Line segment AB has endpoints A(2, -3) and B(-4, 6). What are the coordinates of the midpoint of AB?

Suppose that it costs $16 to place a classified advertisement in the newspaper, plus $6 for each line. Then the cost to place an

nirvana33 [79]

In the question, you were told:

x = # of lines in the ad

y = cost of ad, in dollars

So you're now told 12 lines costs $88.

How do you write that as an ordered pair? Do we write ordered pairs as (x,y) or (y,x)?

4 0

3 years ago

Amir throws a stone off of a bridge into a river. The stone's height (in meters above the water) ttt seconds after Amir throws i

Harrizon [31]

Answer:

1) The vertex form of $h(t) = -5\cdot t^{2}+20\cdot t +160$ is $h -180 = -5\cdot (t-2)^{2}$ , 2) The stone reaches its maximum height 2 seconds after being thrown.

Step-by-step explanation:

1) Given that height of the stone is represented by a second-order polynomial, which depicts a parabola as graph. The best approach to determine the instant when stone reaches its highest is by vertex form, whose form is:

$h-k = C\cdot (t-r)^{2}$

Where:

$r$ , $k$ - Instant and maximum height of the stone, measured in seconds and meters.

$C$ - Vertex constant, which must be negative as there is an absolute maximum, measured in meters per square second.

Let be $h(t) = -5\cdot t^{2}+20\cdot t +160$ , which is transformed into vertex form:

i) $h = -5\cdot t^{2}+20\cdot t +160$ Given

ii) $h = -5\cdot (t^{2}-4\cdot t -32)$ Distributive property/ $(-a)\cdot b = -a\cdot b$

iii) $h = -5\cdot [(t^{2}-4\cdot t +4)+(-36)]$ Existence of additive inverse/Definitions of addition and subtraction

iv) $h = (-5)\cdot (t-2)^{2}+180$ Distributive property/ $(-a)\cdot (-b) = a\cdot b$ /Perfect square binomial

v) $h -180 = -5\cdot (t-2)^{2}$ Compatibility with addition/Existence of additive inverse/Modulative property/Definition of subtraction/Result

The vertex form of $h(t) = -5\cdot t^{2}+20\cdot t +160$ is $h -180 = -5\cdot (t-2)^{2}$ .

2) The time can be extracted from previous results, which indicates that stone reaches its maximum height 2 seconds after being thrown.

4 0

3 years ago

c(1)=−20 c(n)=c(n−1)+10 Find the 2 nd 2 nd 2, start superscript, start text, n, d, end text, end superscript term in the seque

Natalka [10]

Answer:

a(1)=−13

a(n)=a(n−1)+4

Find the

2nd2^{\text{nd}}

2

nd

2, start superscript, start text, n, d, end text, end superscript

term in the sequence⎩

⎪

⎨

⎪

⎧

a(1)=−13

a(n)=a(n−1)+4

Find the

2nd2^{\text{nd}}

2

nd

2, start superscript, start text, n, d, end text, end superscript

term in the sequence

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=4x%20%2B%203y%20%3D%200%20%5C%5C5y%20%2B%2053%20%3D%2011x%20%5C%5C%20" id="TexFormula1" title=

Anvisha [2.4K]

Answer:

x = 3, y = -4

Step-by-step explanation by substitution:

Solve the following system:

{4 x + 3 y = 0 | (equation 1)

5 y + 53 = 11 x | (equation 2)

Express the system in standard form:

{4 x + 3 y = 0 | (equation 1)

-(11 x) + 5 y = -53 | (equation 2)

Swap equation 1 with equation 2:

{-(11 x) + 5 y = -53 | (equation 1)

4 x + 3 y = 0 | (equation 2)

Add 4/11 × (equation 1) to equation 2:

{-(11 x) + 5 y = -53 | (equation 1)

0 x+(53 y)/11 = -212/11 | (equation 2)

Multiply equation 2 by 11/53:

{-(11 x) + 5 y = -53 | (equation 1)

0 x+y = -4 | (equation 2)

Subtract 5 × (equation 2) from equation 1:

{-(11 x)+0 y = -33 | (equation 1)

0 x+y = -4 | (equation 2)

Divide equation 1 by -11:

{x+0 y = 3 | (equation 1)

0 x+y = -4 | (equation 2)

Collect results:

Answer: {x = 3 , y = -4

_____________________________________

Solve the following system:

{4 x + 3 y = 0

5 y + 53 = 11 x

Hint: | Choose an equation and a variable to solve for.

In the first equation, look to solve for x:

{4 x + 3 y = 0

5 y + 53 = 11 x

Hint: | Isolate terms with x to the left hand side.

Subtract 3 y from both sides:

{4 x = -3 y

5 y + 53 = 11 x

Hint: | Solve for x.

Divide both sides by 4:

{x = -(3 y)/4

5 y + 53 = 11 x

Hint: | Perform a substitution.

Substitute x = -(3 y)/4 into the second equation:

{x = -(3 y)/4

5 y + 53 = -(33 y)/4

Hint: | Choose an equation and a variable to solve for.

In the second equation, look to solve for y:

{x = -(3 y)/4

5 y + 53 = -(33 y)/4

Hint: | Isolate y to the left hand side.

Subtract 53 - (33 y)/4 from both sides:

{x = -(3 y)/4

(53 y)/4 = -53

Hint: | Solve for y.

Multiply both sides by 4/53:

{x = -(3 y)/4

y = -4

Hint: | Perform a back substitution.

Substitute y = -4 into the first equation:

Answer: {x = 3 , y = -4

7 0

3 years ago

A plant grows 3 centimeters in 10 days. How much does it grow per day? Write your answer as a fraction in simplest form

ira [324]

10d = 3
d = 3/10
Grows 3/10cm per day

4 0

3 years ago

Read 2 more answers