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Valentin [98]
3 years ago
13

What is a+4=11? please help

Mathematics
2 answers:
Mashutka [201]3 years ago
3 0

Answer:

A = 7

Step by step:

11 - 4 = 7

So,

7 + 4 = 11

Hope this helps!!!

Please be kind to mark Brainliest.☺

Kipish [7]3 years ago
3 0

Answer:

7

Step-by-step explanation:

a+4=11

a=11-4

a=7

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Find the area of a circle of radius 4 cm.​
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Simone has 16 jumps Garcia has 8 jumps how many more jumps did Simone complete than Garcia
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Read 2 more answers
Can you work out G and H for me please. preferably telling me how you did it aswell
Veronika [31]
Abd is a right angled triangle so by pythagoras we know
{g}^{2}  =  {11}^{2}  +  {13}^{2}  \\  {g}^{2}  = 290 \\ g =  \sqrt{290}  \: or \: g = 17.03cm
and to calculate h we need to use the sine rule
\frac{h}{ \sin(22) }  =  \frac{g}{ \sin(32) }  \\  \frac{h}{ \sin(22) }  =  \frac{17.03}{ \sin(32) }  \\ h =  \frac{17.03}{ \sin(32) } \times  \sin(22)   \\ h = 12.04cm
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8 0
3 years ago
At a recent marathon, spectators lined the street near the starting line to cheer for the runners. The crowd lined up 5 feet dee
masha68 [24]

Answer:

29568 people cheered for the runners at the start of the race

Step-by-step explanation:

From the question, the crowd lined up 5 feet deep on both sides of the street for the first mile.

This lined up crowd could be related to a rectangle that is 1 mile long and 5 feet wide.

First, Convert 1 mile to feet

1 mile = 5280 feet

Hence, the length of the rectangle is 5280 feet and the width is 5 feet.

Now, we will determine how many 5 feet by 5 feet square we can get from the 5280 feet by 5 feet rectangle. To do that, we will divide 5280 feet by 5 feet

5280 feet ÷ 5 feet = 1056

Hence, from the rectangle, we can get 1056 5 feet by 5 feet square.

From the question, you estimate that 14 people can comfortably fit in a square that measures 5 feet by 5 feet,

∴ 14 × 1056 people will comfortably fit in the crowed lined up 5 feet deep on one side of the street for the first mile.

14 × 1056 = 14784 people

This is the amount of people that will comfortably fit in the crowed lined up 5 feet deep on one side of the street for the first mile.

Since the crowd lined up on both sides of the street, then

2 × 14784 people will comfortably fit in the crowed lined up 5 feet deep on both sides of the street for the first mile

2 × 14784 = 29568 people

Hence, 29568 people cheered for the runners at the start of the race.

5 0
3 years ago
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