Ask question

Ask question

All categories

3 years ago

8

The first term of an arithmetic progression is 30 and the common difference is -1.5. Find the value of the 25th term

2 answers:

kramer3 years ago

6 0

Answer:

the answer to this is

-6

kap26 [50]3 years ago

4 0

Answer:

$-6$

Step-by-step explanation:

$a_1=30$

$d=-1.5$

$-1.5n+31.5$

$-1.5(25)+31.5$

$-37.5+31.5$

$=-6$

You might be interested in

What is the yintercept of f(x) =(1/2)^x<br> ?

Anestetic [448]

$\text{Find the y-intercept:}\\\\\text{To fidn the y-intercept, we would have to plug in 0 to x. Since the y-intercept}\\\text{exists whenever x is equal to 0}\\\\\text{Solve:}\\\\f(x)=(\frac{1}{2})^0\\\\\text{We know that whenever the exponent is 0, it equals to 1}\\\\f(x)=1\\\\\text{Therefore, this makes the y-intercept 1.}\\\\\text{Answer is stated as a point on the graph}\\\\\boxed{(0,1)}$

4 0

3 years ago

State the property.<br>4(2+3)=(2+3)4

Komok [63]

Hmm, it looks like it could be commutative property of multiplication.

The order of the expression is different.

4(2+3)=

4(5)=20

(2+3)4=

(5)4= 20

8 0

3 years ago

Complete the sentence by selecting the correct word from each drop-down menu. A measure of____ summarizes all of the values in a

zhuklara [117]

I believe it is 1) center 2) single

8 0

3 years ago

Read 2 more answers

K/a = w - v, solve for a

erik [133]

Answer:

a=k-w+v

Step-by-step explanation:

k/a= w-v

a( k/a) = a (w-v)

k -w + v =a

4 0

3 years ago

F(x)=X over x^3-2x^2+5x why will this have no zeros

Mumz [18]

If you evaluate directly this function at x=0, you'll see that you have a zero denominator.

Nevertheless, the only way for a fraction to equal zero is to have a zero numerator, i.e.

$\dfrac{x}{x^3-2x^2+5x}=0\iff x=0$

So, this function can't have zeroes, because the only point that would annihilate the numerator would annihilate the denominator as well.

Moreover, we have

$\displaystyle \lim_{x\to 0} \dfrac{x}{x^3-2x^2+5x} = \lim_{x\to 0} \dfrac{x}{x(x^2-2x+5)} = \lim_{x\to 0} \dfrac{1}{x^2-2x+5} = \dfrac{1}{5}$

So, we can't even extend with continuity this function in such a way that $f(0)=0$

5 0

3 years ago