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HACTEHA [7]
3 years ago
12

What is 1/8 times 2/7 in simplest form

Mathematics
1 answer:
PtichkaEL [24]3 years ago
7 0

Answer:

decimal form: 0.0357

fraction form: 375/10000

Step-by-step explanation:

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Como fala paralelepípedo em inglês?
galben [10]

Answer:

the question it's Portugis ,and i translate to english = "<em>How do you speak cobblestone in English?"</em><em> </em> (if you are still confused you can ask your question again).

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URGENT!!<br><br> given= L1≈L4<br> prove= L2 ≈ L3 <br><br> what are the statements and reasons
Degger [83]

Answer:

<2≅<3   Given

<2≅<1     verbally opposite angle are equal

<3≅<1    since <2≅<3

<4≅<3    verbally opposite angle are equal

<1≅<4      since <3≅<1

Step-by-step explanation:

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8 0
3 years ago
-10(d+1)=-14<br> What is the value of d
IRINA_888 [86]

Answer:

d=\frac{2}{5}

Step-by-step explanation:

-10 (d+1) = -14

-10d -10 = -14

-10d = -14 + 10

-10d = -4

d = 4/10

d= 2/5

5 0
3 years ago
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Consider the curve of the form y(t) = ksin(bt2) . (a) Given that the first critical point of y(t) for positive t occurs at t = 1
mafiozo [28]

Answer:

(a).   y'(1)=0  and    y'(2) = 3

(b).  $y'(t)=kb2t\cos(bt^2)$

(c).  $ b = \frac{\pi}{2} \text{ and}\  k = \frac{3}{2\pi}$

Step-by-step explanation:

(a). Let the curve is,

$y(t)=k \sin (bt^2)$

So the stationary point or the critical point of the differential function of a single real variable , f(x) is the value x_{0}  which lies in the domain of f where the derivative is 0.

Therefore,  y'(1)=0

Also given that the derivative of the function y(t) is 3 at t = 2.

Therefore, y'(2) = 3.

(b).

Given function,    $y(t)=k \sin (bt^2)$

Differentiating the above equation with respect to x, we get

y'(t)=\frac{d}{dt}[k \sin (bt^2)]\\ y'(t)=k\frac{d}{dt}[\sin (bt^2)]

Applying chain rule,

y'(t)=k \cos (bt^2)(\frac{d}{dt}[bt^2])\\ y'(t)=k\cos(bt^2)(b2t)\\ y'(t)= kb2t\cos(bt^2)  

(c).

Finding the exact values of k and b.

As per the above parts in (a) and (b), the initial conditions are

y'(1) = 0 and y'(2) = 3

And the equations were

$y(t)=k \sin (bt^2)$

$y'(t)=kb2t\cos (bt^2)$

Now putting the initial conditions in the equation y'(1)=0

$kb2(1)\cos(b(1)^2)=0$

2kbcos(b) = 0

cos b = 0   (Since, k and b cannot be zero)

$b=\frac{\pi}{2}$

And

y'(2) = 3

$\therefore kb2(2)\cos [b(2)^2]=3$

$4kb\cos (4b)=3$

$4k(\frac{\pi}{2})\cos(\frac{4 \pi}{2})=3$

$2k\pi\cos 2 \pi=3$

2k\pi(1) = 3$  

$k=\frac{3}{2\pi}$

$\therefore b = \frac{\pi}{2} \text{ and}\  k = \frac{3}{2\pi}$

7 0
4 years ago
ian invested an amount of money at 3% per annum compound interest. At the end of 2 years the value of the investment was £2652.2
Scorpion4ik [409]

Answer:

the amount of money Ian invested is P =  £2,500

Step-by-step explanation:

The standard formula for compound interest is given as;

A = P(1+r/n)^{nt} \\P = \frac{A}{(1+r/n)^{nt}} ...........1\\

Where;

A = final amount/value

P = initial amount/value (principal)

r = rate yearly

n = number of times compounded yearly.

t = time of investment in years

For this case, Given that;

A = £2652.25

t = 2 years

n = 1 (semiannually)

r = 3% = 0.03

substituting the given values into equation 1;

P = \frac{A}{(1+r/n)^{nt}} ...........1\\P = \frac{2652.25}{(1+0.03)^{2}} \\P = \frac{2652.25}{(1.03)^{2}} \\

P =  £2,500

the amount of money Ian invested is P =  £2,500

8 0
3 years ago
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