All categories

4 years ago

What is the midpoint (-4,0) and (10,7)?

1 answer:

5 0

We first use midpoint formula
X1+X2 divided by 2 we want x for midpoint.
Y1+Y2 divided by 2 we want Y for midpoint
10-4 divided by 2 =3
7-0 divided by 2 =7/2
therefore midpoint= (3;7/2)

You might be interested in

Click photo for question

nikdorinn [45]

Answer:

see explanation

Step-by-step explanation:

There is a common difference d in the values, that is

9 - 5 = 13 - 9 = 17 - 13 = 21- 17 = 4

To obtain a term in the sequence add 4 to the previous term, thus

t₉ = t₈ + 4 → C

(b)

Given the n th term formula 3n - 8

To find any term, substitute the term position n into the formula

(i) first term (n = 1), then

3(1) - 8 = 3 - 8 = - 5

(ii) tenth term (n = 10, then

3(10) - 8 = 30 - 8 = 22

(iii) 100 th term ( n = 100), then

3(100) - 8 = 300 - 8 = 292

5 0

3 years ago

Write an exponential function in the form y=ab^x that goes through the point (0,17) and (6,1088)

alexandr402 [8]

Answer:

y = 17(2)^x

Step-by-step explanation:

If the graph of y = ab^x goes through (0, 17), then

17 = ab*0, or a = 17

Then the function is y = ab^x with a = 17, or

y = 17*b^x and we must find b.

If the graph of y = 17*b^x also goes through (6, 1088), then the following must be true: 1088 = 17*b^6

which reduces to 64 = b^6

Taking the sixth root of both sides, we get 64^(1/6) = b, and so b = 2

Then the desired exponential function is

y = 17(2)^x

6 0

3 years ago

A cylinder has slant length of 26 and radius of 10 what is the volume

Aleksandr-060686 [28]

Answer:

8168.14

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

The sum of the first two terms of an infinite GP is 6 and each term is 5 times the sum of the succeeding termsthen the second te

kozerog [31]

Answer:

$\dfrac{6}{7}$ .

Step-by-step explanation:

It is given that the sum of first two terms of an infinite GP is 6.

nth term of a GP is

$a_n=ar^{n-1}$

$a+ar=6$ ...(1)

Each term is 5 times the sum of the succeeding terms.

$a_n=5(a_{n+1}+a_{n+2}+...+\infty)$

$ar^{n-1}=5(ar^n+ar^{n+1}+...+\infty)$

$ar^{n-1}=5ar^n(1+r+r^2+...+\infty)$

Divide both sides by a.

$r^{n-1}=5r^n(\dfrac{1}{1-r})$ $[\text{Sum of infinite GP}=\dfrac{a}{1-r}]$

$\dfrac{r^n}{r}=\dfrac{5r^n}{1-r}$

$\dfrac{1}{r}=\dfrac{5}{1-r}$

$1-r=5r$

$1=6r$

$\dfrac{1}{6}=r$

The common ratio is 1/6.

Put r=1/6 in (1).

$a+a(\dfrac{1}{6})=6$

$6a+a=36$

$7a=36$

$a=\dfrac{36}{7}$

Second term $ar=\dfrac{36}{7}\times \dfrac{1}{6}=\dfrac{6}{7}$

Therefore, the second term is $\dfrac{6}{7}$ .

4 0

3 years ago

For the dilation with center (0,0) shown on the graph, your friend says the scale factor is 8/3

olchik [2.2K]

Answer:

It actually depends on where the point of letter is

3 0

3 years ago