All categories

3 years ago

How to solve y-b/x=m for y

2 answers:

8 0

To solve for m we want to get m by itself on one side of the equals sign, with everything else on the other side. First we subtract b from both sides, to get it out of the right side of the equation:
y - b = mx + b - b
y - b = mx
Now we divide both sides by x, to get m completely by itself:
(y-b)/x = (mx)/x
(y-b)/x = m
So m is equal to (y-b)/x.

Marrrta [24]3 years ago

8 0

Answer:

Step-by-step explanation:

y - b/x = m

Adding b/x both sides

y = m+b/x

But i think your question you're trying to ask is (y - b)/x = m

So, for this one,

Multiply both sides by x

y-b = mx

Adding b both sides

To get: y=mx+b

Hope this helps!

You might be interested in

Identify 2 objects you could find in a grocery store that holds more than 100 milliliters

devlian [24]

Milk carton
Perfume bottle

5 0

2 years ago

Read 2 more answers

How do you convert between fractions, decimals, and percentages?

svlad2 [7]

To convert to percent from fraction you divide the numerator by the denominator. from percent to decimal you move two decimal places to the left.

5 0

2 years ago

Read 2 more answers

Jaya is the middle of three siblings whose ages are consecutive even integers. If the sum of their ages is 84,find Jaya’s age

Svetach [21]

12345612345612317890

7 0

1 year ago

Which expression represents the number 2i4−5i3+3i2+−81‾‾‾‾√ rewritten in a+bi form?

vichka [17]

Answer:

The expression $-1+14i$ represents the number $2i^4-5i^3+3i^2+\sqrt{-81}$ rewritten in a+bi form.

Step-by-step explanation:

The value of $i$ is $i=\sqrt{-1}[tex] or [tex]i^{2}=-1[\tex].Now [tex]i^{4}$ in term of $i^{2}[\tex] can be written as, [tex]i^{4}=i^{2}\times i^{2}$

Substituting the value,

$i^{4}=\left(-1\right)\times \left(-1\right)$

Product of two negative numbers is always positive.

$\therefore i^{4}=1$

Now $i^{3}$ in term of $i^{2}[\tex] can be written as, [tex]i^{3}=i^{2}\times i$

Substituting the value,

$i^{3}=\left(-1\right)\times i$

Product of one negative and one positive numbers is always negative.

$\therefore i^{3}=-i$

Now $\sqrt{-81}$ can be written as follows,

$\sqrt{-81}=\sqrt{\left(81\right)\times\left(-1\right)}$

Applying radical multiplication rule,

$\sqrt{ab}={\sqrt{a}}\sqrt{b}$

$\sqrt{\left(81\right)\times\left(-1\right)}={\sqrt{81}}\sqrt{-1}$

Now, $\sqrt{\left(81\right)=9$ and $\sqrt{-1}}=i$

$\therefore \sqrt{\left(81\right)\times\left(-1\right)}=9i$

Now substituting the above values in given expression,

$2i^4-5i^3+3i^2+\sqrt{-81}=2\left(1\right)-5\left(-i\right)+3\left(-1\right)+9i$

Simplifying,

$2+5i-3+9i$

Collecting similar terms,

$2-3+5i+9i$

Combining similar terms,

$-1+14i$

The above expression is in the form of a+bi which is the required expression.

Hence, option number 4 is correct.

5 0

3 years ago

Tanisha cut a triangle with angles of 62°, 30°, and 88° in three pieces so that each piece had one angle of the triangle. She ar

-Dominant- [34]

The right answer for the question that is being asked and shown above is that: "180 - line." Tanisha cut a triangle with angles of 62°, 30°, and 88° in three pieces so that each piece had one angle of the triangle.She arranged the three angles <span>connected together such that the three corners of the triangle were all touching.</span><span> </span>

8 0

2 years ago