PLZZZ CHECK ME Is this correct, if not please help me.

Mathematics

1 answer:

Gwar [14]3 years ago

5 0

Hello! For this question, you solve using order of operations, which you probably already have remembered as PEMDAS. Anytime a number is raised to the 0 power, that is equivalent to 1. 0.5^0 is 1. 250 * 1 is 250. x = 0 gives us the answer of 250. Now, let's solve for the number by the second power. 0.5² is 0.25. 250 * 0.25 is 62.5. x² gives us 62.5. Now, let's subtract the two numbers together. 250 - 62.5 is 187.5. Because the number is decreasing, the slope is negative, so the answer is -187.5. Your answer is correct.

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I'm having a hard time with this question so someone help please

bulgar [2K]

Answer:

7.7 km

Step-by-step explanation:

This question is asking you to find the third side of a triangle in which two sides and an angle not between them are given. The law of sines provides the necessary relationships.

<h3>Law of Sines</h3>

The Law of Sines tells you that triangle side lengths are proportional to the sine of the opposite angle. For this triangle TPJ, the relation is ...

t/sin(T) = p/sin(P) = j/sin(J)

We are given T=68°, p=5.2 km, t=7.5 km, and we are asked to find the length of j, the TP distance.

<h3>Setup</h3>

We know the sum of angles in a triangle is 180°, and the sine of an angle is equal to the sine of its supplement. This lets us fill in the Law of Sines equation like this:

7.5/sin(68°) = 5.2/sin(P) = j/sin(68°+P) . . . . . solve for j

<h3>Solution</h3>

This can be solved in two steps. First, we find angle P, then we use that to find length j.

sin(P) = 5.2/7.5 × sin(68°) ≈ 0.642847

P = arcsin(0.642847) ≈ 40.0045°

Now, we can find j:

j = 7.5 × sin(68° +40.0045°)/sin(68°) ≈ 7.693 . . . km

The distance from the tower to the plane is about 7.7 km.

<em>Additional comment</em>

We have used T, P, and J to refer to the vertices at the tower, plane, and jet.

The triangle can also be solved using the Law of Cosines. In that case, the missing side will be the solution to a quadratic equation with irrational coefficients.

3 0

2 years ago

Draw two polygons that have corresponding angles that are all congruent but are not similar to prove that angle congruence is no

Marina86 [1]

There are infinitely many ways to do this. One such way is to draw a very thin stretched out rectangle (say one that is very tall) and a square. Example: the rectangle is 100 by 2, while the square is 4 by 4.

Both the rectangle and the square have the same corresponding angle measures. All angles are 90 degrees.

However, the figures are not similar. You cannot scale the rectangle to have it line up with the square. The proportions of the sides do not lead to the same ratio

100/4 = 25

2/4 = 0.5

so 100/4 = 2/4 is not a true equation. This numerically proves the figures are not similar.

side note: if you are working with triangles, then all you need are two pairs of congruent corresponding angles. If you have more than three sides for the polygon, then you'll need to confirm the sides are in proportion along with the angles being congruent as well.

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After the 1" year of his $10,000 investment, Ryan had seen an increase of $600. If the investment

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Answer:

Step-by-step explanation:

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