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gizmo_the_mogwai [7]
3 years ago
14

15.3+x=1.3-x so how do I solve it

Mathematics
2 answers:
Nookie1986 [14]3 years ago
5 0

Answer:

0.080

Step-by-step explanation:

I think that's the answer... If not, sorry! Good luck! :)

anzhelika [568]3 years ago
3 0

Answer:

-7

Step-by-step explanation:

Let's try getting rid of the numbers to isolate the x.

Add x on both sides to get rid of the x on the right side and to get 2x on the left side. This leaves us with 15.3 + 2x = 1.3

Subtract 15.3 on both sides to get rid of the 15.3 on the left side.

This would leave us with 2x = -14

Divide 2 on both sides to isolate the x.

The answer is x = -7

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What is the solution to the system? <br><br> (___,___)
Rina8888 [55]
For the first line we have a slope of (y2-y1)/(x2-x1)

(2--2)/(1--1)=4/2=2 so we have:

y=2x+b, now solve for b with either of the points, I'll use: (1,2)

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y=2x

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y=(-3x+14)/2

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6 0
3 years ago
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

a_{11} = 32000 + 10*300

a_{11} = 32000 + 3000

\boxed{\boxed{a_{11} = 35000}}\end{array}}\qquad\checkmark

Respuesta:

In the eleventh year of salary he will earn more than 34700, in the case, this value will be 35000

________________________

¡Espero haberte ayudado, saludos... DexteR! =)

7 0
3 years ago
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