All categories

3 years ago

What does 44/75 equivalent to?????

Mathematics

1 answer:

d1i1m1o1n [39]3 years ago

4 0

Answer:

44 /75

is already in the simplest form. It can be written as 0.586667 in decimal form (rounded to 6 decimal places).

Steps to simplifying fractions

Find the GCD (or HCF) of numerator and denominator

GCD of 44 and 75 is 1

Divide both the numerator and denominator by the GCD

44 ÷ 1

75 ÷ 1

Reduced fraction:

Equivalent fractions: 88/150 132/225 220/375 308/525

You might be interested in

If f(x) = 3x-9 and g(x) = x^2 what is (g °f) (5)

zavuch27 [327]

F(5) = 3(5) - 9 = 6

g o f (5) = g(6) = 6^2 = 36 answer

5 0

3 years ago

Read 2 more answers

How do I find the x? Please and thank you!

maxonik [38]

Answer:

x = 9

Step-by-step explanation:

The two opposite interior angles that are opposite to the exterior angle are congruent as shown in the diagram.

Therefore, based on the exterior angle theorem of a triangle, we have the following equation:

2(7x + 2)° = (17x - 23)

14x + 4 = 17x - 23

Collect like terms

14x - 17x = -4 - 23

-3x = -27

Divide both sides by -3

-3x/-3 = -27/-3

x = 9

3 0

3 years ago

S/4-6.8=-9.8 what does s equal

Hatshy [7]

S/4-6.8=-9.8
S/4= -9.8+6.8
S/4= -3
S=-3*4
S= -12

4 0

3 years ago

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

Deena took six final exams and received these scores:<br> 65, 72, 79, 86, 93, 100.

mash [69]

Answer:for the first exam she made a 65 for the second one she made a 72 for the third one she made a b for the forth one she made a 86 for the fifth one its b and replace it with a c for the last one she made a d

<h2 />

Step-by-step

5 0

2 years ago