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3 years ago

Whats 1 whole 3\8 divided by 3\4 = please show the work

2 answers:

5 0

you need to change 1 3/8 into a mixed fraction, which would be 11/8, but with this fraction, you need to do the reciprocal, which would be 8/11 multiplied by 3/4. You can cross simplify 8 and 4 changing the fractions being multiplied to 2/11 and 3/1. Then you need to multiply straight across, which would get an answer of 6/11.

Yuliya22 [10]3 years ago

4 0

1 3/8 divided by 3/4 is the same as saying multiply 1 3/8 by 4/3 ( reciprocal of 3/4) convert 1 3/8 to 11/8 11/8 x4/3 multiply the numerators and denominators 44/24=11/6 Hope this helps

You might be interested in

70 decreased by 13% =

Pie

If you have a calculator:
1) To find 1%, divide by 100 (70 / 100 = 0.7) 

2) Then, multiply by the % you need (0.7 x 13 = 9.1) 

Therefore 70 - 9.1 = 60.9 

If you don't have a calculator: 
1) Find 50%, 10%, 5%, and 1%. With these, you can make any % you need. 
50% = X / 2 (= 35) 
10% = X / 10 (= 7) 
5% = first, do 10%, then divide that by 2 (= 3.5) 
1% = X / 100 (= 0.7) 
Where X is the number you want to work out % for 

2) Add those % up to get the amount you want 
10% + 1% + 1% +1% = 13% 
7 + 0.7 + 0.7 + 0.7 = 9.1 

Therefore 70 - 9.1 = 60.9
Hope that helps :)

7 0

3 years ago

Please answer this question with explanation!! Thanks

Georgia [21]

The answer would be 12 3/4 because when you continue use the pattern that is used you end up with your answer! Hope this helps

5 0

4 years ago

Dan recently bought a new mountain bike for $550. His friend works

Wittaler [7]

$990
550 x .80 = 440
$550 + $440 = $990

8 0

4 years ago

Pls answer 13 b ill mark u brainlist

gladu [14]

Answer:

7 students

Step-by-step explanation:

1+2+4=7students

5 0

3 years ago

Determine whether the given vectors are orthogonal, parallel or neither. (a) u=[-3,9,6], v=[4,-12,-8,], (b) u=[1,-1,2] v=[2,-1,1

nevsk [136]

Answer:

a) $u v= (-3)*(4) + (9)*(-12)+ (6)*(-8)=-168$

Since the dot product is not equal to zero then the two vectors are not orthogonal.

$|u|= \sqrt{(-3)^2 +(9)^2 +(6)^2}=\sqrt{126}$

$|v| =\sqrt{(4)^2 +(-12)^2 +(-8)^2}=\sqrt{224}$

$cos \theta = \frac{uv}{|u| |v|}$

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

If we replace we got:

$\theta = cos^{-1} (\frac{-168}{\sqrt{126} \sqrt{224}})=cos^{-1} (-1) = \pi$

Since the angle between the two vectors is 180 degrees we can conclude that are parallel

b) $u v= (1)*(2) + (-1)*(-1)+ (2)*(1)=5$

$|u|= \sqrt{(1)^2 +(-1)^2 +(2)^2}=\sqrt{6}$

$|v| =\sqrt{(2)^2 +(-1)^2 +(1)^2}=\sqrt{6}$

$cos \theta = \frac{uv}{|u| |v|}$

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

$\theta = cos^{-1} (\frac{5}{\sqrt{6} \sqrt{6}})=cos^{-1} (\frac{5}{6}) = 33.557$

Since the angle between the two vectors is not 0 or 180 degrees we can conclude that are either.

c) $u v= (a)*(-b) + (b)*(a)+ (c)*(0)=-ab +ba +0 = -ab+ab =0$

Since the dot product is equal to zero then the two vectors are orthogonal.

Step-by-step explanation:

For each case first we need to calculate the dot product of the vectors, and after this if the dot product is not equal to 0 we can calculate the angle between the two vectors in order to see if there are parallel or not.

Part a

u=[-3,9,6], v=[4,-12,-8,]

The dot product on this case is:

$u v= (-3)*(4) + (9)*(-12)+ (6)*(-8)=-168$

Since the dot product is not equal to zero then the two vectors are not orthogonal.

Now we can calculate the magnitude of each vector like this:

$|u|= \sqrt{(-3)^2 +(9)^2 +(6)^2}=\sqrt{126}$

$|v| =\sqrt{(4)^2 +(-12)^2 +(-8)^2}=\sqrt{224}$

And finally we can calculate the angle between the vectors like this:

$cos \theta = \frac{uv}{|u| |v|}$

And the angle is given by:

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

If we replace we got:

$\theta = cos^{-1} (\frac{-168}{\sqrt{126} \sqrt{224}})=cos^{-1} (-1) = \pi$

Since the angle between the two vectors is 180 degrees we can conclude that are parallel

Part b

u=[1,-1,2] v=[2,-1,1]

The dot product on this case is:

$u v= (1)*(2) + (-1)*(-1)+ (2)*(1)=5$

Since the dot product is not equal to zero then the two vectors are not orthogonal.

Now we can calculate the magnitude of each vector like this:

$|u|= \sqrt{(1)^2 +(-1)^2 +(2)^2}=\sqrt{6}$

$|v| =\sqrt{(2)^2 +(-1)^2 +(1)^2}=\sqrt{6}$

And finally we can calculate the angle between the vectors like this:

$cos \theta = \frac{uv}{|u| |v|}$

And the angle is given by:

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

If we replace we got:

$\theta = cos^{-1} (\frac{5}{\sqrt{6} \sqrt{6}})=cos^{-1} (\frac{5}{6}) = 33.557$

Since the angle between the two vectors is not 0 or 180 degrees we can conclude that are either.

Part c

u=[a,b,c] v=[-b,a,0]

The dot product on this case is:

$u v= (a)*(-b) + (b)*(a)+ (c)*(0)=-ab +ba +0 = -ab+ab =0$

Since the dot product is equal to zero then the two vectors are orthogonal.

5 0

4 years ago

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