(\cos y-\cos\left(y+\dfrac\pi3\right)\right)\,\mathrm dy$

$=\sin\dfrac{5\pi}3-\sin\left(\dfrac{5\pi}3+\dfrac\pi3\right) - \sin0+\sin\dfrac\pi3=0$

So the average value of $f$ over this region is 0.

b. The area of $D$ is $\dfrac{2\pi}3\cdot\dfrac{7\pi}6=\dfrac{7\pi^2}9$ . The integral of $f$ is

$\displaystyle\int_0^{7\pi/6}\int_0^{2\pi/3}\sin(x+y)\,\mathrm dx\,\mathrm dy$

$=\displaystyle\int_0^{7\pi/6}\left(\cos y-\cos\left(y+\frac{2\pi}3\right)\right)\,\mathrm dy$

$=\sin\dfrac{7\pi}6-\sin\left(\dfrac{7\pi}6+\dfrac{2\pi}3\right) - \sin0+\sin\dfrac{2\pi}3=\dfrac{\sqrt3}2$

The average vale of $f$ over this region is then $\dfrac{\frac{\sqrt3}2}{\frac{7\pi^2}9}=\dfrac{9\sqrt3}{14\pi^2}$ .

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3 years ago