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2 months ago

S=5r^2t solve for T please

2 answers:

7 0

Hello!
The equation is S = 5r^2 * T
To isolate T, we aim to have it alone on one side of the equation.
We can achieve this by dividing both sides by 5r^2.
So, our equation becomes S/5r^2 = T, and this is our final answer.

I hope this is useful!

tester [12.3K]2 months ago

6 0

Begin by taking the square root of both sides
RootS=5r rootT
Next, divide both sides by 5r
RootS/5r=root T
Then, square both sides
S/5r^2=T Alternatively, you could simply divide both sides by 5r^2 to reach the same result.[[TAG_6]]

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Find the perimeter of a quadrilateral with vertices at C (−2, 1), D (2, 4), E (5, 0), and F (1, −3). Round your answer to the ne

tester [12383]

Answer: 20 units

Step-by-step explanation:

The formula for calculating the distance between two coordinates P(a,b) and Q(c,d) is as follows:-

$PQ=\sqrt{(d-b)^2+(c-a)^2}$

Initially provided: The coordinates of the quadrilateral's vertices are C (−2, 1), D (2, 4), E (5, 0), and F (1, −3).

The distance between vertices C (−2, 1) and D (2, 4) is as follows:-

$CD=\sqrt{(4-1)^2+(2-(-2))^2}\\\\\Rightarrow\ CD=\sqrt{(3)^2+(4)^2}\\\\\Rightarrow\ CD=\sqrt{9+16}=\sqrt{25}\\\\\Rightarrow\ CD=5\text{ units}$

Distance between points D (2, 4) and E (5, 0) is calculated as:-

$DE=\sqrt{(4-0)^2+(2-5)^2}\\\\\Rightarrow\ DE=\sqrt{(4)^2+(-3)^2}\\\\\Rightarrow\ DE=\sqrt{16+9}=\sqrt{25}\\\\\Rightarrow\ DE=5\text{ units}$

Distance between coordinates E (5, 0) and F (1, −3) is given by:-

$EF=\sqrt{(-3-0)^2+(1-5)^2}\\\\\Rightarrow\ EF=\sqrt{(-3)^2+(-4)^2}\\\\\Rightarrow\ EF=\sqrt{9+16}=\sqrt{25}\\\\\Rightarrow\ EF=5\text{ units}$

Distance from C (−2, 1) to F (1, −3) is as follows:-

$FC=\sqrt{(-3-1))^2+(1-(-2))^2}\\\\\Rightarrow\ FC=\sqrt{(-4)^2+(3)^2}\\\\\Rightarrow\ FC=\sqrt{16+9}=\sqrt{25}\\\\\Rightarrow\ FC=5\text{ units}$

The total perimeter of the quadrilateral is calculated as = CD + DE + EF + FC

$=5+5+5+5=20\text{ units}$

3 0

2 months ago

Read 2 more answers

At 7:00 A.M., Alicia pours a cup of tea whose temperature is 200°F. The tea starts to cool to room temperature (72°F). At 7:02 A

zzz [12365]

Response:

Part A

The best choice is;

b. y = 128(0.989)x + 72

Part B

Ella can consume the tea post 7:22

Detailed clarification:

Here, we observe the temperature variation with time represented in an exponential equation as follows;

Let the temperature at time x (measured in minutes) be y

Thus, y = a × mˣ + b

Where:

b = Curve shift or the limiting value of the decreasing exponential function as x → ∞

When y = 200, x = 0

This leads to 200 = a × m⁰ + c = a + c

Here, c represents the graph's shift, which is the temperature increase = final temperature = 72°F

Thus, a = 200 - 72 = 128°F

When y = 197, at x = 2 minutes

Thus, 197 = 128·m² + 72 =

m² = (197 - 72)/128 = 125/128

m = √(125/128) = 0.98821

Therefore, the exponential cooling equation is given by;

y = 128 × (0.98821)ˣ + 72

Hence the best choice is b. y = 128(0.989)x + 72

Part B

When the tea's temperature reaches 172°F, we have;

172 = 128 × (0.989)ˣ + 72

Thus, (0.989)ˣ = (172 - 72)/128 = 100/128 = 25/32

log(0.989)ˣ = log(25/32)

x·log(0.989) = log(25/32)

x = log(25/32)/log(0.989) = 22.32 minutes

Thus, Ella can consume the tea post 7:22.

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