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eimsori [14]
3 years ago
13

How to solve 2.75 + .003+ .158

Mathematics
2 answers:
natita [175]3 years ago
8 0
2.75 + .003 + .158 \\ \\ 2.911 \\ \\ Answer: \fbox {2.911}

This problem can be done using long addition.
bogdanovich [222]3 years ago
3 0
2.75 + .003+ .158 = \boxed{\bf{2.911}}
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The perimeter of a rectangle garden is 48 feet. The length is 5 times the width.
VARVARA [1.3K]
I'll draw a diagram for you, so that it's easier to understand. 
First you have to understand that the perimeter is all the sides add up together.

Let's set up an equation, using l for the length, and w for the width.

5w + w = 48
Simplify
6w = 48
Divide by 6 on both sides
w = 8

Now the value we got is actually both widths (the green sides), so the width is actually 4. Which means that the length is 20. 

The answer is 4ft by 20 ft.

Hopefully this helps! If you have any more questions or don't understand, feel free to DM me, and I'll get back to you ASAP! :)

3 0
3 years ago
Read 2 more answers
Help this is really hard
scoundrel [369]

Answer:

x = 18, y = 45

Step-by-step explanation:

4x - 7 + a = 180 (linear pair)

But, a = 6x + 7 (alternate interior angles)

=> 4x - 7 + 6x + 7 = 180

=> 10x = 180

=> x = 18

Now, 3y - 20 = 6x + 7 (vertically opposite angles)

=> 3y - 20 = 6(18) + 7

=> 3y = 108 + 7 + 20

=> 3y = 135

=> y = 45

Hope it helps :)

Please mark my answer as the brainliest

4 0
2 years ago
Read 2 more answers
Pls help me with either one
blagie [28]

7. 65 + 75 = 140

180 - 140 = 40

( I think this is the answer)

6 0
2 years ago
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What is 5738 cents simplified?
galina1969 [7]
It is $57.38 you just multiply it by 1.00 and you get $57.38.
5 0
3 years ago
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The circle given by : x^2 + y^2 - 6y - 12 = 0 can be written as:
deff fn [24]

Step-by-step explanation:

Given equation of circle is,

{x}^{2}  +  {y}^{2}  - 6y - 12 = 0 \\ by \: compairing \: it \: with \:   \\ {x}^{2}  +  {y}^{2}   + 2gx + 2fy = 0 \: we \: get \\2 g = 0 \: or \: g = 0 \\ 2f=  - 6 \\ or \: f=  - 3 \\ c =  - 12

again the another form of circle is,

{x}^{2}  +  {(y - k)}^{2}  = 21 \\ or \:  {x}^{2}  + {y}^{2}  - 2ky +  {k}^{2}  - 21 = 0 \\ by \: compairing \:it \: with \:  \\ {x}^{2}  +  {y}^{2}   + 2gx + 2fy = 0 \: we \: get  \\ g = 0 \\f =  - k \\ c =  {k}^{2}  - 21

now equating the values of f in both equations,

-k=-3

i.e. k=3

therefore k=3

4 0
3 years ago
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