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DENIUS [597]
3 years ago
9

What is the reciprocal of 1 and 4/5?

Mathematics
1 answer:
Mashutka [201]3 years ago
7 0

Answer:

4/5=5/4

1=1

:) hope this helped!

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PLEASE EXPLAIN YOUR REASONING!!
xxTIMURxx [149]

Answer:

$120.00

Step-by-step explanation:

2000/100 = $20 = 1%

20x12 = $240 = annual intrest = 12m

June 15 - December 15 = 6m

240/2 = $120 for half a year intrest!

3 0
3 years ago
Can someone help me out with this question??
dybincka [34]

Answer:

No of cookies baked by Eva=24

No of cookies baked by Jamir=?

Total cookies=56

Solution

24+p=56

p=56-24

p=32

8 0
3 years ago
Plz help, really would appreciate it!!! =]
Nonamiya [84]

Answer:

144 x 7 = 1008

The train can hold 1008 people

Mark brainlest

Step-by-step explanation:

4 0
3 years ago
guys it's really important I'm begging you to answer as many as possible!!!!! it's a practice paper due tomorrow please answer a
alexira [117]

Step-by-step explanation:

5b.

P (6, -1), Q (1, 3), R (x, 8)

the distance between points is found via Pythagoras for right-angled triangles, as the distance is the Hypotenuse c, and the coordinate differences of x and y are the "legs".

for PQ

c² = (6-1)² + (-1 - 3)² = 5² + (-4)² = 25 + 16 = 41

now we need find the x, so that we get 41 as the square of the distance for QR

41 = (1-x)² + (3-8)² = (1-x)² + (-5)² = (1-x)² + 25

aha ! the 41 and 25 terms are the same as above.

that means (1-x)² must be equal to 16.

=>

1-x = 4 => x = -3

or

1-x = -4 => x = 5

so, R could be either (-3, 8) or (5, 8).

5c.

(4 + sqrt(7))/(2×sqrt(7) + 5) = a + b×sqrt(7)

let's use the old trick of

(a+b)(a-b) = a² - b²

to get rid of the sqrt(7) in the denominator.

we multiply with

(2×sqrt(7) - 5)/(2×sqrt(7) - 5)

=>

(4 + sqrt(7))×(2×sqrt(7) - 5) / (2×sqrt(7) + 5)×(2×sqrt(7) - 5) =

= (8×sqrt(7) - 20 +2×7 - 5×sqrt(7)) / (4×7 - 25) =

= (3×sqrt(7) - 6) / 3 = sqrt(7) - 2

now, we see

sqrt(7) - 2 = a + b×sqrt(7)

a = the sum of every term without sqrt(7) = -2

b = the sum of all factors of sqrt(7) = 1

6a.

similar triangles.

EBA and DFA are similar triangles due to the parallel baselines and the sharing of the same "legs" and therefore using the same angles.

and that means that all the lengths of lines in one triangle are the result of the lengths of the corresponding lines in the other triangle multiplied by the same factor.

since E is the midpoint of AD, that means this factor is 2 :

AE×2 = AD.

therefore, FA = AB×2 and AB = FA/2

that proves that B is the midpoint of FA.

that also means that L is the midpoint of DF.

and DF = EB×2 or EB = DF/2

as L is the midpoint of DF, it means that LF = DF/2 = EB.

6b.

x - 2/x = 5

i.

let's square everything.

(x - 2/x)² = 25

x² - 2×2x/x + 4/x² = 25

x² - 4 + 4/x² = 25

x² + 4/x² = 29

ii.

let's multiply with (x - 2/x) again.

(x² + 4/x²)×(x - 2/x) = 29×5

x³ - 2x²/x + 4x/x² - 8/x³ = 145

x³ - 2x + 4/x - 8/x³ = 145

x³ - 8/x³ - 2×(x - 2/x) = 145

x³ - 8/x³ - 2×5 = 145

x³ - 8/x³ = 155

6c.

I can't draw here, but let's transform the equations into regular line equations :

2y - x = 8

2y = x + 8

y = (1/2)x + 4

when assuming x = 0 for one point. and y = 0 for a second point we get (0, 4) and (-8, 0)

y - 2x = 1

y = 2x + 1

same trick as above with x = 0 and y = 0 :

(0, 1) and (-1/2, 0). but we could also use x=1 to get an integer as x coordinate for the second point: (1, 3)

7a.

i.

the radius is half the diameter = (18+8)/2 = 13 cm

ii.

the shortest chord through M creates a right-angled triangle with the radius from 0 to the point on the circle, where the chord ends as the Hypotenuse. one leg is M0, which is

13 - 8 = 5 cm

the other leg is half of the length of the chord, as the chord goes up and down of AB.

so,

13² = 5² + x²

169 = 25 + x²

144 = x²

x = 12 cm

so, the whole chord is 2×12 = 24cm long.

7b.

3/(sqrt(6) + sqrt(3)) rationalized is

3×(sqrt(6) - sqrt(3)) / (sqrt(6) + sqrt(3))×(sqrt(6) - sqrt(3))

3×(sqrt(6) - sqrt(3)) / (6 - 3) =

3×(sqrt(6) - sqrt(3)) / 3 = sqrt(6) - sqrt(3)

1/(sqrt(3) + sqrt(2)) rationalized is

(sqrt(3) - sqrt(2)) / (sqrt(3) + sqrt(2))×(sqrt(3) - sqrt(2))

(sqrt(3) - sqrt(2)) / (3 - 2) = sqrt(3) - sqrt(2)

4/(sqrt(6) + sqrt(2)) rationalized is

4×(sqrt(6) - sqrt(2)) / (sqrt(6) + sqrt(2))×(sqrt(6) - sqrt(2))

4×(sqrt(6) - sqrt(2)) / (6 - 2) =

4×(sqrt(6) - sqrt(2)) / 4 = sqrt(6) - sqrt(2)

so, we have in total

sqrt(6) - sqrt(3) + sqrt(3) - sqrt(2) - sqrt(6) + sqrt(2) = 0

the whole expression is eliminating every term and is simply 0.

3 0
2 years ago
A Customer Telephone Center receives 1,200 calls in a 24-hour period. Of these calls, 75% occur between 9:30 a.m. and 3:30 p.m.,
Volgvan

Divide Total time into two Time Periods. A regular Demand and a High Demand Time Period. Since 75% of the 1,200 calls occur between 9:30 am and 3:30 pm. We multiply .75 x 1,200 to get 900. We get an average of 900 calls every day between the hours of 9:30 am and 3:30 pm – Our net time for this time period is 6 hours.

Therefore, 6 Hours/900 becomes our quotient

Again, since the denominator is larger we invert it to 900 calls/6 Hours

To get a demand of 150 calls per hour.

We need to be able to handle 150 calls per hour.

So how Many Call Representatives are needed?

Again, our historical data tells us that each person can handle 10 calls an hour.

Therefore 150 calls per hour /10 Minutes = 15 Customer Service Representatives are needed during Peak Time!

Now, Subtract the Peak Hours from the 21 Hours Net Time Per day, gives us 15 Non-Peak hours we have to staff.

8 0
3 years ago
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