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3 years ago

If the first term in an arithmetic sequence is –3 and the tenth term is 15, what is the common difference?

Mathematics

1 answer:

Triss [41]3 years ago

6 0

Given that the First term of the Arithmetic sequence = -3

⇒ a = -3

We know that nth term of a Arithmetic sequence is : Tn = a + (n - 1)d

where a is the first term and d is the common difference

Given that 10th term is 15

⇒ T₁₀ = -3 + (10 - 1)d

⇒ -3 + (10 - 1)d = 15

⇒ 9d = 18

⇒ d = 2

⇒ Common difference of given Arithmetic Sequence is 2

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Pls help me with this

DedPeter [7]

Answer:

the volume of the cone would be approximately 490.09 m^3.

Step-by-step explanation:

The volume of a cone is pir^2h/3

So it would be pi*6^2*13/3

=pi*36*13/3

=pi*156

=490.09

Therefore, the volume of the cone would be approximately 490.09 m^3.

I hope this helped and have a good rest of your day!

7 0

3 years ago

What happens to the mode of the data set shown below if the number 8 is added to the data set?

Dmitrij [34]

Answer:

Step-by-step explanation:

right now 15 is the mode because there are 3 of them

if we made it so that there was another 8 there would also be 3 of them

this would mean that there are two modes

6 0

2 years ago

Evaluate <br><br> 14y ÷ 5 - 4.9 for y = 6

Lubov Fominskaja [6]

Answer:

The final answer is 11.9

Step-by-step explanation:

14 x 6 = 84

84/5 = 16.8

16.8 - 4.9 = 11.9

Can I have brainliest? It would help me out, if not thanks anyways! Hope this helped and have a nice day!

8 0

3 years ago

Show that every triangle formed by the coordinate axes and a tangent line to y = 1/x ( for x > 0)

vfiekz [6]

Answer:

Step-by-step explanation:

given a point $(x_0,y_0)$ the equation of a line with slope m that passes through the given point is

$y-y_0 = m(x-x_0)$ or equivalently

$y = mx+(y_0-mx_0)$ .

Recall that a line of the form $y=mx+b$ , the y intercept is b and the x intercept is $\frac{-b}{m}$ .

So, in our case, the y intercept is $(y_0-mx_0)$ and the x intercept is $\frac{mx_0-y_0}{m}$ .

In our case, we know that the line is tangent to the graph of 1/x. So consider a point over the graph $(x_0,\frac{1}{x_0})$ . Which means that $y_0=\frac{1}{x_0}$

The slope of the tangent line is given by the derivative of the function evaluated at $x_0$ . Using the properties of derivatives, we get

$y' = \frac{-1}{x^2}$ . So evaluated at $x_0$ we get $m = \frac{-1}{x_0^2}$

Replacing the values in our previous findings we get that the y intercept is

$(y_0-mx_0) = (\frac{1}{x_0}-(\frac{-1}{x_0^2}x_0)) = \frac{2}{x_0}$

The x intercept is

$\frac{mx_0-y_0}{m} = \frac{\frac{-1}{x_0^2}x_0-\frac{1}{x_0}}{\frac{-1}{x_0^2}} = 2x_0$

The triangle in consideration has height $\frac{2}{x_0}$ and base $2x_0$ . So the area is

$\frac{1}{2}\frac{2}{x_0}\cdot 2x_0=2$

So regardless of the point we take on the graph, the area of the triangle is always 2.

6 0

3 years ago

help me on these 5 math problems, answer them correctly and explain your answers, please and thank you! <3 i have trouble on

BARSIC [14]

Answer: H

Step-by-step explanation: move the dilations

6 0

2 years ago

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