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3 years ago

1. 1/2 x 4/7 =

Mathematics

2 answers:

zmey [24]3 years ago

8 0

Answer:

1. 1/2 x 4/7 = 2/7

2. 2/3 x 1/6 = 1/9

3. 1/4 divided by 3/8 = 2/3

Step-by-step explanation:

1/2 x 4/7

All you have to do, is multiply from across

1*4/ 2*7

4/14 = 2/7 when it's simplified ( i divided them by 2, the numerator and denominator)

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2/3 x 1/6

same as usual, multiply from across

2*1/3*6

2/18 = 1/9 when it's simplified ( i divided them by 2, both the numerator and denominator)

__________________________________________________________

1/4 ÷ 3/8

KCF = Keep Change flip

Keep the 1/4

Change the division side to multiplication

Flip the 3/8 to it's reciprocal ( 8/3)

1/4 × 8/3

Now multiply from across

1*8/4*3

8/12 = 2/3 when simplified ( divided them by 4, the numerator and denominator)

Hope this helped!

otez555 [7]3 years ago

5 0

Answer:

1=2/7 ,2=1/9 ,3=2/3

Step-by-step explanation:

answer

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$\bf \stackrel{\textit{sum of 22 and \underline{g}ail's height}}{22+g}~~\stackrel{is}{=}~~74\implies 22+g=74$

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3 years ago

The half-life of cesium-137 is 30 years. Suppose we have a 180-mg sample. (a) Find the mass that remains after t years. (b) How

DedPeter [7]

Answer:

a) $Q(t) = 180e^{-0.023t}$

b) 11.4mg of cesium-137 remains after 120 years.

c) 225.8 years.

Step-by-step explanation:

The following equation is used to calculate the amount of cesium-137:

$Q(t) = Q(0)e^{-rt}$

In which Q(t) is the amount after t years, Q(0) is the initial amount, and r is the rate at which the amount decreses.

(a) Find the mass that remains after t years.

The half-life of cesium-137 is 30 years.

This means that Q(30) = 0.5Q(0). We apply this information to the equation to find the value of r.

$Q(t) = Q(0)e^{-rt}$

$0.5Q(0) = Q(0)e^{-30r}$

$e^{-30r} = 0.5$

Applying ln to both sides of the equality.

$\ln{e^{-30r}} = \ln{0.5}$

$-30r = \ln{0.5}$

$r = \frac{\ln{0.5}}{-30}$

$r = 0.023$

$Q(t) = Q(0)e^{-0.023t}$

180-mg sample, so Q(0) = 180

$Q(t) = 180e^{-0.023t}$

(b) How much of the sample remains after 120 years?

This is Q(120).

$Q(t) = 180e^{-0.023t}$

$Q(120) = 180e^{-0.023*120}$

$Q(120) = 11.4$

11.4mg of cesium-137 remains after 120 years.

(c) After how long will only 1 mg remain?

This is t when Q(t) = 1. So

$Q(t) = 180e^{-0.023t}$

$1 = 180e^{-0.023t}$

$e^{-0.023t} = \frac{1}{180}$

$e^{-0.023t} = 0.00556$

Applying ln to both sides

$\ln{e^{-0.023t}} = \ln{0.00556}$

$-0.023t = \ln{0.00556}$

$t = \frac{\ln{0.00556}}{-0.023}$

$t = 225.8$

225.8 years.

8 0

3 years ago

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kogti [31]

Answer:

2.5 pounds

Step-by-step explanation:

6 0

1 year ago

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The Parker family drove a total of 180 miles on their road trip. This distance 1.5 times the distance they drove on the first da

Usimov [2.4K]

180 ÷ 1.5 = 120

Check your answer - 120 × 1.5 = 180

Final answer: 120 miles.

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3 years ago

Find the equation of the tangent line to the curve 2exy=x+y at (0,2).

DiKsa [7]

Answer:

The equation of the tangent line to the curve

3 x - y = 2

Step-by-step explanation:

Step(i):-

Given function = f(x,y) = $2e^{xy} -x-y=0$ ...(i)

Differentiating equation (i) with respective to 'x' , we get

$2 e^{xy} \frac{d}{d x} (x y) -1 -\frac{dy}{dx} =0$

apply formula

$\frac{d}{dx} (UV) = UV^{l} +V U^{l}$

step(ii):-

⇒ $2 e^{xy} (x(\frac{d}{d x} ( y))+y(1)) -1 -\frac{dy}{dx} =0$

⇒ $2 e^{xy} (x(\frac{d}{d x} ( y))+ 2e^{xy} y(1)) -1 -\frac{dy}{dx} =0$

Taking common d y/d x

$(2 e^{xy} (x) -1)\frac{dy}{dx} =1- 2e^{x y} y(1))$

$\frac{dy}{dx} =\frac{1- 2e^{x y} y(1))}{(2 e^{xy} (x) -1)}$

put At (0,2)

$\frac{dy}{dx} =\frac{1- 2e^{0} 2(1))}{(2 e^{0} (0) -1)}=\frac{1-4}{-1} =3$

slope of the curve m = 3

Step(iii):-

The equation of the tangent line to the curve

$y-y_{1} =m(x-x_{1} )$

y - 2 = 3 ( x - 0 )

3 x - y = 2

Final answer:-

The equation of the tangent line to the curve

3 x - y = 2

8 0

3 years ago