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Serga [27]
3 years ago
6

1. Divide

Mathematics
1 answer:
BartSMP [9]3 years ago
6 0

Answer:

656×756÷765+656+65=

Step-by-step explanation:

it was soo easy now i get what it is talking about no

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please help if you know! lesson name: L 10.5: Using the Distributive Property here is the pic i don't understand
Andrei [34K]

Answer:

Area of the shaded region = 1.72r²

Step-by-step explanation:

Radius of the circle = r

Length of the rectangle = 4r

Width of the rectangle = 2r

Area of the circle = πr²

Area of the rectangle = Length × Width

= 4r × 2r

= 8r²

Area of the shaded region = Area of the rectangle - 2 × Area of the circle

= 8r² - 2 × πr²

= 8r² - 2πr²

= r²(8 - 2π)

= r²(8 - 2 × 3.14)

= r²(8 - 6.28)

= 1.72r²

4 0
3 years ago
Ammonia (NH3) readily dissolves in water to yield a basic solution
natita [175]
I think it’s Arrhenius acid I’m not so sure
8 0
3 years ago
Let’s think about another type of scenario. What if you were told that a bracelet requires 10 beads and 10 minutes to make while
Georgia [21]

Answer:

$425

Step-by-step explanation:

Let x represent the number of bracelets made and let y represent the number of necklace made.

Since the craftsman has 1000 beads to work with, hence:

10x + 20y ≤ 1000    (1)

Also, the craftsman has 1600 minutes, hence:

10x + 40y ≤ 1600   (2)

From ploting equations 1 and 2 on the geogebra online graphing, we can see that the solution to the problem is (40, 30).

Since the bracelet costs $5 and a necklace costs $7.50, hence the maximum revenue is:

Revenue = 5x + 7.5y = 5(40) + 7.5(30) = $425

4 0
3 years ago
Please help me! I need this answered fast!
zysi [14]

Answer:

Number 5 is D

Number 6 is A

3 0
3 years ago
The population of bacteria in a culture grows at a rate proportional to the number of bacteria present at time t. After 3 hours
Arada [10]

Answer:

The number of bacteria at initial = 187

Step-by-step explanation:

Given that the population of bacteria in a culture grows at a rate proportional to the number of bacteria present at time t.

\frac{dN}{dt} = k N

\frac{dN}{N}  = k dt

Integrating both side we get

㏑ N = k t + C ------- (1)

Now given that after 3 hours it is observed that 500 bacteria are present and after 10 hours 5000 bacteria are present.

⇒ ㏑ 500 = 3 k + C -------- (2)

⇒ ㏑ 5000 = 10 k + C ------ (3)

⇒ ㏑ 5000 -  ㏑ 500 = 7 k

⇒ ㏑\frac{5000}{500} = 7 k

⇒  ㏑ 10 = 7 k

⇒ k = 0.329

Put this value of k in equation (2),

⇒ ㏑ 500 = 3 × 0.329 + C

⇒ C = 5.23

Put this value of C in equation 1 we get,

⇒ ㏑ N = k t + 5.23

Initially when t = 0 , then

⇒ ㏑ N = 5.23

⇒ N =  187

Thus the number of bacteria at initial = 187

8 0
4 years ago
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