the long leg of a 30-60-90 triangle is 11√3. what is the length of the hypotenuse?

The two triangles are not similar because y=60 so it is impossible to have three congruent angles.

A triangle is a three sided polygon, which has three vertices and three angles which has the sum 180 degrees.

Given that;

In △ABC has interior angles with measures x°, 100°, and 70°,

And, △DEF has interior angles with measures y°, 70°, and 50°.

Hence, Measure of x is,

x + 100 + 70 = 180

x + 170 = 180

x = 180 - 170

x = 10

And, Measure of y is,

y + 70 + 50 = 180

y + 120 = 180

y = 180 - 120

y = 60

Thus, The two triangles are not similar because y=60 so it is impossible to have three congruent angles.

Learn more about the triangle visit;

brainly.com/question/1058720

#SPJ3

Option B.The height of a cone is 15 inches.

To find the height of the cone, we will use the formula for the volume of a cone:

V = (1/3) * π * r² * h

We know the volume (V) is 180 cubic inches and the radius (r) is 6 inches.
We need to solve for the height (h).
First, let's plug in the known values:

180 = (1/3) * π * (6)² * h

180 = (1/3) * π * 36 * h

540 = π * 36 * h

540 / 36π = h

We can simplify this further, h = 15/π. Since we do not use π for the answer:

By rational observation of given options, B. 15 in. seems accurate.

Elephants have ivory tusks, which are elongated incisor teeth, primarily for digging, stripping bark, and defense, and they may have evolved for sexual selection and competitive interactions among males.

Elephants have ivory tusks because they are actually elongated incisor teeth. These tusks are composed of dentine and are covered by a layer of enamel, just like our teeth. Elephants use their tusks for various purposes, such as digging for water, stripping bark from trees, and as a weapon for defense. However, the primary reason they have tusks is not entirely understood. It's believed that tusks might have evolved for sexual selection and display, as well as for competitive interactions among male elephants.

Using proportions, it is found that the correct sentence is:

Decreased 10% from last quarter's goal of 250 consumers retained, hence, the number of customers is:

[tex]0.9(250) = 225[/tex]

The current rate has to be increased by 16%, hence, the goal is:

[tex]1.16(225) = 261[/tex]

Hence, the sentence is:

"That means our target is 261 consumers retained."

To learn more about proportions, you can take a look at https://brainly.com/question/24372153

Final answer:

Greg started with 20 quarters. The equation for the number of quarters saved after x days is y = 20 + 3x. With Greg's savings rate of 3 quarters per day, he cannot have exactly 100 quarters on any day since 100 is not a multiple of 3 plus the 20 starting quarters.

Explanation:

To determine how many quarters Greg started with, let 's' be the number quarters Greg started with, and 'd' be the number of quarters he saves each day. After 2 days, he had a total of 26 quarters saved, and after 5 days, he had a total of 35 quarters saved. These can be written as two equations:

Subtracting the first equation from the second gives:

3d = 9

From which we find that d = 3 (Greg saves 3 quarters a day). Plugging this value back into the first equation:

s + 2(3) = 26

s + 6 = 26

s = 20

Greg started with 20 quarters.

B. The equation to model the number of quarters Greg has saved, y, after x days is:

y = 20 + 3x

C. With the savings rate of 3 quarters a day, Greg's total will always be a multiple of 3 plus the 20 he started with. 100 is not a multiple of 3, meaning he can never save exactly 100 quarters by the end of any given day since 100 minus the 20 quarters he started with leaves 80, which is not divisible by 3.

Final answer:

Greg started with 20 quarters and saves 3 quarters every day afterward. The equation to model the number of quarters saved after x days is y = 20 + 3x. It is impossible for Greg to have exactly 100 quarters saved by the end of any given day because 100 cannot be reached by adding multiples of 3 to the starting amount of 20 quarters.

Explanation:

To solve for the number of quarters Greg started with and the number of quarters he saves each day, we can set up a system of equations. Let's designate q as the number of quarters Greg started with and d as the number of quarters he saves each day. From the problem, we have two points of data: On day 2, Greg has 26 quarters, and on day 5, he has 35 quarters.

The equations representing these two data points are:

Subtracting the first equation from the second gives us:

3d = 9

Dividing both sides by 3 gives us:

d = 3

Now that we have the value for d, we can substitute it back into the first equation:

q + 2(3) = 26

q + 6 = 26

Subtracting 6 from both sides gives us:

q = 20

Greg started with 20 quarters.

For the equation to model the number of quarters saved, y, after x days, we have:

y = q + dx

This simplifies to:

y = 20 + 3x

To address part C, let's analyze the possibility of Greg having exactly 100 quarters saved. If we set y to 100 in the equation y = 20 + 3x and solve for x, we get:

100 = 20 + 3x

80 = 3x

x = 80/3

x ≈ 26.67

Since x must be an integer because Greg cannot save a fraction of a day, he cannot have exactly 100 quarters by the end of any given day. This is because 100 is not a multiple of 3 (the daily amount Greg saves) when starting from 20. Thus, it's impossible for Greg to have exactly 100 quarters saved by the end of any given day.

Answer:

Step-by-step explanation:

Quadratic equation is given by x²-24x+c

This is in the form expansion of (a-b)², that is a²-2ab+b²

Comparing

            a² = x²,    a = x

            -2ab = -24x

             -2xb = -24x

                  b = 12

           c = b² = 12² =144

   x²-24x+144 = (x-12)²

Option D is the correct answer

The second root will be a + √b the missing sign is positive because the roots are always in the pair.

Polynomial is the combination of variables and constants systematically with "n" number of power in ascending or descending order.

[tex]\rm a_1x+a_2x^2+a_3x^3+a_4x^4..........a_nx^n[/tex]

The quadratic formula:

[tex]\rm x = \dfrac{-b \pm\sqrt{b^2-4ac}}{2a}[/tex]

Here the roots are in pairs:

One root is a - √b

The second root will be a + √b

Thus, the second root will be a + √b the missing sign is positive because the roots are always in the pair.

Learn more about Polynomial here:

brainly.com/question/17822016

#SPJ2

The median is the better measure of center when dealing with data sets that have outliers or extreme values.

The median is the better measure of center when a data set contains several outliers or extreme values. It is less sensitive to measurements with extreme values, providing a more robust estimation of central tendency. The median can be thought of as the middle number in a series, where 50% of the measurements are above it and 50% are below.

Final answer:

Use the law of sines when you have two angles and one side, or two sides and a non-inclusive angle. Use the law of cosines when you have two sides and the included angle or all three sides. Apply trigonometric identities for vector addition and resolving forces.

Explanation:

Deciding when to use the law of sines or the law of cosines depends on the information given in a triangle problem. The law of sines is most useful when you know either two angles and one side (AAS or ASA) or two sides and a non-enclosed angle (SSA). The law states that the ratio of the length of a side to the sine of its opposing angle is consistent for all sides and angles in a triangle, often written as a/sin A = b/sin B = c/sin C.

The law of cosines is handy when you have either two sides and the included angle (SAS) or when you have all three sides (SSS). This law relates the lengths of the sides of a triangle to the cosine of one of its angles, following the formula c² = a² + b² - 2ab cos C.

For vector addition problems or when decomposing forces into components, trigonometric identities involving sine and cosine are employed. If you're resolving a weight into components, you may use sine to find the component opposite the angle and cosine for the component adjacent to the angle.