5. Solve the system of equations.<br> 2x+3y=0<br> x-2y=-7<br> 1-5y =

Please help me there can only be one answer

12345 [234]

Answer:-
I think it’s the last option where x and y both are zero.
2x + 2y = 0
Here x = 0 and y = 0.
Hope it helps if u hv any questions feel free to ask...hv a grt day :)

4 0

3 years ago

Joey has 5 collector stamps.He receives 3 more stamps each week that he completes community service hours.X=the number of weeks

kirza4 [7]

Answer:

Therefore the required equation is

Y=5+3X

Step-by-step explanation:

Given that , Joey has 5 collector stamps.

He gets 3 more stamps each week.

The number of week is X.

1st week he got 3 stamps

2nd week he got (3+3)=2.3 stamps

3rd week he got (3+3+3)=3.3 stamps

Stamp that he got=( the number of week ×3)

Therefore at end of Xth week he got =(X × 3) =3X stamps

Total number of stamps that collected after x week

=5+3X

Therefore the required equation is

Y=5+3X

6 0

4 years ago

2 (t-4) + 1 math I ready

Roman55 [17]

Answer:

2t-7

Step-by-step explanation:

Not sure what you need ;( hope this helps!

6 0

3 years ago

Read 2 more answers

Question regarding logarithms.

Eddi Din [679]

$5^{x-2}-7^{x-3}=7^{x-5}+11\cdot5^{x-4}\\\\5^{x-2}-7^{x-2-1}=7^{x-2-3}+11\cdot5^{x-2-2}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\5^{x-2}-\dfrac{7^{x-2}}{7^1}=\dfrac{7^{x-2}}{7^3}+11\cdot\dfrac{5^{x-2}}{5^2}\\\\5^{x-2}-\dfrac{1}{7}\cdot7^{x-2}=\dfrac{1}{343}\cdot7^{x-2}+\dfrac{11}{25}\cdot5^{x-2}\\\\-\dfrac{1}{7}\cdot7^{x-2}-\dfrac{1}{343}\cdot7^{x-2}=\dfrac{11}{25}\cdot5^{x-2}-5^{x-2}\\\\\left(-\dfrac{1}{7}-\dfrac{1}{343}\right)\cdot7^{x-2}=\left(\dfrac{11}{25}-1\right)\cdot5^{x-2}$

$\left(-\dfrac{49}{343}-\dfrac{1}{343}\right)\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\\\\-\dfrac{50}{343}\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\qquad\text{multiply both sides by}\ \left(-\dfrac{25}{14}\right)\\\\\dfrac{50\cdot25}{343\cdot14}\cdot7^{x-2}=5^{x-2}\qquad\text{divide both sides by}\ 7^{x-2}\\\\\dfrac{25\cdot25}{343\cdot7}=\dfrac{5^{x-2}}{7^{x-2}}\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}$

$\dfrac{5^2\cdot5^2}{7^3\cdot7}=\left(\dfrac{5}{7}\right)^{x-2}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\\dfrac{5^4}{7^4}=\left(\dfrac{5}{7}\right)^{x-2}\\\\\left(\dfrac{5}{7}\right)^4=\left(\dfrac{5}{7}\right)^{x-2}\iff x-2=4\qquad\text{add 2 to both sides}\\\\\boxed{x=6}$

6 0

3 years ago

Read 2 more answers

The arithmetic sequence from the previous problem is 8. 7, 7. 3, 5. 9, 4. 5, 3. 1,. . . . What is the common difference, d, of t

o-na [289]

I think 8 is again maybe its true

3 0

3 years ago

5. Solve the system of equations. 2x+3y=0 x-2y=-7 1-5y =