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Ludmilka [50]
3 years ago
12

In parallelogram DEFG, DH = x + 2, HF = 2y, GH = 3x – 3, and HE = 5y + 1. Find the values of x and y. x = 10, y = 18 x = 18, y =

10 x = 20, y = 11 x = 11, y = 20
Mathematics
2 answers:
strojnjashka [21]3 years ago
8 0

Answer:

I think The answer to your question is 10

Step-by-step explanation:

amid [387]3 years ago
3 0

Answer:

I am not sure about this but the answer is 10

Step-by-step explanation:

if x + 2 wouldn't it be 10 + 2 and i also think y = 10 be 2(10) = 20 and 2 + 10 = 12

if you want me to add it 20 + 12 = 32

  • Hope this helps
  • Hope this is right
  • I hope i get brainliest
  • I hope you get this right
  • Have a nice day
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Given: f(x) = 3x-7, g(x)=2x2 -3x+1, h(x)=4x+1, k(x) =-x2+3 Find: k(x) +g(x)​
Afina-wow [57]

Answer:

k(x) + g(x) = x² - 3x + 4

General Formulas and Concepts:

<u>Alg I</u>

  • Combining like terms

Step-by-step explanation:

<u>Step 1: Define</u>

f(x) = 3x - 7

g(x) = 2x² - 3x + 1

h(x) = 4x + 1

k(x) = -x² + 3

<u>Step 2: Find k(x) + g(x)</u>

  1. Substitute:                              k(x) + g(x) = -x² + 3 + 2x² - 3x + 1
  2. Combine like terms (x²):        k(x) + g(x) = x² + 3 - 3x + 1
  3. Combine like terms (Z):         k(x) + g(x) = x² - 3x + 4
7 0
3 years ago
Complete the following equations . Round each value to the nearest thousandth.
juin [17]
Answered.                            ...........
5 0
3 years ago
Can someone please help me please
Assoli18 [71]
A
Volume of the Cylinder
Givens
H = 60 yards.
Diameter = 20 yards
pi = 3.14

Formula
V = pi * r^2 * h

Calculations
r = d/2
r = 32/2
r = 16

V =  3.14 * 16^2 * 60 
V = 48230 cubic yards [Cylinder's Volume]

Cone
<em>Formula</em> 
V = 1/3 pi r^2 H

<em>Givens</em>
pi = 3.14
r = 16 yards
h = 20 yards

<em>Sub and solve</em>
V = 1/3 3.14 * 16^2 * 20
V = 5359 cubic yards.

<em>Total Volume of the structure</em>
48230 + 5359 = 53589 Cubic Yards

<em>Water Content</em>
The answer to this part requires a proportion.

1 Cubic yard will hold 201.97 gallons.
53589 yd^3 = x

1/201.97 = 53589 /x  [ You should get a pretty big answer]
x = 201.87 * 53589
x = 10 819 092 gallons can be held by the tank. 

10 819 092 gallons <<<< answer

B 
If the height of both the cylinder and the cone remain the same. If the radius doubles in both the cylinder and the cone then the tank will hold 4 times as much. 

Total volume before doubling the radius = pi * r^2 h + 1/3 pi r^2 h
New Total Volume = pi * (2*r)^2 h + 1/3 pi * (2r)^2 h
New Total volume = pi * 4r^2 h + 1/3 pi *4 r^2 h
New Total Volume = 4 [pi r^2 h + 1/3 pi r^2 h]
but pi r^2 h + 1/3 pi r^2 h is the total volume before doubling the radius

New volume = 4 original volume. <<<<< answer to part B
6 0
3 years ago
Kayla and latasha work for a department store the amount of sales they made over the last year was collected monthly in the data
tresset_1 [31]

Based on the information represented by the boxplot ;

  • Latasha's lowest sale amount = 50

  • Kayla's median is between 200 and 300

  • Latasha has a greater spread due to higher IQR value

1.) <em><u>The Lowest amount of sale made by Latasha in one month </u></em>

  • The minimum value is denoted by the starting position of the lower whisker on a boxplot.

  • Lowest amount of sale made by Latasha = 50

2.) <em><u>50</u></em><em><u>%</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>sales</u></em><em><u> </u></em><em><u>made</u></em><em><u> </u></em><em><u>by</u></em><em><u> </u></em><em><u>Kayla</u></em><em><u> </u></em><em><u>:</u></em>

  • 50% of sales made marks the median value in a boxplot, it is denoted by the vertical line in between the box.

  • 50% of sales made by Kayla is between 200 and 300

  • With median sale value being 250

3.) <em><u>Spread</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>middle</u></em><em><u> </u></em><em><u>50</u></em><em><u>%</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>sales</u></em><em><u> </u></em><em><u>:</u></em>

  • The measure of spread of the middle 50% of a distribution on a boxplot is the Interquartile range (IQR) of the distribution

  • IQR = Upper Quartile (Q3) - Lower quartile(Q1)

<u>For Latasha</u> :

  • Q3 = 450 (Endpoint of the box)
  • Q1 = 150 (starting point of the box)

  • IQR = 450 - 150 = 300

<u>For</u><u> </u><u>Kayla</u><u> </u><u>:</u><u> </u>

  • Q3 = 375 (Endpoint of the box)
  • Q1 = 100 (starting point of the box)
  • IQR = 375 - 100 = 275

  • Since, Latasha's IQR is greater than Kayla's, then Latasha has a greater mid 50% spread than Kayla.

Learn more :brainly.com/question/24582786

4 0
2 years ago
Plz help i beg i will give brainlist and 20 points
defon

Answer:

His equation is correct but his calculations are incorrect.

Step-by-step explanation:

180-40=140 so y=140

Hope this helps!

6 0
2 years ago
Read 2 more answers
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