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3 years ago

What is -5x + 3y= 12 in slope intercept form?

Mathematics

1 answer:

babymother [125]3 years ago

8 0

Answer:

y = (5/3)x + 4

Step-by-step explanation:

-5x + 3y = 12 (What you start with)

3y = 12 + 5x (Adding 5x to both sides)

y = (5/3)x + 4 (Dividing the equation by 3)

You might be interested in

12% 0f 72 is what number

earnstyle [38]

12% of 72 is:

=12×72/100

=8.64

So the 12% of 72 is 8.64 is your answer.

7 0

3 years ago

While Preparing For A Morning Conference, Principal Corsetti Is Laying Out 8 Dozen Bagels On Square Plates. Each Plate Can Hold

USPshnik [31]

We have 8 dozen bagels, or 8*12=96 bagels. Each plate can hold 14 bagels, so we have enough bagels to fill 96/14=about 6.86 plates. However, we cannot have a fraction of a plate, so we round up to have a total of seven plates. To fill all seven plates fully, 7*14=98 bagels would be needed, which is two more than we have.

To summarize, Mr. Corsetti has seven plates of bagels, and would need two more bagels to fill the last one up.

6 0

3 years ago

What is 5/6 divided by 5

dimulka [17.4K]

.166 and with 6 repeating

7 0

3 years ago

Read 2 more answers

Which value is NOT a solution of 8x3 – 1 = 0?

Tpy6a [65]

Note: As you may have unintentionally missed to add the value choices. But, I would make sure to explain the concept so that you may improve your understanding in terms of solving these type of questions.

Answer:

Any value other than the values $x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4}$ will not be a solution of $8x^3\:-\:1\:=\:0$ .

Step-by-step explanation:

Considering the equation

$8x^3\:-\:1\:=\:0$

Steps to solve the equation

$8x^3-1=0$

$\mathrm{Add\:}1\mathrm{\:to\:both\:sides}$

$8x^3-1+1=0+1$

$\mathrm{Simplify}$

$x^3=\frac{1}{8}$

$\mathrm{Divide\:both\:sides\:by\:}8$

$\frac{8x^3}{8}=\frac{1}{8}$

$\mathrm{Simplify}$

$x^3=\frac{1}{8}$

$\mathrm{For\:}x^3=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt[3]{f\left(a\right)},\:\sqrt[3]{f\left(a\right)}\frac{-1-\sqrt{3}i}{2},\:\sqrt[3]{f\left(a\right)}\frac{-1+\sqrt{3}i}{2}$

$x=\sqrt[3]{\frac{1}{8}},\:x=\sqrt[3]{\frac{1}{8}}\frac{-1+\sqrt{3}i}{2},\:x=\sqrt[3]{\frac{1}{8}}\frac{-1-\sqrt{3}i}{2}$

So,

$x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4}$

Therefore,

Any value other than the values $x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4}$ will not be a solution of $8x^3\:-\:1\:=\:0$ .

Keywords: solution, value

Learn more about equation solution from brainly.com/question/1679491

#learnwithBrainly

7 0

3 years ago

If the sum of the zereos of the quadratic polynomial is 3x^2-(3k-2)x-(k-6) is equal to the product of the zereos, then find k?

lys-0071 [83]

Answer:

Step-by-step explanation:

So I'm going to use vieta's formula.

Let u and v the zeros of the given quadratic in ax^2+bx+c form.

By vieta's formula: