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agasfer [191]
4 years ago
13

What are the x-intercepts of the graph

Mathematics
1 answer:
faltersainse [42]4 years ago
7 0

there is no graph here to answer this

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PLEASE HELP ASAP WILL GIVE BRAINLIEST!!!!!!!!! DO NOT JUST GIVE THE ANSWER PLEASE HELP ME WORK THROUGH THE PROBLEM!!!!!!!!!!!!!!
S_A_V [24]
Convert the mixed numbers into improper fractions...

-19/3 - (-13/3)

The LCD is 9

-19/9 + 13 * 3/9

-19 + 3 * 13/9

20/9

to mixed number is 2 2/9

Answer is A
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3 years ago
Compute 7 2/3 + 8 1/2 + 4 1/6. which is the sum in simplified form\
kotegsom [21]
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Read 2 more answers
A. 22<br> B. 30<br> C. 40<br> D. 65
dolphi86 [110]
This answer to this question is b
3 0
3 years ago
Multiply the following: show your SOLUTIONS
abruzzese [7]
The answer for the problems are below in the picture let me know if that makes sense.

5 0
3 years ago
Last year Tyler used 60 feet of fencing to enclose a rectangular garden with an area of 200 ft². This year Tyler plans to use 12
Alik [6]

Answer:

Area of new garden = 800 ft²

Step-by-step explanation:

Given:

Old fencing = Parameter of old garden = 60 ft

Area of old garden = 200 ft

Find:

Area of new garden if Parameter of new garden is 120 ft

Computation:

Assume;

Length of old garden = l

Width of old garden = b

So,

2[l + b] = 60

l + b = 30

l = 30 - b

Area of old garden = 200 ft

l x b = 200

So,

(30 - b)b = 200

b² - 30b + 200 = 0

b² - 20b - 10b + 200

b(b-20) - 10(b-20)

(b-20)(b-10)

So,

Width of old garden = 10 ft

Length of old garden = 20 ft

New parameter is double

So,

Width of new garden = 10 x 2 = 20 feet

Length of old garden = 20 x 2 = 40 feet

Area of new garden = l x b

Area of new garden = 20 x 40

Area of new garden = 800 ft²

4 0
3 years ago
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