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2 years ago

Half life can be used to solve for the amount (A) of a substance left after t time by using the following equation:

Mathematics

2 answers:

Annette [7]2 years ago

7 0

Answer: Its C

Step-by-step explanation:

KengaRu [80]2 years ago

3 0

Answer is in the link below
Anwerkilk.com

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24 - 11 = 13 what is the relative additions sentence

aliya0001 [1]

It's just a math equation.

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2 years ago

Adam is comparing the graphs of y=4x and y=8x. Which of the following statements is TRUE?

Inga [223]

The first one is true your welcome hope thi helps

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2 years ago

A ball is thrown straight up with a initial velocity of 56 feet per second . The height , h , of the ball t seconds after it is

Tju [1.3M]

Formula: H(t) = 56t – 16t^2

H(t) = - 16t^2 + 56t

A. What is the height of the ball after 1 second? H (1) = 56(1) – 16(1) ^2 = 40 pt.

B. What is the maximum height? X = - (56)/2(- 16) = 1.75 sec h (1.75) = 56(1.75) – 16(1.75) ^2 h (1.75) = 49ft.

C. After how many seconds will it return to the ground? – 16t^2 + 56t = 0 - 8t =0 t = 0

- 8t (2 + - 7) = 0 2t – 7 = 0 t = 7/2 Ans: 3.5 seconds

3 0

3 years ago

Jim's cat, c, weighs 7 pounds. His dog weighs six times plus 3 pounds. more than his cat. Which equation represent the dogs weig

Alex17521 [72]

C =7
d is 6 times c+3
d=6c+3

d=6d+3 is equation
(if you sub, you would get 45 is weight)

5 0

3 years ago

In ΔABC, the lengths of a, b, and c are 22.5 centimeters, 18 centimeters, and 13.6 centimeters, respectively.

Irina-Kira [14]

Given the values of the three sides of the triangle, we can apply the Cosine Law to find the angles of the triangle. Recall that for we can express the value of c through the equation below.

$c^{2} = a^{2} + b^{2} - 2abcosC$

Rearranging this equation, we can find the value ∠C as shown below.

$\cos C = \frac{a^{2}+b^{2}-c^{2}}{2ab}$
$C = cos^{-1} (\frac{a^{2}+b^{2}-c^{2}}{2ab})$

We can apply the same reasoning for finding the value of ∠B as shown.

$B = cos^{-1} (\frac{a^{2}+c^{2}-b^{2}}{2ac})$

Plugging in the values of the sides (see image attached) from the given. It will now be straightforward to compute for ∠B and ∠C.

$C = cos^{-1} (\frac{22.5^{2}+18^{2}-13.6^{2}}{2(22.5)(18)})$
$C \approx 37.19$

$B = cos^{-1} (\frac{22.5^{2}+13.6^{2}-18^{2}}{2(22.5)(13.6)})$
$B \approx 53.13$

Answer: ∠C = 37.19° and ∠B = 53.13°

7 0

3 years ago

Read 2 more answers