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Rus_ich [418]
3 years ago
6

Determine the area. Can some guide me how to solve this?

Mathematics
1 answer:
san4es73 [151]3 years ago
5 0
Split it up into compound shapes, calculate the areas individually and then add and subtract the relevant quantities.

For instance, the shape in part a can be split up into a quarter circle, a trapezium and a half circle. I have included a diagram that may be of assistance in understanding what I mean but I leave the calculations to you. (They're quite easy when you split the shape up anyway)

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Which is greater? 670.9 or 670.008
Free_Kalibri [48]

Answer:

670.9

Step-by-step explanation:

because 670.9 is bigger and 670.008 has more 0's and 0's aren't greater

5 0
3 years ago
There are 25 quarts of water in a bucket. If you want to bottle it in 1/2 quart bottles, how many bottles do you have
nika2105 [10]
The answer will be 50 quarts using proportion and ratio
6 0
3 years ago
Read 2 more answers
A water sprinkler sprays in a circular pattern a distance of 10 ft from the sprinkler to the edge. What is the circumference of
vfiekz [6]

Answer:

<h2>62.8 ft</h2>

Step-by-step explanation:

7 0
3 years ago
Find a quadratic polynomial the sum and product of whose zeroes are -8 and 12 respectively. Hence find the zeroes.
Olenka [21]

Step-by-step explanation:

We khow the sum and the product of the zeroes of this quadratic polynomial

Here is a trick :

when we khow the sum S and the product P ofvtwo numbers we can find them by solving :

x²-Sx+p=0

here S= -8 and P=12

so:

x²+8x+12=0

Let Δ be the discrminant of this equation: a= 1 , b= 8 and c=12

Δ= 8²-4*12 =16

the zeros are:

(-8-4)/2= -6

(-8+4)/2 = -2

verify:

-6+(-2)= -8

-2*(-6)= 12

now the polynomial quadratic is:

(x+6)(x+2)

7 0
3 years ago
Find all solutions in the interval [0, 2π) sin2x + sin x = 0
Molodets [167]
Sin 2 x + sin x = 0
( sin 2 x = 2 sin x cos x )
2 sin x cos x + sin x  = 0
sin x ( 2 cos x + 1 ) = 0
sin x = 0 
x 1 = 0
x 2 = π 
or: 2 cos x + 1 = 0
2 cos x = - 1
cos x = -1/2
x 3 = 2π/3
x 4 = 4π/3
3 0
3 years ago
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