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wel
3 years ago
5

On a coordinate grid, the coordinates of vertices P and Q for polygon RQRS are P(1, 2) and Q(-1, 2)

Mathematics
1 answer:
Nata [24]3 years ago
8 0
We need to find the length of side PQ of the polygon.
Distance formula:

Using distance formula we get




The length of side PQ of the polygon is 5 units.
I think this is what you were asking..?
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there a 158 beats per minute in a certain taylor swift song. if the song is 3 minutes and 30 seconds long, how many total beats
sergiy2304 [10]
Answer is 553 beasts
Step by step:
Multiply 158 by 3 to get 474
Divide 158 by 2 to get 79
Add 474 with 79 and get 553
4 0
3 years ago
240 is 60% of what number?
Brums [2.3K]
Hi there! To find the answer to that question, we can write and solve a proportion. Set it up like this: 240/x = 60/100. This is because we're looking for the whole and we already know the part, which is 240. Cross multiply the values in order to get 24,000 = 60x. Now, divide each side by 60 to isolate the variable. When you do, you get x = 400. You can check this by dividing and then multiplying that quotient by 100. 240/400 is 0.6 and 0.6 * 100 is 60. That makes 60%. There. 240 is 60% of 400.
3 0
3 years ago
Read 2 more answers
Find the volume of a rectangular prism with a base that has an area of 22 square centimeter and a height of 8 centimeters.
Jlenok [28]

Answer:

176cm³

Step-by-step explanation:

Multiply area of base times height

22cm² × 8cm = 176cm³

3 0
3 years ago
Determine the values of xfor which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001.(Enter
MrMuchimi

Answer:

The values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001 is 0 < x < 0.3936.

Step-by-step explanation:

Note: This question is not complete. The complete question is therefore provided before answering the question as follows:

Determine the values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001. f(x) = e^x ≈ 1 + x + x²/2! + x³/3!, x < 0

The explanation of the answer is now provided as follows:

Given:

f(x) = e^x ≈ 1 + x + x²/2! + x³/3!, x < 0 …………….. (1)

R_{3} = (x) = (e^z /4!)x^4

Since the aim is R_{3}(x) < 0.001, this implies that:

(e^z /4!)x^4 < 0.0001 ………………………………….. (2)

Multiply both sided of equation (2) by (1), we have:

e^4x^4 < 0.024 ……………………….......……………. (4)

Taking 4th root of both sided of equation (4), we have:

|xe^(z/4) < 0.3936 ……………………..........…………(5)

Dividing both sides of equation (5) by e^(z/4) gives us:

|x| < 0.3936 / e^(z/4) ……………….................…… (6)

In equation (6), when z > 0, e^(z/4) > 1. Therefore, we have:

|x| < 0.3936 -----> 0 < x < 0.3936

Therefore, the values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001 is 0 < x < 0.3936.

3 0
3 years ago
Divide. 0.0075 ÷ (1.5  x  10−6) write your answer in scientific notation.
Alborosie
0.0075 ÷ (1.5 x 10^-6) = (7.5 x 10^-3) ÷ (1.5 x 10^-6) = (7.5 ÷ 1.5) x (10^(-3 - (-6))) = 5 x 10^(-3 + 6) = 5 x 10^3
4 0
3 years ago
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