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alisha [4.7K]
2 years ago
15

Hi, Guys This is Math 5th Grade I was Doing my homeWork when I did not know a question and I decided maybe 1 of you guys can hel

p me I do not understand the question pls explain it to me as well as give me the answer I will Mark as Brainlist But remember if you steal my points I can simply Steal your points back by reporting you Pls Help me I used all my points for this 1 questions PLSSS

Mathematics
1 answer:
alexgriva [62]2 years ago
6 0

Answer:

Circle A's attributes were used to be organized because both shapes have a <u>right angle</u>. Circle B's attributes were used to be organized because both shapes have <u>2 sets of parallel lines</u> that are across from each other. The middle was organized because both shapes have <u>2 sets of parallel lines</u> and <u>4 right angles</u>.

Step-by-step explanation:

Let me know if this helps!

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Initially, there are 40 grams of A and 50 grams of B, and for each gram of B, 2 grams of A is used. It is observed that 15 grams
hram777 [196]

Answer:

X(16)=25.71grams

Step-by-step explanation:

let X(t) denote grams of C formed in  t mins.

For X grams of C we have:

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Amounts of A,B remaining at any given time is expressed as:

40-\frac{2}{3}Xg of A and  50-\frac{1}{3}Xg  of B

Rate at which C is formed satisfies:

\frac{dX}{dt} \infty(40-\frac{2}{3}X)(50-\frac{1}{3}X)->\frac{dX}{dt}=k(90-X)\\\therefore \frac{dX}{(90-X)^2}=kdt->\int{\frac{dX}{(90-X)^2}} \, =\int {k} \, dt  \\\therefore \frac{1}{90-X}=kt+c->90-X=\frac{1}{kt+c}\\\\X(t)=90-\frac{1}{kt+c}

Apply the initial condition,X(0)=0 ,to the expression above

0=90-\frac{1}{c} \ \ ->c=\frac{1}{90}\\\therefore\\X(t)=90-\frac{1}{kt+\frac{1}{90}} \ \ ->X(t)=90-\frac{90}{90kt+c}

Now at X(8)=15:

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Substitute  in X(t) to get

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3 years ago
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Step-by-step explanation:

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