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larisa86 [58]
3 years ago
12

Thank you so much for helping me with the first question. I have and another question that I am a little confused on also, could

you please help me with this one too?
Y-8=3/7 (X-6) I got y= 7/3x +38/7 Is this correct?
Mathematics
1 answer:
Bogdan [553]3 years ago
5 0
Y=3/7x+38/7 is correct answer
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Find the circumference and area of a circle with radius 14m
liubo4ka [24]
Circumference of a circle

C = 2πr

since r = 14 then

C = 2π14

C =28π.

Area of a circle

A = πr²

A = π14²

A = 196π
8 0
3 years ago
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Work out the following questions
Likurg_2 [28]

Answer:

a)8/21

b)1/9

Step-by-step explanation:

because when we divide fraction we resprocal the divisor

4 0
2 years ago
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Between 0 degrees Celsius and 30 degrees Celsius, the volume V (in cubic centimeters) of 1 kg of water at a temperature T is giv
Ira Lisetskai [31]

Answer:

T = 3.967 C

Step-by-step explanation:

Density = mass / volume

Use the mass = 1kg and volume as the equation given V, we will come up with the following equation

D = 1 / 999.87−0.06426T+0.0085043T^2−0.0000679T^3

   = (999.87−0.06426T+0.0085043T^2−0.0000679T^3)^-1

Find the first derivative of D with respect to temperature T

dD/dT = \dfrac{70000000\left(291T^2-24298T+91800\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^2}

Let dD/dT = 0 to find the critical value we will get

\dfrac{70000000\left(291T^2-24298T+91800\right)}  = 0

Using formula of quadratic, we get the roots:

T =  79.53 and T = 3.967

Since the temperature is only between 0 and 30, pick T = 3.967

Find 2nd derivative to check whether the equation will have maximum value:

-\dfrac{140000000\left(395178T^4-65993368T^3+3286558821T^2+2886200857800T-121415215620000\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^3}

Substituting the value with T=3.967,

d2D/dT2 = -1.54 x 10^(-8)    a negative value. Hence It is a maximum value

Substitute T =3.967 into equation V, we get V = 0.001 i.e. the volume when the the density is the highest is at 0.001 m3 with density of

D = 1/0.001 = 1000 kg/m3

Therefore T = 3.967 C

3 0
3 years ago
What is the length of the unknown leg in the right triangle? A right triangle has a side with length 20 meters, hypotenuse with
galben [10]

The length of the unknown leg of the triangle is 15 m.

<u>Step-by-step explanation:</u>

Length of one leg = 20 m

Length of the hypotenuse= 25m

As it is a right angled triangle we can use pythogoras theorem.

Let the unknown length be y

(20) (20)  + y(y) = (25) (25)

400 + y(y) = 625

y(y) = 225

y = √225

y = 15

The length of the unknown leg is 15 m.

6 0
2 years ago
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4) The width of a rectangle is 8 inches. The length is 4 times its
jeka57 [31]

Answer:

Length = 4 * 8 = 32

Area = width * length = 8 * 32 = 256 square inches

Step-by-step explanation:

4 0
3 years ago
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