All categories

3 years ago

Solve for y. 2.4x -1.5y=3

Mathematics

2 answers:

Reil [10]3 years ago

8 0

Is there an x? Or do I need to solve for both?

labwork [276]3 years ago

6 0

Answer:

y = 1.6x - 2

Step-by-step explanation:

2.4x - 1.5y = 3

→2.4x - 1.5y + -2.4x = 3 + -2.4x ( first add -2.4x on both sides) = -1.5y = -2.4x + 3

→ $\frac{-1.5}{-1.5}$ = $\frac{-2,4x + 3}{-1.5}$ ( divide both sides by egative 1.5)

→ y = 1.6x -2 or -2 + 1.6

HOPE IT HELPED

You might be interested in

Prove the trigonometric identity

Annette [7]

Answer:

Proved See below

Step-by-step explanation:

Man this one is a world of its own :D Just a quick question are you a fellow Add Math student in O levels i remember this question from back in the day :D Anyhow Lets get started

For this question we need to know the following identities:

$1+tan^{2}x=sec^2x\\\\1+cot^2x=cosec^2x\\\\sin^2x+cos^2x=1$

Lets solve the bottom most part first:

$1-\frac{1}{1-sec^2x} \\\\$

Take LCM

$1-\frac{1}{1-sec^2x} \\\\\frac{1-sec^2x-1}{1-sec^2x} \\\\\frac{-sec^2x}{1-sec^2x} \\\\\frac{-(1+tan^2x)}{-tan^2x}$

now break the LCM

$\frac{-1}{-tan^2x}+\frac{-tan^2x}{-tan^2x}\\\\\frac{1}{tan^2x}+1\\\\cot^2x+1$

because 1/tan = cot x

and furthermore,

$cot^2x+1\\cosec^2x$

now we solve the above part and replace the bottom most part that we solved with $cosec^2x$

$\frac{1}{1-\frac{1}{cosec^2x} } \\\\\frac{1}{1-sin^2x} \\\\\frac{1}{cos^2x}\\\\sec^2x$

Hence proved! :D

4 0

3 years ago

Anyone help me please

Whitepunk [10]

Assuming you are solving for x...

5 0

3 years ago

Jeanie wants to hang a circular mirror in a frame in her front wall. Determine the area of the largest mirror Jeanie can han if

AnnyKZ [126]

56=2(pi)(r)
r=8.9127
pi(8.9127)^2=249.56cm^2
≈245cm^2 (3s.f.)

3 0

3 years ago

a cashier at the supermarket has $685 in 4 different denomination bills in her register at the end of the day. the number of $5

netineya [11]

9514 1404 393

Answer:

85 ones
70 fives
15 tens
2 fifties

Step-by-step explanation:

Let a, b, c, d represent the numbers of $1, $5, $10, and $50 bills, respectively. The problem statement tells us ...

a +5b +10c +50d = 685 . . . . . total amount of cash

b = 10 +4c . . . . . . . . . . . . the number of fives is 10 more than 4 times tens

a = b+c . . . . . . . . . . . as many ones as fives and tens combined

d = 2 . . . . . . . . . . the register contained 2 fifty-dollar bills

Substituting for 'a', then for 'b', we have ...

(b+c) +5b +10c +50d = 685

6(10 +4c) +11c +50d = 685

60 +35c +50d = 685

Substituting d=2 and subtracting 160 gives ...

35c = 525

c = 15

b = 10 +4c = 10 +4(15) = 70

a = b+c = 70 +15 = 85

The register contained ...

85 ones
70 fives
15 tens
2 fifties

5 0

2 years ago

Read 2 more answers

The answer to this Pythagorean problem

Karolina [17]

The correct answer is 20 because if you solve A=hbb 1=5 8 2=20 Hope this helps! ;D

3 0

3 years ago