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Licemer1 [7]
3 years ago
15

Write 5 5/8 as a improper fraction A. 33/8 B. 45/8 C. 40/8 D. 35/8

Mathematics
2 answers:
Vikentia [17]3 years ago
8 0

Answer:

I believe this would be B

Step-by-step explanation:

The way I do improper fractions is called the 'boot'

It is where you take the whole number and the denominator and circle both of them so it kind of looks like a boot and you multiply those together.

So 5 * 8= 40 then you add your numerator to that which gets you 45, and you keep the denominator the same

So you end up with 45/8

Sorry if its a little confusing

Hope this helps :)

iris [78.8K]3 years ago
4 0

Answer:

B

Step-by-step explanation:

5 is the whole number, so when turning the 5 into fraction of eighths...8*5=40 --> 40/8

Then, we will add the 5/8 to the 40/8. (40/8)+(5/8) = (45/8)

The answer is B. 45/8

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If it cost 4 tokens to park for 1 hour how many tokesns will it cost for to park for 22 hours
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It will cost 88 tokens to park for 22 hours because you will have to multiply the 4 by the 22 to get the 88 tokens so therefore the answer is 88 tokens
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Victoria went shopping for ingredients to make a stew.The table shows the weight and cost of each of the ingredients that she bo
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Answer:

a) the beef cost the most

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7 0
3 years ago
Find an equation of the plane that contains the points p(5,−1,1),q(9,1,5),and r(8,−6,0)p(5,−1,1),q(9,1,5),and r(8,−6,0).
topjm [15]
Given plane passes through:
p(5,-1,1), q(9,1,5), r(8,-6,0)

We need to find a plane that is parallel to the plane through all three points, we form the vectors of any two sides of the triangle pqr:
pq=p-q=<5-9,-1-1,1-5>=<-4,-2,-4>
pr=p-r=<5-8,-1-6,1-0>=<-3,5,1>

The vector product pq x pr gives a vector perpendicular to both pq and pr.  This vector is the normal vector of a plane passing through all three points
pq x pr
=
  i   j   k
-4 -2 -4
-3  5  1
=<-2+20,12+4,-20-6>
=<18,16,-26>

Since the length of the normal vector does not change the direction, we simplify the normal vector as
N = <9,8,-13>

The required plane must pass through all three points.
We know that the normal vector is perpendicular to the plane through the three points, so we just need to make sure the plane passes through one of the three points, say q(9,1,5).

The equation of the required plane is therefore
Π :  9(x-9)+8(y-1)-13(z-5)=0
expand and simplify, we get the equation
Π  :  9x+8y-13z=24

Check to see that the plane passes through all three points:
at p: 9(5)+8(-1)-13(1)=45-8-13=24
at q: 9(9)+8(1)-13(5)=81+9-65=24
at r: 9(8)+8(-6)-13(0)=72-48-0=24
So plane passes through all three points, as required.

3 0
3 years ago
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