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3 years ago

Gary jogs 2/3 of a mile in 1/3 of an hour. how many mipes can he jog in 1 hour?

Mathematics

1 answer:

cricket20 [7]3 years ago

8 0

2 miles.
This is because 2/3 times 3 equals 2.

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PLEASE HELP

Andrej [43]

Answer: -3 and 0

Step-by-step explanation: f(x) = g(x) where they intersect. Two lines intersect when they have the same x and y values.

5 0

3 years ago

Joe wants to place plants around each of four sides of a square yard. If he’s planned for each of the four sides to show six pla

blondinia [14]

Answer:

Step-by-step explanation:

1 side = 6 plants (including one in the corner)

3 sides = 5 plants each

6+15 = 21

7 0

3 years ago

A cellphone company charges $60 for a cellphone and a monthly rate to use the phone.Regina will pay $1,140 for a 24-month cellph

Alinara [238K]

The monthly rate is $45 for this cellphone plan.

Step-by-step explanation:

Given,

Amount charged for cell phone = $60

Amount paid by Regina for 24 month plan = $1140

Number of months = 24

Let,

m be the monthly rate for cellphone plan

Number of months * Monthly rate + cellphone charges = Amount paid

$24m+60=1140\\24m=1140-60\\24m=1080$

Dividing both sides by 24

$\frac{24m}{24}=\frac{1080}{24}\\m=45$

The monthly rate is $45 for this cellphone plan.

Keywords: subtraction, division

Learn more about subtraction at:

#LearnwithBrainly

7 0

3 years ago

a contractor is mixing concrete and mixed 3 parts concrete and 9 parts water. How much water is needed for 12 parts concrete?

weeeeeb [17]

Answer:

36 parts water

Step-by-step explanation:

3/9 = 1/3 = for every 1 part concrete, it's 3 parts water.

12/x = 1/3; x = 36

so, 12 x 3 = 36

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

3 years ago