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dolphi86 [110]
3 years ago
10

Find the perimeter of △ABC with vertices A(−4, 3), B(1, −5), and C(−4, −5). Round your answer to the nearest hundredth.

Mathematics
1 answer:
miss Akunina [59]3 years ago
8 0
Let's use the distance formula: \sqrt{ (x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}

So it is:

AB : \sqrt{(-5)^2 + 8^2} = 9.43
BC: \sqrt{5^2 + 0^2 } = 5
CA: \sqrt{0^2 + 8^2} = 8

So 9.43 + 5 + 8 =22.43
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Solve the equation.
Levart [38]

Answer:

x=34

Step-by-step explanation:

6 - ( x-7) ^ 1/3 = 3

Subtract 6 from each side

6-6 - ( x-7) ^ 1/3 = 3-6

- ( x-7) ^ 1/3 = -3

Divide each side by a negative

 ( x-7) ^ 1/3 = 3

Cube each side

 ( x-7) ^ 1/3 ^3 = (3)^3

x-7 = 27

Add 7 to each side

x-7+7 = 27+7

x = 34

Check

6 - ( 34-7) ^ 1/3 = 3

6 - (27^1/3 = 3

6 -3 =3

3=3

Good solution

8 0
3 years ago
Plsssssssss help me in this i need it asap ill give brainly
egoroff_w [7]

Answer:

16

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
What is the value of x in this triangle
polet [3.4K]

Answer:

A

Step-by-step explanation:

if x is the least it only makes common sense

HOPE THIS HELPS!

5 0
3 years ago
Read 2 more answers
8 multiplied by 3/4 divided by 3/3
bezimeni [28]
8 * 3/4

Multiply 8 by 4 to get 32 then divide 32 by 3 to get 3.6 repeating

3/3 is basically 1 so its still 3.6 

3.6 is  3 3/5

-XCalypso

~ Dont forget to rate, thank and vote brainliest answer! :D ~
5 0
3 years ago
Complete the table for the radioactive iotope. (Round your anwer to 2 decimal place. Iotope : 239 Pu
Marizza181 [45]

If the radio active isotope has the half life of 24100 years, then the initial quantity is 2.16 grams

The half life in years = 24100

Consider the quantity of the radio active isotope remaining

y = Ce^{kt}

When t = 1000 the y = 1.2

y = C/2 when t = 1599

Substitute the values in the equation

C/2 = Ce^{24100k}

Cancel the C in both side

1/2 = e^{24100k}

Here we have to apply ln to eliminate the e terms

ln (1/2) = 24100k

k = ln(1/2) / 24100

k = -2.87× 10^-5

To find the initial value we have to substitute the value of k and y in the equation

1.2 = Ce^{1000 ×  -2.87× 10^-5}

C = 1.2 / e^(-0.0287)

C = 2.16 gram

Hence, the initial quantity of the radioactive isotope is 2.16 gram

Learn more about half life here

brainly.com/question/4318844

#SPJ4

7 0
1 year ago
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