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3 years ago

What is the Lcm of 28 and 32

Mathematics

1 answer:

lesantik [10]3 years ago

6 0

Answer:

The lcm of 28 is 4

The lcm for 32 is also 4

Step-by-step explanation:

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4m^2-3m^6+5m^44m

mamaluj [8]

6th degree polynomial. You look at the largest value of the superscript

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2 years ago

Read 2 more answers

Given: ABCD is a trapezoid, AB = CD, BK ⊥ AD, AK = 10, KD = 20 Find: BC AD

alexdok [17]

Answer:

 BC = 10 and AD = 30

Step-by-step explanation:

In figure-1 , AB = CD ,BK ⊥ AD, AK = 10, KD = 20.

Since, line AD is sum of AK and KD, then

AD = AK + KD

AD = 10 + 20

AD = 30

Since, BC ║AD and BK ⊥ AD then similarly we construct CL ⊥ AD

so, BC = KL and AK = LD

KL = AD - LD

KL = 20 - 10

KL = 10

Since, BC = KL then BC = 10

Hence,  BC = 10 and AD = 30

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3 years ago

Verify identity: (sec(x)-csc(x))/(sec(x)+csc(x))=(tan(x)-1)/(tan(x)+1)

Nikitich [7]

So hmmm let's do the left-hand-side first

$\bf \cfrac{sec(x)-csc(x)}{sec(x)+csc(x)}\implies \cfrac{\frac{1}{cos(x)}-\frac{1}{sin(x)}}{\frac{1}{cos(x)}+\frac{1}{sin(x)}}\implies 
\cfrac{\frac{sin(x)-cos(x)}{cos(x)sin(x)}}{\frac{sin(x)+cos(x)}{cos(x)sin(x)}}
\\\\\\
\cfrac{sin(x)-cos(x)}{cos(x)sin(x)}\cdot \cfrac{cos(x)sin(x)}{sin(x)+cos(x)}\implies \boxed{\cfrac{sin(x)-cos(x)}{sin(x)+cos(x)}}$

now, let's do the right-hand-side then

$\bf \cfrac{tan(x)-1}{tan(x)+1}\implies \cfrac{\frac{sin(x)}{cos(x)}-1}{\frac{sin(x)}{cos(x)}+1}\implies \cfrac{\frac{sin(x)-cos(x)}{cos(x)}}{\frac{sin(x)+cos(x)}{cos(x)}}
\\\\\\
\cfrac{sin(x)-cos(x)}{cos(x)}\cdot \cfrac{cos(x)}{sin(x)+cos(x)}\implies \boxed{\cfrac{sin(x)-cos(x)}{sin(x)+cos(x)}}$

7 0

3 years ago

−5x+2y=9 y=7x solve the system of equation

egoroff_w [7]

Here to solve the equation we need to plug in the value y as given. -5x+2(7x)=9. -5x+14x=9. 9x=9. X=1.

6 0

2 years ago

A recent survey based on a random sample of n=470 voters, predicted that independent candidate for the mayoral election will get

julsineya [31]

I have an expression

$\sigma = \sqrt{p(1-p)/n}$

floating around in my head; let's see if it makes sense.

The variance of binary valued random variable b that comes up 1 with probability p (so has mean p) is

$E( (b-p)^2 ) = (-p)^2(1-p) + (1-p)^2p = p(1-p)$

That's for an individual sample. For the observed average we divide by n, and for the standard deviation we take the square root:

$\sigma = \sqrt{p(1-p)/n}$

Plugging in the numbers,

$\sigma = \sqrt{.24(1-.24)/490} = 0.019 = 1.9\%$

One standard deviation of the average is almost 2% so a 27% outcome was 3/1.9 = 1.6 standard deviations from the mean, corresponding to a two sided probability of a bit bigger than 10% of happening by chance.

So this is borderline suspect; most surveys will include a two sigma margin of error, say plus or minus 4 percent here, and the results were within those bounds.

7 0

3 years ago