What is the distance between points (-4, 5) and (8, 40) on the coordinate plane?

Mathematics

1 answer:

Zolol [24]3 years ago

8 0

The distance is 37
I attached a picture

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the perimeter of the rectangle below is 98 units. Find the length of CD. Write your answer without variables. cd=4x+3 cb=5x+1

zimovet [89]

$2(4x + 3) + 2(5x + 1) = 98 \\ 8x + 6 + 10x + 2 = 98 \\ 18x + 8 = 98 \\ 18x = 90 \\ x = 5$
Substitute x with 5,

CD =
$4x + 3 \\ = 4(5) + 3 \\ = 20 + 3 \\ = 23$
CD = 23 units

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Hey can you help me answer this question by giving me the answer?

Leto [7]

The value of f(a)=4-2a+6 $a^{2}$ , f(a+h) is $6a^{2} +6h^{2} -2a-2h+12ah$ , [f(a+h)-f(a)]/h is 6h+12a-2 in the function f(x)=4-2x+6 $x^{2}$ .

Given a function f(x)=4-2x+6 $x^{2}$ .

We are told to find out the value of f(a), f(a+h) and [f(a+h)-f(a)]/hwhere h≠0.

Function is like a relationship between two or more variables expressed in equal to form.The value which we entered in the function is known as domain and the value which we get after entering the values is known as codomain or range.

f(a)=4-2a+6 $a^{2}$ (By just putting x=a).

f(a+h)== $4-2(a+h)+6(a+h)^{2}$

=4-2a-2h+6( $a^{2} +h^{2} +2ah$ )

=4-2a-2h+6 $a^{2} +6h^{2} +12ah$

= $6a^{2} +6h^{2}-2a-2h+12ah$

[f(a+h)-f(a)]/h=[ $6a^{2} +6h^{2}-2a-2h+12ah$ -(4-2a+6 $a^{2}$ )]/h

= $(6a^{2} +6h^{2} -2a-2h+12ah)/h$

= $(6h^{2} -2h+12ah)/h$

=6h+12a-2.

Hence the value of function f(a)=4-2a+6 $a^{2}$ , f(a+h) is $6a^{2} +6h^{2} -2a-2h+12ah$ , [f(a+h)-f(a)]/h is 6h+12a-2 in the function f(x)=4-2x+6 $x^{2}$ .

Learn more about function at brainly.com/question/10439235

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