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choli [55]
3 years ago
7

A boy walks meter in 6 min. how many meter can he walk in 1 hour​

Mathematics
2 answers:
AVprozaik [17]3 years ago
7 0

Answer:

10 meters

Step-by-step explanation:

1hr(60min)

60÷6=10

Scilla [17]3 years ago
3 0

Answer:

mama mo blue

Step-by-step explanation:

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Which quotient matches the description of solving with partial quotients for 240÷15
Illusion [34]

Answer:

C. 16

Step-by-step explanation:

240/15

Step one: 10x15= 150

              240-150= 90  

Partial quotient: 10

Step two: 6x15= 90

                 90-90= 0

Partial quotient: 6

Step 3: 10+6= 16

Answer check: 240/15= 16

3 0
2 years ago
El taller de baile tiene tres veces la cantidad de inscritos que el taller de arte el taller de arte tiene 98 escritos menos que
Eduardwww [97]

Answer:

The number of registrants in art workshop is 49.

Step-by-step explanation:

The dance workshop has three times the number of registrants than the art workshop the art workshop has 98 less writings than the dance workshop how many registrations the art workshop has

Let the registrants in the art workshop is p.

The registrants in the dance work shop is 3 p.

According to question,

p = 3p - 98

2 p = 98

p = 49

So, the number of registrants in art workshop is 49.  

8 0
2 years ago
When you subtract one square number from another the answer is 3 what are the two square numbers
dimulka [17.4K]
The numbers are 1 and 2

1 squared = 1 because 1 x 1 = 1
2 squared = 4 because 2 x 2 = 4

4 - 1 = 3
5 0
3 years ago
Kaliska is jumping rope. The vertical height of the center of her rope off the ground R(t) (in cm) as a function of time t (in s
xz_007 [3.2K]

Answer:

R (t) = 60 - 60 cos (6t)

Step-by-step explanation:

Given that:

R(t) = acos (bt) + d

at t= 0

R(0) = 0

0 = acos (0) + d

a + d = 0 ----- (1)

After \dfrac{\pi}{12} seconds it reaches a height of 60 cm from the ground.

i.e

R ( \dfrac{\pi}{12}) = 60

60 = acos (\dfrac{b \pi}{12}) +d --- (2)

Recall from the question that:

At t = 0, R(0) = 0 which is the minimum

as such it is only  when a is  negative can acos (bt ) + d can get to minimum at t= 0

Similarly; 60 × 2 = maximum

R'(t) = -ab sin (bt) =0

bt = k π

here;

k  is the integer

making t the subject of the formula, we have:

t = \dfrac{k \pi}{b}

replacing the derived equation of k into R(t) = acos (bt) + d

R (\dfrac{k \pi}{b}) = d+a cos (k \pi) = \left \{ {{a+d  \ for \ k \ odd} \atop {-a+d \ for k \ even}} \right.

Since we known a < 0 (negative)

then d-a will be maximum

d-a = 60  × 2

d-a = 120 ----- (3)

Relating to equation (1) and (3)

a = -60 and d = 60

∴ R(t) = 60 - 60 cos (bt)

Similarly;

For R ( \dfrac{\pi}{12})

R ( \dfrac{\pi}{12}) = 60 -60 \ cos (\dfrac{\pi b}{12}) =60

where ;

cos (\dfrac{\pi b}{12}) =0

Then b = 6

∴

R (t) = 60 - 60 cos (6t)

7 0
2 years ago
Mathematics Mh Slope
Mkey [24]

Answer:

Step-by-step explanation:

m = 4 ; (1,-2)

y - y₁ = m (x - x₁)

y - [-2] = 4 (x - 1)

y + 2 = 4x - 4

y = 4x - 4 - 2

y = 4x - 6

4 0
3 years ago
Read 2 more answers
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