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2 years ago

4 1/4 divided by 3 and 1/2

Mathematics

1 answer:

kenny6666 [7]2 years ago

5 0

Answer:

1 3/14

Step-by-step explanation:

1. You want to turn the fractions into mixed numbers.

2. 4 1/4 is equal to 17/4

3. 3 1/2 is equal to 7/2

4. I have a fraction calculator and got 1 and 6/28.

5. To solve it yourself, flip 7/2 to make it 2/7 and change from division to multiplication.

Looked like: $\frac{17}{4}$ ÷ $\frac{7}{2}$

Looks like: $\frac{17}{4}$ · $\frac{2}{7}$

17 times 2 is 34 and 4 times 7 is 28. Leaving us with 34/28.

Now just simplify and get an answer of 1 and 3/14.

Hope this helps! Please vote me as Brainliest. Let me know if you need more of an explanation. Remember, cheating is never the answer. Try your best before you seek answers. Hope it's not too late. Good luck with whatever you are doing! Have an amazing day!

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Measurement of angle 1

OLEGan [10]

Answer:

55 degrees

Step-by-step explanation:

Given that a circle and inside two chords with same arc length.

We are to find the angle between the two chords.

Given that two arcs subtend angle 125 degrees at the centre.

Let us join the two ends of chords to make the figure as a triangle inside a circle.

The triangle is isosceles as two arcs and hence chords are equal.

By central angle theorem we have the two equal angles as 1/2 (125) = 62.5

Hence we have a triangle with two equal angles 62.5 and another angle 1.

By triangle sum of angles theorem

angle 1+62.5+62.5 = 180

Hence angle A = 180-62.5-62.5 = 55 degrees.

5 0

2 years ago

Multiply. <br><br> 4 × 0.002 =

trapecia [35]

The answer to your question is 0.008

7 0

3 years ago

2.The number of grams A of a certain radioactive substance present at time, in yearsfrom the present, t is given by the formulaA

professor190 [17]

Answer:

Given that,

The number of grams A of a certain radioactive substance present at time, in years

from the present, t is given by the formula

$A=45e^{-0.0045(t)}$

a) To find the initial amount of this substance

At t=0, we get

$A=45e^{-0.0045(0)}$ $A=45e^0$

We know that e^0=1 ( anything to the power zero is 1)

we get,

$A=45$

The initial amount of the substance is 45 grams

b)To find thehalf-life of this substance

To find t when the substance becames half the amount.

A=45/2

Substitute this we get,

$\frac{45}{2}=45e^{-0.0045(t)}$

$\frac{1}{2}=e^{-0.0045(t)}$

Taking natural logarithm on both sides we get,

$\ln (\frac{1}{2})=-0.0045(t)^{}$ $(-1)\ln (\frac{1}{2})=0.0045(t)$ $\ln (\frac{1}{2})^{-1}=0.0045(t)$ $\ln (2)=0.0045(t)$ $0.6931=0.0045(t)$ $t=\frac{0.6931}{0.0045}$ $t=154.02$

Half-life of this substance is 154.02

c) To find the amount of substance will be present around in 2500 years

Put t=2500

we get,