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Sophie [7]
3 years ago
12

A landscaper builds a regular hexagonal patio in a circular garden. The area not covered by the patio will be covered in grass.

The radius of the garden is 1.5 m. Find the area covered by grass in the garden.
Mathematics
1 answer:
Taya2010 [7]3 years ago
7 0

The area that's covered by the grass will be 7.065m²

The area of a circle is found by using the formula: = πr²

where, r = radius = 1.5m

π = 3.14

Therefore, to solve the question, we'll slot the value of the radius into the formula and this will be:

Area = πr².

Area = 3.14 × 1.5²

Area = 3.14 × 1.5 × 1.5

Area = 7.065m²

In conclusion, the area is 7.065m²

Read related link on:

brainly.com/question/24705421

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Gnesinka [82]
6x^2+7x-3=\ \ \ \ | write\  7x\ as\ -2x+9x\\\\ 6x^2-2x+9x-3=0\\\\ 2x(3x-1)+3(3x-1)=0\\\\\boxed{(2x+3)(3x-1)}=0
4 0
3 years ago
1. Which expression is equal to 815-97?
lara [203]

Answer:

the answer is C.815+(-97)

Step-by-step explanation:

5 0
2 years ago
Find the solution set to each inequality. Express the solution in set notation.
daser333 [38]

The solution to the inequality 6m + 2 > -27 is  m > -4.33

The solution to the inequality 8(p-6)>4(p-4) is  p > 8

The given inequality is:

6m  +  2  >  - 27

Subtract 2 to both sides of the inequality

6m  +  2 - 2  >  -27  -  2

6m   >  -29

Divide both sides by 6

\frac{6m}{6}> \frac{-29}{6}  \\m  > \frac{-29}{6}\\m > -4.83

For the inequality  8(p-6)>4(p-4)

Expand the inequality using the distributive rule

8p  -  48   >  4p  -  16

Collect like terms

8p  -  4p  >  -16  +  48

4p    >  32

Divide both sides of the inequality 4

\frac{4p}{4} > \frac{32}{4}\\p > 8

The solution to the inequality 6m + 2 > -27 is m > -4.33

The solution to the inequality 8(p-6)>4(p-4) is p > 8

Learn more here: brainly.com/question/15816805

3 0
3 years ago
Hi! Pls help me with this question:
zimovet [89]

Answer:

PR=8

ST=5

Step-by-step explanation:

2x=8

x=4

PR=2x=8

5z=2z+3

3z=3

z=1

ST=5z=5

8 0
3 years ago
(c). Under a set of controlled laboratory conditions, the size of the population P of a certain bacteria culture at time t (in s
Bezzdna [24]

(i) Since P(t) gives the population of the culture after t seconds, the population after 1 second is

P(1) = 3•1² + 3e¹ + 10 = 13 + 3e ≈ 21.155

In Mathematica, it's convenient to define a function:

P[t_] := 3t^2 + 3E^t + 10

(E is case-sensitive and must be capitalized. Alternatively, you could use Exp[t]. You can also specify that the argument t must be non-negative by entering a condition via P[t_ ;/ t >= 0], but that's not necessary.)

Then just evaluate P[1], or N[P[1]] or N <at symbol> P[1] or P[1] // N to get a numerical result.

(ii) The average rate of change of P(t) over an interval [a, b} is

(P(b) - P(a))/(b - a)

Then the ARoC between t = 2 and t = 6 is

(P(6) - P(2))/(6 - 2) ≈ 321.030

In M,

(P[6] - P[2])/(6 - 2)

and you can also include N just as before.

(iii) You want the instantaneous rate of change of P when t = 60 (since 1 minute = 60 seconds). Differentiate P :

P'(t) = 6t + 3e^t

Evaluate the derivative at t = 60 :

P'(60) = 6•60 + 3e⁶⁰ = 360 + 3e⁶⁰

The approximate value is quite large, so I'll just leave its exact value.

In M, the quickest way would be P'[60], or you can differentiate and replace (via ReplaceAll or /.) t with 60 as in D[P[t], t] /. t -> 60.

5 0
2 years ago
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