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3 years ago

The nth term of a sequence is 6 - n work out the first three terms and the 20th term of the sequence.

Mathematics

1 answer:

timurjin [86]3 years ago

3 0

Step-by-step explanation:

a_n=6-n

➜a_1=6-1

➜,a_2=6-2

➜a_3=6-3

➜a_20=6-20

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The width of jays original photo is 8 inches. The length of the original photos is 10 inches. He prints a smaller version that h

aleksandrvk [35]

I'm guessing the answer is 6in.
because the original width of the photo is 8in-4in
which is 4in.
and 8 to 10 is a difference of 2
so add 2 to 4
its 6

8 0

3 years ago

find the equation of straight line which passes through the point of intersection of straight lines.x+2y+3=0 and 3x+4y=7 and par

erastovalidia [21]

Answer:

Equation of the line in question: $y = x - 21$ .

Step-by-step explanation:

Start by finding the intersection of the two straight lines. The equation for both lines shall hold at their intersection. (Using the idea of the Gaussian Elimination.)

$\left\{\begin{aligned}&x + 2y =-3\\&3x + 4y=7\end{aligned}\right.$ .

Add -3 times the first equation to the second:

$\left\{\begin{aligned}&x + 2y =-3\\& -2y=16\end{aligned}\right.$ .

Add the second equation $-2y=16$ to the first:

$\left\{\begin{aligned}&x = 13\\&y=-8\end{aligned}\right.$ .

Hence the intersection of the two lines will be $(13, -8)$ .

Now, find the slope of that straight line. $y - x = 8$ is equivalent to $y = x +8$ . The slope of that line is equal to 1. So will be the slope of the line in question.

Apply the point-slope form of a line on a Cartesian plane:

Point: $(13, -8)$ ,
Slope: $1$ .

Equation of the line:

$(y - (-8)) = (x - 13)$ .

Simplify to obtain:

$y = x -21$ .

7 0

4 years ago

The equation y 4.9t 2 3.5t 2.4 relates the height y in meters to the elapsed time t in seconds for a ball thrown downward at 3.5

tamaranim1 [39]

Answer:

$t \approx 0.43\,s$

Step-by-step explanation:

The vertical displacement function is $y(t) = -4.9\cdot t^{2}-3.5\cdot t + 2.4$ , where $y(t)$ is measured in meters and $t$ in seconds. Ball hits the ground when $y(t) = 0$ . That is: