All categories

3 years ago

In the system of equations, which variable would it be easiest to solve for?

Mathematics

2 answers:

Mnenie [13.5K]3 years ago

5 0

Answer:

Step-by-step explanation:

The easiest to solve for is y in the second equation. Just subtract 3x from both sides, obtaining y = -3x + 9.

Likurg_2 [28]3 years ago

4 0

Answer:

y in the second equation

Step-by-step explanation:

y in the second equation because +1 is the understood coefficient which means you will have one step and that is it to solve for y in 3x+y=9

You might be interested in

What number increased by 1/2 of itself equals 54?

skelet666 [1.2K]

I think it's 108. But I'm not really good at math so...

7 0

3 years ago

Read 2 more answers

Jamal ran 8 kilometers, sprinted 500 meters, and climbed bleachers for a distance of 1.5 kilometers during track practice. How m

stiv31 [10]

Answer: $10,000\ meters$

Step-by-step explanation:

In order to calculate the total meters Jamal traveled during the practice, you need to make the conversions from kilometers to meters.

You need to remember that:

$1\ km=1,000\ m$

Therefore:

$(8\ km)(\frac{1,000\ m}{1\ km})=8,000\ m$

$(1.5\ km)(\frac{1,000\ m}{1\ km})=1,500\ m$

Finally, you can solve the following addition in order to get the total meters Jamal traveled during the practice. The sum is:

$total=8,000\ m+500\ m+1,500\ m\\\\total=10,000\ m$

4 0

3 years ago

Read 2 more answers

Which is equivalent to sqaure root 10 ^ 3/4x

Digiron [165]

Answe√1000/2x

Step-by-step explanation:

√10³/4x=√1000/2x

7 0

3 years ago

Find the simple interest and the amount when: p=rs 2650 rate=8%p.a. And time =2 1/2years

DochEvi [55]

SI=PRT/100
= (2650*8*2.5)/100
= 53000/100
= 530
Simple Interest= 530

3 0

3 years ago

5.1. Disprove the statement: If a and b are any two real numbers, then log(ab) = log(a) + log(b).

Ket [755]

Answer:

If (a and b )≤ 0 then $log(ab)=log(a)+log(b)$ is disproved

Step-by-step explanation:

If a and b are positive real numbers then:

$log(ab)=log(a)+log(b)$

But if a and b are negative then this axiom is not true as log is not defined

$log_{c}(x)$ = undefined $for \quad x\leq 0$

So if (a and b )≤ 0 then $log_{c}(a)$ and $log_{c}(b)$ are undefined but $log_{c}(-a*-b)$ is defined.

5 0

3 years ago