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3 years ago

What are the coordinates of the midpoint of the line segment with endpoints (7,2) and (3,4)

Mathematics

1 answer:

emmasim [6.3K]3 years ago

5 0

Answer:

The answer is

Step-by-step explanation:

The midpoint M of two endpoints of a line segment can be found by using the formula

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(7,2) and (3,4)

The midpoint is

We have the final answer as

Hope this helps you

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3 years ago

The population of a local species of dragonfly can be found using an infinite geometric series where a1 = 42 and the common rati

Mashutka [201]

we are given

first term is

$a_1=42$

common ratio is

$r=\frac{3}{4}$

now, we can find nth term

$a_i=a_1(r)^{i-1}$

now, we can plug values

$a_i=42(\frac{3}{4})^{i-1}$

now, we can write in sigma form

$sum=\sum _{i=1}^{\infty }\:42(\frac{3}{4})^{i-1}$

now, we can find sum

we can use formula

$sum=\frac{a}{1-r}$

now, we can plug values

we get

$sum=\frac{42}{1-\frac{3}{4}}$

$sum=168$

so, option-D.................Answer

3 0

4 years ago

A rectangle room is 12 1/2 feet long and 10 1/3 feet wide. what is the area of the room

Veronika [31]

Answer:

129 1/6 feet

Step-by-step explanation:

Formula for area is L*W

1. convert the mixed numbers to improper fractions

12 1/2 feet = 25/2

10 1/3 feet = 31/3

2. multiply numerator by numerator (25*31 = 775

3. multiply denominator by denominator (2*3=6

4. convert improper fraction 775/6 into a mixed number by dividing 775 by 6.

5. 775/6 = 129 r. 1 or 129 and 1/6

4 0

3 years ago

Answer quick please look at attachment.

Margarita [4]

Answer:

c) KJ and JH

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Suppose that the functions p and q are defined as follows.

Brut [27]

<h2>Answer:</h2>

Answer:

(r o q)(-1) = 20

(q o r)(-1) = -11

Step-by-step explanation:

Given

q(x) = -2x + 1q(x)=−2x+1

r(x) = 2x^2 + 2r(x)=2x2+2

Solving (a): (r o q)(-1)

In function:

(r o q)(x) = r(q(x))

So, first we calculate q(-1)

q(x) = -2x + 1q(x)=−2x+1

q(-1) = -2(-1) + 1q(−1)=−2(−1)+1

q(-1) = 2 + 1q(−1)=2+1

q(-1) = 3q(−1)=3

Next, we calculate r(q(-1))

Substitute 3 for q(-1)in r(q(-1))

r(q(-1)) = r(3)

This gives:

r(x) = 2x^2 + 2r(x)=2x2+2

r(3) = 2(3)^2 + 2r(3)=2(3)2+2

r(-1) = 2*9 + 2r(−1)=2∗9+2

r(-1) = 20r(−1)=20

Hence:

(r o q)(-1) = 20

Solving (b): (q o r)(-1)

So, first we calculate r(-1)

r(x) = 2x^2 + 2r(x)=2x2+2

r(-1) = 2(-1)^2 + 2r(−1)=2(−1)2+2

r(-1) = 2*1 + 2r(−1)=2∗1+2

\begin{gathered}r(-1) = 6\\\end{gathered}r(−1)=6

Next, we calculate r(q(-1))

Substitute 6 for r(-1)in q(r(-1))

q(r(-1)) = q(6)

q(x) = -2x + 1q(x)=−2x+1

q(6) = -2(6) + 1q(6)=−2(6)+1

q(6) =- 12 + 1q(6)=−12+1

q(6) = -11q(6)=−11

Hence:

(q o r)(-1) = -11

8 0

3 years ago