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elena55 [62]
3 years ago
12

A line contains the point (2 , 1) and has a slope of -2. What is the equation of the line?

Mathematics
2 answers:
erik [133]3 years ago
5 0

Answer:

The answer would be y=-2x+5

Step-by-step explanation:

hope this helped :)

gogolik [260]3 years ago
3 0

Answer: y=-2x+5

Step-by-step explanation:

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Write all integers between:<br><br>a. -17 and -12<br><br>b. -1 and -6<br><br>​
egoroff_w [7]

Answer:

look at explanation

Step-by-step explanation:

a) -13,-14,-15,-16

b) -2,-3,-4,-5

5 0
2 years ago
Which point on the number line represents the approximate volume of a cone with the radius of 3 units and a height of 4 units? U
alekssr [168]

Answer: Third red dot after 30 on the number line represents the approximate volume of a cone.

Step-by-step explanation:

Since we have have given that

Radius of cone = 3 units

Height of cone = 4 units

So, Volume of cone is given by

Volume=\frac{1}{3}\pi^2\times h\\\\Volume=\frac{1}{3}\times 3.14\times 3\times 3\times 4\\\\Volume=37.68\ units^3

Hence, Third red dot after 30 on the number line represents the approximate volume of a cone.

5 0
3 years ago
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IM 7.1.10 Cool Down
Liono4ka [1.6K]

Answer:

Calculating the Actual Distance using the Scale

If the scale is 1 : x, then multiply the map distance by x to calculate the actual distance.

Step-by-step explanation:

4 0
3 years ago
Which is equivalent to 64 Superscript one-fourth?
Schach [20]

Answer:

2 RootIndex 4 StartRoot 4 EndRoot

Step-by-step explanation:

we have

64^{\frac{1}{4}}

Decompose the number 64 in prime factors

64=2^{6}=2^{4}2^{2}

substitute

64^{\frac{1}{4}}=(2^{4}2^{2})^{\frac{1}{4}}=2^{\frac{4}{4}}2^{\frac{2}{4}}=2\sqrt[4]{4}

7 0
3 years ago
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Find the half range Fourier sine series of the function
worty [1.4K]
The full range is -\pi (length 2L=2\pi), so the half range is L=\pi. The half range sine series would then be given by

f(x)=\displaystyle\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L=\sum_{n\ge1}b_n\sin nx

where

b_n=\displaystyle\frac2L\int_0^Lf(x)\sin\dfrac{n\pi x}L\,\mathrm dx=\frac2\pi\int_0^\pi(\pi-x)\sin nx\,\mathrm dx

Essentially, this is the same as finding the Fourier series for the function

\begin{cases}g(x)=\begin{cases}\pi-x&\text{for }0

Integrating by parts yields

b_n=\dfrac2\pi\left(\dfrac\pi n-\dfrac{\sin n\pi}{n^2}\right)=\dfrac2n

So the half range sine series for this function is simply

f(x)=\displaystyle\sum_{n\ge1}\frac{2\sin nx}n
5 0
3 years ago
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