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Taya2010 [7]
3 years ago
6

Perry and Jacob each improved their yards by planting hostas and ivy. They bought their supplies from the same store. Perry spen

t $122 on 7 hostas and 5 pots of ivy. Jacob spent $100 on 5 hostas and 5 pots of ivy. What is the cost of one hosta and the cost of one pot of ivy. ENTER your answer in the format hosta=## ivy=##
Mathematics
1 answer:
marissa [1.9K]3 years ago
3 0

Answer:

hosta=11 ivy=9

Step-by-step explanation:

Logically thinking... $20 for 1 h and 1 i

So I would say $11 H and $9 I

You might be interested in
Please help:
8_murik_8 [283]
7 7/9 clergs.
wergs= w
nergs= n
clergs= c

3w= 4n, so 5w= 6 2/3n
Since 6n=7c, then 6 2/3n= 7 7/9n


Hope that answered your question!!
5 0
3 years ago
82, 62, 95, 81, 89, 51, 72, 56, 97, 98, 79, 85 order from least to greatest find the The median of the lower half of the data an
Studentka2010 [4]

Answer:

The median of the lower half of the data was determined to be 67 while the median of the upper half of the data was determined to be 92.

Step-by-step explanation:

<u>Step 1:  Order from least to greatest</u>

51, 56, 62, 72, 79, 81, 82, 85, 89, 95, 97, 98

<u>Step 2:  Determine the median of the lower half of the data</u>

There are 12 numbers which means that there is going to be 6 numbers on both sides of the data.  So using the first 6 numbers we will be able to determine the median of the lower half of the data.

51, 56, 62, 72, 79, 81

Since there is an even amount of numbers, that means that we will need to find the middle between the 3rd and 4th number.  We can do that by adding them up together and dividing by 2.

(62 + 72) / 2 = 67

Step 3:  <u>Determine the median of the upper half of the data</u>

So using the first 6 numbers we will be able to determine the median of the upper half of the data.

82, 85, 89, 95, 97, 98

Since there is an even amount of numbers, that means that we will need to find the middle between the 3rd and 4th number.  We can do that by adding them up together and dividing by 2.

(89 + 95) / 2 = 92

Answer: The median of the lower half of the data was determined to be 67 while the median of the upper half of the data was determined to be 92.

5 0
2 years ago
1.) What is the equation of the path of firework #1? Write your equation in general form.
valina [46]

1. I'm assumig that the paths are perfect parabolas

this means that their general forms can be written in y=ax^2+bx+c

it's easier to find vertex form first then expand to get general form

vertex form is y=a(x-h)^2+k where the vertex is (h,k) and a is a constant


firework #1

vertex is (10,50), so (h,k)=(10,50) and h=10, k=50

h_1=a(t-10)^2+50

to find the value of a, subsitute another point

(0,0)

0=a(0-10)^2+50)

0=100a+50

a=\frac{-1}{2}

so the equation in vertex form is h_1=\frac{-1}{2}(t-10)^2+50

expand to get general form

h_1=\frac{-1}{2}(t^2-20t+100)+50

h_1=\frac{-1}{2}t^2+10t-50+50

h_1=\frac{-1}{2}t^2+10t





2.

same as last time

vertex is (10,72) so (h,k)=(10,72) so h=10 and k=72

equation is h_2=a(t-10)^2+72

find a

use another point

(0,22)

22=a(0-10)^2+72

22=100a+72

-50=100a

a=\frac{-1}{2}

so the equation in vertex form is h_2=\frac{-1}{2}(t-10)^2+72




3.

range is the numbers that h is allowed to be

think about what h represents. it represents the height of the rocket

from the graph, we can see that the lowest possible height is 0yd and the highest height is 50yd

so range is 0 to 50 or 0≤h≤50


domain is the numbers that t is allowed to be

think about what t represents. it represents how long the rocket has been flying

it will stop flying when it hits the ground or at t=20

it starts flying at t=0

so domain is from 0 to 20 or 0≤t≤20

3 0
3 years ago
carol bought 8 identical boxes for a total of 3.60$. how much did each box cost? which statement would you use?
Tju [1.3M]
You just divide the 3.60 by the eight and then you get 45 cents for each box.
7 0
3 years ago
Read 2 more answers
If someone could help me with this, that would be great.
Snowcat [4.5K]
I’m 99% sure the answer is:

c: 2t - d = 260
t - d = 320
5 0
3 years ago
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