How do I find the value of x

Mathematics

1 answer:

juin [17]3 years ago

8 0

You set (x+15) equal to 40 and solve for x.

x+15=40
-15 -15

x=25

Hope this helps!

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Can someone help me with this?

Anna35 [415]

Answer:

Let 'a' be the first term, 'r' be the common ratio and 'n' be the number of terms

Series = 2+6+18.......= 2+2•3¹+ 2•3².......= 728

Now,

$Sum = \frac{a( {r}^{n} - 1) }{(r - 1)} \\$

So,

$\frac{a( {r}^{n} - 1)}{(r - 1)} = 728 \\ \frac{2( {3}^{n} - 1) }{(3 - 1)} = 728 \\ \frac{2( {3}^{n} - 1) }{2} = 728 \\ {3}^{n} - 1 = 728 \\ {3}^{n}=728+1\\ {3}^{n} = 729 \\ {3}^{n} = {3}^{6} \\ \boxed{ n = 6}$

Therefore, number of terms is 6

6 is the right answer.

3 0

2 years ago

What is the equation of a line, in general form, that passes though points (-1,2) and (5,2)?

Tom [10]

Answer:

Step-by-step explanation:

To write the equation of the line. first calculate the slope using the slope formula.

$m = \frac{y_2-y_1}{x_2-x_1} = \frac{2-2}{5--1}= \frac{0}{6} = 0$

Since the slope of this line is 0, it is horizontal and has the form y=b where b is the y-coordinate. So y = 2 is the equation. In general form, it would be y-2 = 0

3 0

3 years ago

Determine the values of x on which the function f(x)=2x^2-x-15/4x^2-12x is discontinuous and verify the type of discontinuity at

melomori [17]

<span>Determine the values of x on which the function f(x)=2x^2-x-15/4x^2-12x is discontinuous and verify the type of discontinuity at each point.

A.There is a vertical asymptote at 0 and a hole at 3.
B.There are vertical asymptotes at -5/2 & 0 and a hole at 3.
C.There is a vertical asymptote at 3 and a hole at 0.
D.There are vertical asymptotes at 0 & 3.
</span>
Result:
(-7/4)x^2-13x

The roots are:
x= -52/7
x=0

See attached pictures.