Solve the equations for x and y

A company is offering a job with a salary of $48,000 for the first year, then a raise of 2% each year after that. Write a formul

Romashka-Z-Leto [24]

Answer:

a.48,000+2=48,002 b.30×2=60% +2=62%+48,0000=48,062

3 0

4 years ago

(c ) Find the values of x and y in the vector equation (5÷3)+2(x÷y)-(1÷-7)=0

german

Answer:

y= -21x/19 x≠0

Step-by-step explanation:

(I just reviewed this!) Combined like terms and used Pemdas for Order of Operations. Canceled out -21x with -21 and we know that x is not equal to 0. If you need more help there are some sites like Khan Academy.

7 0

3 years ago

Can someone help me out with some pesky math quiz :) I put a pic

Sindrei [870]

1. Simplifying ∛54 gives 3∛2

2. 27^1/3 * (27)^3 is equal to 27

3. The expression (2j4k)^4/5 can be written as $\sqrt[4]{32j^{20}k^{5} }^{5}$ making the first option the correct option

4. simplifying $\sqrt[4]{16x^{4} y^{12} }$ will result to 2x∛y

5. simplifying $\sqrt[5]{2940x^{13} y^{7} }$ will result to $14x^{6} y^{3} \sqrt{15xy}$ making the last option the correct option

<h3>How to simplify $\sqrt[5]{2940x^{13} y^{7} }$ </h3>

<u>given data</u>

$\sqrt[5]{2940x^{13} y^{7} }$

= ( 2940 * x^13 * y^7 )^1/5

$= ( 2^2 * 5 * 3 * 7^2 * x^{12} * x * y^6 * y )^{1/5}$

$= ( 2^2 * 5 * 3 * 7^2 * x^{12} * x * y^6 * y )^{1/5*5/2}$

$= ( 2^2 * 5 * 3 * 7^2 * x^{12} * x * y^6 * y )^{5/5*1/2}$

multiplying the values inside the parenthesis by ^1/2

$= ( 2 * 5^{1/2} * 3^{1/2} * 7 * x^{6} * x^{1/2} * y^3 * y^{1/2} )$

$= ( 14* 5^{1/2} * 3^{1/2} * x^{6} * x^{1/2} * y^3 * y^{1/2} )$

$= ( 14 * x^6 * y^3 ) ( 5^{1/2} * 3^{1/2} * x^{1/2} * y^{1/2} )$

$= ( 14x^6y^3 ) ( 15^{1/2} * x^{1/2} * y^{1/2} )$

$= ( 14x^6y^3 ) ( 15xy)^{1/2}$

$= ( 14x^6y^3 ) \sqrt{( 15xy)}$

we can therefore say that the last option is the answer

Read more on equations here: brainly.com/question/22688504

#SPJ1

3 0

1 year ago

Circle O is inscribed in ABC. What is the perimeter of ABC

xenn [34]

Since O is inscribed in triangle ABC, you know that AB, AC, and BC are all tangent to the circle. The tangent segments theorem asserts that AD is congruent to A F; BD is congruent to BE; and CF is congruent to CE. (NOTE: E is the point where the circle touches BC - I've taken the liberty of labeling it as such.)

Then the perimeter $P$ of triangle ABC will be twice the sum of the labeled edges:

$P=2(12\,\mathrm{cm}+16\,\mathrm{cm}+6\,\mathrm{cm})=72\,\mathrm{cm}$

7 0

3 years ago

Read 2 more answers

PLEASE HELP WILL MARK BRAINLIEST (sketch the graph of each function)

MA_775_DIABLO [31]

The answer is in the enclosed diagram

6 0

3 years ago