Solve the system by the elimination method.. 2a + 3b = 6 5a + 2b

I need help on this plz i don’t understand the question

AnnZ [28]

We know that the area of a circle in terms of π will be πr². However the area with respect to the diameter will be a different story. The first step here is to find a function relating the area and diameter of any circle --- ( 1 )

For any circle the diameter is 2 times the radius,

d = 2r

Therefore r = d / 2, which gives us the following formula through substitution.

A = π(d / 2)² = πd² / 4

Hence the area of a circle as the function of it's diameter is A = πd² / 4. You can also say f(d) = πd² / 4.

Now we can substitute " d " as 4, solving for the area ( A ) or f(4) --- ( 2 )

f(4) = π(4)² / 4 = 16π / 4 = 4π - This makes the area of circle present with a diameter of 4 inches, 4π.

3 0

2 years ago

N

Feliz [49]

Umm I’m guessing 390

7 0

2 years ago

What is the base of the triangle if the area is 56 square meters and the height is 7 m?

attashe74 [19]

Answer:

the answer is b

Step-by-step explanation:

the answer is B 16 m

5 0

2 years ago

2 times 5 times 10 to the 5th power

xxMikexx [17]

10 to the 5th power is 100,000. 2 X 5 is 10.
10 X 100,000 is one million.

the answer is 1,000,000. (one million)

5 0

2 years ago

Read 2 more answers

A metal beam was brought from the outside cold into a machine shop where the temperature was held at 65degreesF. After 5 min, t

ivolga24 [154]

Answer:

The beam initial temperature is 5 °F.

Step-by-step explanation:

If T(t) is the temperature of the beam after t minutes, then we know, by Newton’s Law of Cooling, that

$T(t)=T_a+(T_0-T_a)e^{-kt}$

where $T_a$ is the ambient temperature, $T_0$ is the initial temperature, $t$ is the time and $k$ is a constant yet to be determined.

The goal is to determine the initial temperature of the beam, which is to say $T_0$

We know that the ambient temperature is $T_a=65$ , so

$T(t)=65+(T_0-65)e^{-kt}$

We also know that when $t=5 \:min$ the temperature is $T(5)=35$ and when $t=10 \:min$ the temperature is $T(10)=50$ which gives:

$T(5)=65+(T_0-65)e^{k5}\\35=65+(T_0-65)e^{-k5}$

$T(10)=65+(T_0-65)e^{k10}\\50=65+(T_0-65)e^{-k10}$

Rearranging,

$35=65+(T_0-65)e^{-k5}\\35-65=(T_0-65)e^{-k5}\\-30=(T_0-65)e^{-k5}$

$50=65+(T_0-65)e^{-k10}\\50-65=(T_0-65)e^{-k10}\\-15=(T_0-65)e^{-k10}$

If we divide these two equations we get

$\frac{-30}{-15}=\frac{(T_0-65)e^{-k5}}{(T_0-65)e^{-k10}}$

$\frac{-30}{-15}=\frac{e^{-k5}}{e^{-k10}}\\2=e^{5k}\\\ln \left(2\right)=\ln \left(e^{5k}\right)\\\ln \left(2\right)=5k\ln \left(e\right)\\\ln \left(2\right)=5k\\k=\frac{\ln \left(2\right)}{5}$

Now, that we know the value of $k$ we can use it to find the initial temperature of the beam,

$35=65+(T_0-65)e^{-(\frac{\ln \left(2\right)}{5})5}\\\\65+\left(T_0-65\right)e^{-\left(\frac{\ln \left(2\right)}{5}\right)\cdot \:5}=35\\\\65+\frac{T_0-65}{e^{\ln \left(2\right)}}=35\\\\\frac{T_0-65}{e^{\ln \left(2\right)}}=-30\\\\\frac{\left(T_0-65\right)e^{\ln \left(2\right)}}{e^{\ln \left(2\right)}}=\left(-30\right)e^{\ln \left(2\right)}\\\\T_0=5$

so the beam started out at 5 °F.

6 0

3 years ago

Solve the system by the elimination method.. 2a + 3b = 6 5a + 2b - 4 = 0