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Nezavi [6.7K]
3 years ago
7

How do u figure -11/4+1/2

Mathematics
1 answer:
worty [1.4K]3 years ago
4 0

Answer:

-2 1/4 or -9/4, depending if the question says simplified or not.

Step-by-step explanation:

-11/4 + 1/2

-11/4 + 2/4=-9/4

-9/4= -2 1/4

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assume that at+tu=su the ratio between the lengths of st and tu is 1:3 if su has a length of 148 find the value of tu
g100num [7]
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6 0
3 years ago
Point A and point B are placed on a number line. Point A is located at -20 and point B is 5 less than point A. Which statement a
dusya [7]
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3 0
3 years ago
What is the slope-intercept equation of the line below?
laila [671]

Answer:

There is no image.

Step-by-step explanation:

Can you add a image or something because You did not give a "line below"

7 0
3 years ago
for freinds share 32 tokens at the arcade .if they share them equally ,what fraction of tokens will each person get?            
Anvisha [2.4K]
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6 0
3 years ago
Read 2 more answers
A training field is formed by joining a rectangle and two semicircles, as shown below. The rectangle is 96m long and 74m wide.
Oduvanchick [21]

Answer:

11402.66 m^{2}

Step-by-step explanation:

The width of rectangle is the diameter of the semi-circle part

Area of one semicircle is given by \frac {0.5\pi d^{2}}{4}

Total area of semi circle will be 2\times\frac {0.5\pi d^{2}}{4}

Substituting 74 m for d and \pi as 3.14 we obtain

Total area semi-circle=2\times\frac {0.5*3.14\times 74^{2}}{4}=4298.66 m^{2}

Area of rectangle is given by the product of length and width

Rectangular area=96 m*74 m=7104 m^{2}

Total area of rectangular and semi-circles will be

4298.66 m^{2}+7104 m^{2}=11402.66 m^{2}

Therefore, area of training field is 11402.66 m^{2}

6 0
3 years ago
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