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2 years ago

Please help with this math

Mathematics

2 answers:

Nezavi [6.7K]2 years ago

6 0

Answer:

jqajgabaNJabba

Step-by-step explanation:

bjHBabBJajabanaa

liq [111]2 years ago

3 0

$x=2+½*\sqrt{10}$

$x=2-½*\sqrt{10}$

Answer:

Solution given:

y=2x²-5x+6.....[1]

y=3x+3.........….[2]

solving equation 1&2

2x²-5x+6=3x+3

2x²-5x-3x+6-3=0

2x²-8x+3=0

comparing above equation with ax²+bx+c=0 we get

a=2

b=-8

c=3

by using quadratic equation

$x=\frac{+8±\sqrt{(-8)²-4*2*3}}{2*2}$

$x=\frac{+8±2\sqrt{10}}{4}$

taking positive

$x=\frac{+8+2\sqrt{10}}{4}$

$x=2+½*\sqrt{10}$

taking negative

$x=2-½*\sqrt{10}$

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Answer:

so this would be the equation

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Radioactive Decay:

Vadim26 [7]

The question is incomplete, here is the complete question:

The half-life of a certain radioactive substance is 46 days. There are 12.6 g present initially.

When will there be less than 1 g remaining?

<u>Answer:</u> The time required for a radioactive substance to remain less than 1 gram is 168.27 days.

<u>Step-by-step explanation:</u>

All radioactive decay processes follow first order reaction.

To calculate the rate constant by given half life of the reaction, we use the equation:

$k=\frac{0.693}{t_{1/2}}$

where,

$t_{1/2}$ = half life period of the reaction = 46 days

k = rate constant = ?

Putting values in above equation, we get:

$k=\frac{0.693}{46days}\\\\k=0.01506days^{-1}$

The formula used to calculate the time period for a first order reaction follows:

$t=\frac{2.303}{k}\log \frac{a}{(a-x)}$

where,

k = rate constant = $0.01506days^{-1}$

t = time period = ? days

a = initial concentration of the reactant = 12.6 g

a - x = concentration of reactant left after time 't' = 1 g

Putting values in above equation, we get:

$t=\frac{2.303}{0.01506days^{-1}}\log \frac{12.6g}{1g}\\\\t=168.27days$

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2 years ago

What is the area of the composite figure

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Answer:

Step-by-step explanation:

To find the area of a composite figure, separate it into the regular parts which make the irregular shape. This shape is composed of a semi-circle and a rectangle. Find the area by finding the area of each shape.

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The semi-circle has a diameter of 2 + 4 + 2 = 8. The area this figure uses the radius which is half the diameter. The radius is 4. To find the area substitute r = 4 into $\pi r^2 = \pi 4^2 = 16\pi$ . However the semi-circle has a smaller circle cut out of it with radius 2. The area of the smaller circle is $\pir^2 = \pi 2^2=4\pi$ . The semi circle in the shape is the areas subtracted which equals 12π.

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2 years ago

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Talja [164]

Answer:

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Step-by-step explanation:

Using the Quotient Rule>

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