Answer:
RL=5x+28 and
RO=8X-11
diagonal of square bisect equally the side
:.5x+28=8x-11
11+28=8x-5x
39=3x
x=39/3=13
<u>RY</u><u>=</u><u>RL</u><u>=</u><u>5</u><u>x</u><u>+</u><u>2</u><u>8</u><u>=</u><u>5</u><u>×</u><u>1</u><u>3</u><u>+</u><u>2</u><u>8</u><u>=</u><u>9</u><u>3</u><u>If the answer is 93, move to </u><u>answer</u><u>.</u>
Answer:
64√2 or 64 StartRoot 2 EndRoot
Step-by-step explanation:
A 45-45-90 traingle is a special traingle. Let's say one of the leg of the triangle is x. The other one is also x because of the isosocles triangle theorem. Therefore, using the pytagorean theorem, you find that x^2+x^2=c^2. 2(x)^2=c^2. You then square root both sides and get c= x√2.
Therefore, the two legs are x and the hypotenuse is x√2. x√2=128 because the question says that the hypotenuse is 128. Solve for x by dividing both sides by √2. X=128/√2. You rationalize it by multiplying the numberator and denominator of the fraction by √2. √2*√2= 2.
X=(128√2)/2= 64√2 cm.
Since X is the leg, the answer would be 64√2
Answer: 14/25 = 56/100
11/20 = 55/100
14/25 > 11/20
Step-by-step explanation:
The least common multiple of 20 and 25 is 100.
(100 = 20*5, 25*4)
So we can take the common denominator 100.
14*4/25*4 =56/100. ∴14/25 = 56/100
11*5/20*5 =55/100. ∴11/20 = 55/100
56>55
∴56/100 > 55/100
∴14/25 > 11/20
Hope it's good;)
Answer:
1.
2.543.6
Step-by-step explanation:
We are given that
y(0)=200
Let y be the number of bacteria at any time
=Number of bacteria per unit time


Where k=Proportionality constant
2.
,y'(0)=100
Integrating on both sides then, we get

We have y(0)=200
Substitute the values then , we get


Substitute the value of C then we get





Differentiate w.r.t

Substitute the given condition then, we get



Substitute t=2
Then, we get 

e=2.718
Hence, the number of bacteria after 2 hours=543.6