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Soloha48 [4]
3 years ago
10

Perform the indicated operation 1/3 divide 3/8

Mathematics
2 answers:
Fittoniya [83]3 years ago
8 0
1 × 8            8              8
-------    =   -------    =    ---
3 × 3         3 × 3          9


So, your answer is   \frac{8}{9}
ella [17]3 years ago
7 0

Answer:

\frac{8}{9}

Step-by-step explanation:

perform the indicated operation 1/3 divide 3/8

divide the fraction 1/3  and 3/8

When we divide the fractions, we flip the second fraction and multiply with first fraction.

\frac{\frac{1}{3} }{\frac{3}{8} }

When we flip 3/8 it becomes 8/3

\frac{1}{3} \cdot \frac{8}{3}

Multiply numerator with numerator and denominator with denominator

1 times 8 is 8

3 times 3 is 9

So final answer is \frac{8}{9}

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Answer:

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Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Given: KLMN is a trapezoid m∠N = m∠KML ME ⊥ KN , ME = 3√5 , KE = 8, LM:KN = 3:5 Find: KM, LM, KN, Area of KLMN
lora16 [44]
Q1)Find KM
As ME is perpendicular to KN, ∠KEM is a right angle
Therefore ΔKEM is a right angled triangle 
KE is given and and ME is also given, we need to find KM
for this we can use Pythogoras' theorem where the square of hypotenuse is equal to the sum of the squares of the adjacent sides.
KM² = KE² + ME²
KM² = 8² + (3√5)²
       = 64 + 9x5
KM = √109
KM = 10.44

Q2)Find LM
It is said that ratio of LM:KN is 3:5
Therefore if we take the length of one unit as x
length of LM is 3x
and the length of KN is 5x
KN is greater than LM by 2 units 
If we take the figure ∠K and ∠N are equal. 
Since the angles on opposite sides of the bases are equal then this is called an isosceles trapezoid. So if we draw a line from L that cuts KN perpendicularly at D, ΔKEM and ΔLDN are congruent therefore KE = DN
since KN is greater than LM due to KE and DN , the extra 2 units of KN correspond to 16 units 
KN = LM + 2x 
2x = KE + DN
2x = 8+8
x = 8
LM = 3x = 3*8 = 24

Q3)Find KN 
Since ∠K and ∠N are equal, when we take the 2 triangles KEM and LDN, they both have;
same height ME = LD perpendicular distance between the 2 parallel sides 
same right angle when the perpendicular lines cut KN
∠K = ∠N 
when 2 angles and one side of one triangle is equal to two angles and one side on another triangle then the 2 triangles are congruent according to AAS theorem (AAS). Therefore KE = DN 
the distance ED = LM
Therefore KN = KE + ED + DN
 since ED = LM = 24
and KE + DN = 16
KN = 16 + 24 = 40
another way is since KN = 5x and x = 8
KN = 5 * 8 = 40

Q4)Find area KLMN
Area of trapezium can be calculated using the following general equation 
Area = 1/2 x perpendicular height between parallel lines x (sum of the parallel sides)
where perpendicular height - ME
2 parallel sides are KN and LM
substituting values into the general equation
Area = 1/2 * ME * (KN+ LM) 
         = 1/2 * 3√5 * (40 + 24)
         = 1/2 * 3√5 * 64
         = 3 x 2.23 * 32
         = 214.66 units²


8 0
3 years ago
Read 2 more answers
Find the sum and express it in simplest form. (ab+4a-6)+(ab+6)
riadik2000 [5.3K]
= 2ab + 4a + 6 - 6

= 2ab + 4a

you can factor this  to 2a(b + 2)
3 0
3 years ago
Read 2 more answers
How would I solve this problem in steps pls
gregori [183]
In order to solve for a, we must isolate it. To do this, we should add c to both sides, making a the only term on the left.

Final answer: a=c+d-r
3 0
3 years ago
true or false If x represents a random variable with mean 114 and standard deviation 40, then the standard deviation of the samp
Goshia [24]

Answer:

\mu = 114, \sigma = 40

We also know that we select a sample size of n =100 and on this case since the sample size is higher than 30 we can apply the central limit theorem and the distribution for the sample mean would be given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the standard deviation for the sampling distribution would be:

\sigma_{\bar X}= \frac{40}{\sqrt{100}}= 4

So then the answer is TRUE

Step-by-step explanation:

Let X the random variable of interest and we know that the true mean and deviation for this case are given by:

\mu = 114, \sigma = 40

We also know that we select a sample size of n =100 and on this case since the sample size is higher than 30 we can apply the central limit theorem and the distribution for the sample mean would be given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the standard deviation for the sampling distribution would be:

\sigma_{\bar X}= \frac{40}{\sqrt{100}}= 4

So then the answer is TRUE

6 0
3 years ago
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