Need help with this question

Mathematics

1 answer:

solniwko [45]3 years ago

6 0

A₁ = -8 This is the first term in an arithmetic sequence.
d = 3, the common difference

a(n) = a₁ + d(n-1)

a₇ = -8 + 3(7-1)
a₇ = -8 + 3(6)
a₇ = -8 + 18
a₇ = 10

To check:
a₁ = -8
a₂ = -8 + 3 = -5
a₃ = -5 + 3 = -2
a₄ = -2 + 3 = 1
a₅ = 1 + 3 = 4
a₆ = 4 + 3 = 7
a₇ = 7 + 3 = 10

You might be interested in

What property is illustrated by this problem?

gladu [14]

Answer:

the answer is zero property

4 0

3 years ago

Read 2 more answers

Graph the line that passes through the points ( 0 , − 4) and (−5,2) and determine the equation of the line.

Eduardwww [97]

Answer:

i am highly sorry but there is no graph for me to help you with on this site :(

but to help you the X line always comes first for example the (-5,2) the -5 would be on the X line. I really hope that this help. :)

7 0

3 years ago

Write an equation for a circle with a diameter that has endpoints at (–4, –7) and (–2, –5). Round to the nearest tenth if necess

Zinaida [17]

since we know the endpoints of the circle, we know then that distance from one to another is really the diameter, and half of that is its radius.

we can also find the midpoint of those two endpoints and we'll be landing right on the center of the circle.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-2-(-4)]^2+[-5-(-7)]^2}\implies d=\sqrt{(-2+4)^2+(-5+7)^2} \\\\\\ d=\sqrt{2^2+2^2}\implies d=\sqrt{2\cdot 2^2}\implies d=2\sqrt{2}~\hfill \stackrel{~\hfill radius}{\cfrac{2\sqrt{2}}{2}\implies\boxed{ \sqrt{2}}} \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-2-4}{2}~~,~~\cfrac{-5-7}{2} \right)\implies \left( \cfrac{-6}{2}~,~\cfrac{-12}{2} \right)\implies \stackrel{center}{\boxed{(-3,-6)}} \\\\[-0.35em] ~\dotfill$

$\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{-6}{ k})\qquad \qquad radius=\stackrel{\sqrt{2}}{ r} \\[2em] [x-(-3)]^2+[y-(-6)]^2=(\sqrt{2})^2\implies (x+3)^2+(y+6)^2=2$

4 0

3 years ago

Read 2 more answers

Find the equation of the line that passes through the points (-5,7) and (2,3)

Olegator [25]

Answer:

$y=-\frac{4}{7}x+4\frac{1}{7}$

Step-by-step explanation:

So, in order to solve this problem, I started off by drawing it out. On my graph that I have attached below, I first started out by locating the points (-5,7) and (2,3). Now, this is an optional step, but I highly encourage practicing your graphing skills by solving this problem on graph paper as well. Next, I connected the two points that I just graphed. This is the line that passes through (-5,7) and (2,3).

Now, here is where the actual solving starts. If you haven't already been taught this yet, I will introduce it to you now. I am going to find the equation of this line by filling in what I know in the equation y=mx+b, where m= the slope of the line, and b= y intercept.

Slope of the line: m= $\frac{y_{1} - y_{2} }{x_{1} - x_{2} } = \frac{7-3}{-5-2} = \frac{4}{-7}= -\frac{4}{7}$

If you haven't been taught how to find the slope of a line I recommend you find out.

Substitute the slope into the equation.

$y=-\frac{4}{7} x+b\\$

Now, we will solve for the 'b,' or y intercept.

We already have x and y values to use: (-5,7) or (2,3). I'll use x=2 and y=3 to solve for the y intercept.

$y=-\frac{4}{7} x+b\\\\3=-\frac{4}{7} *2+b\\\\3=-\frac{8}{7} +b\\b=3+\frac{8}{7} \\b=\frac{21}{7}+\frac{8}{7}=\frac{29}{7} =4\frac{1}{7} \\b=4\frac{1}{7}$