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2 years ago

Can i get help on this one ? im almost running our of points i would give 50 but i need more ;-; anyway

Mathematics

1 answer:

JulsSmile [24]2 years ago

4 0

Answer:

Step-by-step explanation:

Let x = third side

Using the Triangle Inequality theorem which states that the sum of two sides of a triangle must be longer than the third side and the difference of the two sides is the lower limit of the third side, the answer to your question is that the third side must be between 3 and 13, or written using inequalities, 3 < third side (or x) < 13 is the range.

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X^2-3=5x simplify this

Alex Ar [27]

To simplify x^2-3 =5x

You would first subtract 5x from both sides

X^3-5x-3

Next you would plug this into the quadratic equation:

X=5+/- the square root of: -5^2 -4(1)(-3)/2(1)

Then simplify,
5+/- the squareroot of 24 + 12 / 2

5+/- The squareroot of 36 /2

5+/- 6 / 2

Then solve if it’s an addition or subtract

5+6/2

X=11/2 or 5 1/2

5-6/2

X= -1/2

The answer is x=-1/2 or x=11/2

Hope this helps

7 0

2 years ago

Which is the next term in the pattern 1/12, 4/15, 9/18, 16/21

photoshop1234 [79]

I’m guessing it would be 21/24

5 0

3 years ago

Can someone please help me? 100% sure please

Nataliya [291]

Part A

Distribute the one-half.

1.24g + 0.87m + 0.98z

Part B

(29.68 - x) + 14.45

Or simplified to 44.13 - x

Part C

a = number of times at the store
b = gloves' price
c = hat's price

6 0

2 years ago

Read 2 more answers

mr.browns salary is 32,000 and imcreases by $300 each year, write a sequence showing the salary for the first five years when wi

chubhunter [2.5K]

Hello!

We have the following data:

a1 (first term or first year salary) = 32000

r (ratio or annual increase) = 300

n (number of terms or each year worked)

We apply the data in the Formula of the General Term of an Arithmetic Progression, to find in sequence the salary increases until it exceeds 34700, let us see:

formula:

$a_n = a_1 + (n-1)*r$

* second year salary

$a_2 = a_1 + (2-1)*300$

$a_2 = 32000 + 1*300$

$a_2 = 32000 + 300$

$\boxed{a_2 = 32300}$

* third year salary

$a_3 = a_1 + (3-1)*300$

$a_3 = 32000 + 2*300$

$a_3 = 32000 + 600$

$\boxed{a_3 = 32600}$

* fourth year salary

$a_4 = a_1 + (4-1)*300$

$a_4 = 32000 + 3*300$

$a_4 = 32000 + 900$

$\boxed{a_4 = 32900}$

* fifth year salary

$a_5 = a_1 + (5-1)*300$

$a_5 = 32000 + 4*300$

$a_5 = 32000 + 1200$

$\boxed{a_5 = 33200}$

We note that after the first five years, Mr. Browns' salary has not yet surpassed 34700, let's see when he will exceed the value:

* sixth year salary

$a_6 = a_1 + (6-1)*300$

$a_6 = 32000 + 5*300$

$a_6 = 32000 + 1500$

$\boxed{a_6 = 33500}$

* seventh year salary

$a_7 = a_1 + (7-1)*300$

$a_7 = 32000 + 6*300$

$a_7 = 32000 + 1800$

$\boxed{a_7 = 33800}$

* eighth year salary

$a_8 = a_1 + (8-1)*300$

$a_8 = 32000 + 7*300$

$a_8 = 32000 + 2100$

$\boxed{a_8 = 34100}$

* ninth year salary

$a_9 = a_1 + (9-1)*300$

$a_9 = 32000 + 8*300$

$a_9 = 32000 + 2400$

$\boxed{a_9 = 34400}$

* tenth year salary

$a_{10} = a_1 + (10-1)*300$

$a_{10} = 32000 + 9*300$

$a_{10} = 32000 + 2700$

$\boxed{a_{10} = 34700}$

we note that in the tenth year of salary the value equals but has not yet exceeded the stipulated value, only in the eleventh year will such value be surpassed, let us see:

* eleventh year salary

$a_{11} = a_1 + (11-1)*300$