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MrRissso [65]
3 years ago
9

How many minutes are there in 12.5 hours?

Mathematics
2 answers:
neonofarm [45]3 years ago
7 0

Answer:

750 minutes

Step-by-step explanation:

DerKrebs [107]3 years ago
5 0

Answer:

750 minutes would be your answer

Step-by-step explanation:

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A = b and c = d, then a x c = b x d<br> what kind of reasong is used here
Phoenix [80]
A = b is given to us

Multiply both sides by c to get
a = b
a*c = b*c
The rule to multiply both sides by c is the multiplicative property of equality

Then on the right side, replace c with d. This replacement is possible due to the fact that c = d. This is using the substitution property.

So we go from
a*c = b*c
to this
a*c = b*d
5 0
3 years ago
Find the value of f(–3) and g(3) if f(x) = –6x + 3 and g(x) = 3x + 21x–3.
ryzh [129]

Answer:

f(-3) = 21

g(3) = 69

Step-by-step explanation:

We have been given the following functions;

f(x) = –6x + 3

g(x) = 3x + 21x–3.

To find f(-3) , we simply substitute x = -3 in the function of f(x);

f(-3) = -6(-3) + 3

      = 21

To find g(3) , substitute x = 3 in the function of g(x);

g(3) = 3(3) +21(3) - 3

      = 69

3 0
3 years ago
(20pts) the monthly average new york city temperature in fahrenheit are given below. Mont h jan feb mar apr may jun jul aug sep
HACTEHA [7]

Mean of the low is approximately 47.7  Median of the low is 45 and 50 or 47.5

3 0
3 years ago
The boiling point of jet fuel is 329°
natka813 [3]

Since, the boiling point of jet fuel  = 329 °F

We have to convert the boiling point of jet fuel in ^\circ C.

The expression which is used to convert ^\circ F to ^\circ C is given by:

C = \frac{5}{9}(F-32)

= \frac{5}{9}(329-32)

= \frac{5}{9}(297)

= 165^\circ C

Therefore, the boiling point of jet fuel is 165 ^\circ C.

8 0
3 years ago
Read 2 more answers
.
ryzh [129]

Answer:

(1 ) Inner curved surface area of the well is  109.9 sq. meters.

(2) The cost of plastering the total curved surface area is  4396.

Step-by-step explanation:

The inner diameter = 3.5 m

Depth of the well =  10 m

Now, Diameter = 2 x Radius

⇒R = D/ 2  = 3.5/2 = 1.75

or, the inner radius of the well = 1.75 m

CURVED SURFACE AREA of cylinder = 2πr h

⇒The inner curved surface area =  2πr h  = 2 ( 3.14) (1.75)(10)

                                                                      = 109 sq. meters

Hence, the inner curved surface area of the well is  109.9 sq. meters.

Now, the cost of plastering the curved area is 40 per sq meters

So, the cost of total plastering total area = 109.9 x(Cost per meter sq.)

                                                                    =  109.9 x (40)

                                                                   =   4396

Hence, the cost of plastering the total curved surface area is  4396.

8 0
3 years ago
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