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Akimi4 [234]
3 years ago
5

A certain rectangle has a perimeter of 22 units and an area of 30 square units. two of the vertices have coordinates at (2,2) an

d (2,7). find the two missing coordinates. use the coordinate plane to support your answer
Mathematics
1 answer:
Zigmanuir [339]3 years ago
5 0
2* (width + length) = 22 which can be restated

A) 2w + 2L = 22

Dividing "A" by 2 we get
A) w + L = 11
B) width * length = 30
length = 30 / width and we put this into equation "A"
 w + 30/width = 11
Multiplying both sides by width
width^2 + 30 = 11w

w^2 -11w + 30 =0
width = 5 and/or width =6
Therefore length = 5 and/or length =6
(2,2) (2,7)
Difference in x = 0 and difference in y = 5
So, we'll make x=6 and the 2 other coordinates are
(8, 2) and (8, 7)

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In the diagram below, f(x)=x^3+2x^2 is graphed. Also graphed is g(x), the result of a translation of f(x). Determine an equation
Art [367]

Answer:

g(x)=x^3+2x^2 - 4

Step-by-step explanation:

See attached worksheet

6 0
2 years ago
:
Anit [1.1K]
B is the correct answer to this question
4 0
3 years ago
The number is three digits, the number is even, the tens digit is half the hundreds digit, the sum of the digit is 14 what is th
ser-zykov [4K]

Let x represent the digit in the hundreds place, y represent the digit in the tens place, and z represent the digit in the ones place. What do we know about each digit?

x: twice y (2, 4, 6, or 8)

y: half of x (1, 2, 3, or 4)

z: an even number (0, 2, 4, 6, or 8).

The sume of the digits is 14. So, x + y + z = 14

Choose one of the digits and look at your options (if any).

when x=2, then y = 1, which means z has to be 11 (NOT VALID)

when x=4, then y = 2, which means z has to be 8 (428 works!)

For fun, let's see if any other numbers work:

when x=6, then y = 3, which means z has to be 5 (NOT VALID)

when x=8, then y = 4, which means z has to be 2 (842 works!)

Answer: 428 and 842 both work



5 0
3 years ago
In a diagram, X is the midpoint of Segment VZ. VW =5, and VY =20. Find the coordinates of W,X, and Y.
guapka [62]

solution:

x(a_{x},b_{x})is midpoint of vz\\
a_{x}=\frac{a_{v}+a_{z}}{2}=\frac{-12+22}{2}=\frac{10}{2}=5\\
b_{x}=\frac{b_{v}+b_{z}}{2}=\frac{0+0}{2}=0\\
and,\\
x(5,0)\\
w(a_{w},b_{w}),v_{w} and w is on the right of v.\\
a_{w}-a_{v}=5\\
a_{w}-(-12)=5\\
a_{w}+12=5\\
a_{w}=5-12=-7\\
w(-7,0)\\
y(a_{y},b_{y})=20 and y is on the right of v.\\
a_{y}-a_{v}=20\\
a_{y}-(-12)=20\\
a_{y}+12=20\\
a_{y}=20-12=8\\
y=(8,0)

7 0
3 years ago
Seventy percent of children who go to the doctor have fevers. Of those with fevers, 30% also have a rash. Of those without fever
Allisa [31]

Answer:

The probability that a child with a rash does not have a fever is 22%

Step-by-step explanation:

1. Probability of having fever:

P(fever)=0.70

2. Probability of not having fever:

P(not fever)=1-P(fever)\\P(not fever)=1-0.70\\P(not fever)=0.30

3. Probability of fave fevers and a rash:

P(fever and rash)=(0.70)(0.30)=0.21

4. Probability of having a rash but not a fever:

P(rash and not fever)=(0.30)(0.20)=0.60

5. Probability of having a rash:

P(rash)=P(rash and fever)+P(rash and no fever)\\P(rash)=0.21+0.06=0.27

6. Probability a child with a rash does not have a fever

P=\frac{P(rash and not fever)}{P(rash)} =\frac{0.06}{0.27} =0.22

22% of the child at the doctor's office with a rash does not have a fever.

5 0
3 years ago
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