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3 years ago

PLEASE HELP MEEE

Mathematics

2 answers:

Softa [21]3 years ago

8 0

Answer:

The answer is 18. (A)

Step-by-step explanation:

Let me know if this helps

antoniya [11.8K]3 years ago

6 0

Answer:

(6a-4)°=(2a+68)°

6a-2a=68+4

4a=72

a=72/4

a=18

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What are the coordinates of the circumcenter of a triangle with vertices A(−1, 1) , B(5, 1) , and C(−1, −1) ?

rewona [7]

The answer to this is (2,0)

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3 years ago

Pleaseee helpppp answer correctly !!!!!!!!!!!!!!!! Will mark Brianliest !!!!!!!!!!!!!!!!!

Anit [1.1K]

Answer:

540°

General Formulas and Concepts:

Math

Counting

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Geometry

Sum of Angles: 180(n - 2)°

Step-by-step explanation:

Step 1: Define

We are given a 5-sided polygon (irregular pentagon)

n = 5

Step 2: Find Sum

Substitute in n [Sum of Angles]: 180(5 - 2)°
(Parenthesis) Subtract: 180(3)°
Multiply: 540°

5 0

3 years ago

Read 2 more answers

Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li

krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$