Find the 11th term from the end of the A. P 10,7,4 -62.

-32

Step-by-step explanation:

a=10, d = 7-10 = -3

First find which term is -62

an = -62

a = (n-1) d= -62

Putting values

10 + (n-1)(-3) = -62

10 + (n(-3) -1 (-3) = -62

10 - 3n + 3 = -62

13 - 3n = -62

-3n = -62-13

n = 75/3 = 25

We need to find out 11th term from the last term ie., 25 - 10 = 15th term

So we need to find a14

an = a +(n-1) d

a15 = 10+ (15-1) (-3)

= 10 + 14 (-3)

= 10 -42

= -32

So the 11th term from the 15th term from the first = -32

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

answer: solve 5\cdot 2^x=2405⋅2  

x

=2405, dot, 2, start superscript, x, end superscript, equals, 240.

solution

to solve for xxx, we must first isolate the exponential part. to do this, divide both sides by 555 as shown below. we do not multiply the 555 and the 222 as this goes against the order of operations!

\begin{aligned} 5\cdot 2^x& =240 2^x& =48 \\ \end{aligned}  

5⋅2  

x

2  

x

​    

=240

=48

​

step-by-step explanation: