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4 years ago

Find an equation of the line that

Mathematics

1 answer:

Viefleur [7K]4 years ago

7 0

Answer:

y = (-11/12)x - 3

Step-by-step explanation:

Recall that for a linear equation in the form y = mx + b,

m = slope and b = y-intercept.

hence to satisfy parts (a) and (b) of the questions, we need to extract the y-intercept from the equation in part (a) and the slope from part (b)

For part (a)

given: y+7x+3=0 (rearrange this to get slope-intercept form)

y+7x+3=0 (subtract 3 from both sides)

y + 7x = -3 (subtract 7x from both sides)

y = -7x - 3

we can see that the y-intercept is -3.

For part (b)

given: -11x -12y=-7 (rearrange this to get slope-intercept form)

-11x -12y=-7 (add 11x toboth sides)

-12y = 11x - 7 (divide both sides by -12)

y = (-11/12) - (7/12)

we can see that the slope is (-11/12).

combining the y-intercept from part (a) and the slope from part (b) into the general equation given in the first line above,

y = (-11/12)x - 3 (answer)

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4a-2b+40 when a=2 and b=7

hammer [34]

Answer:

Step-by-step explanation:

4(2)-2(7)+40

8-14+40

-6+40=34

7 0

3 years ago

Read 2 more answers

Which expression is equivalent?

babymother [125]

Answer:

Third choice from the top is the one you want

Step-by-step explanation:

This whole concept relies on the fact that if the index of a radical exactly matches the power under the radical, both the radical and the power cancel each other out. For example:

$\sqrt[6]{x^6} =x$ and another example:

$\sqrt[12]{2^{12}}=2$

Let's take this step by step. First we will rewrite both the numerator and the denominator in rational exponential equivalencies:

$\frac{\sqrt[4]{6} }{\sqrt[3]{2} }=\frac{6^{\frac{1}{4} }}{2^{\frac{1}{3} }}$

In order to do anything with this, we need to make the index (ie. the denominators of each of those rational exponents) the same number. The LCM of 3 and 4 is 12. So we rewrite as

$\frac{6^{\frac{3}{12} }}{2^{\frac{4}{12} }}$

Now we will put it back into radical form so we can rationalize the denominator:

$\frac{\sqrt[12]{6^3} }{\sqrt[12]{2^4} }$

In order to rationalize the denominator, we need the power on the 2 to be a 12. Right now it's a 4, so we are "missing" 8. The rule for multiplying like bases is that you add the exponents. Therefore,

$2^4*2^8=2^{12}$

We will rationalize by multiplying in a unit multiplier equal to 1 in the form of

$\frac{\sqrt[12]{2^8} }{\sqrt[12]{2^8} }$

That looks like this:

$\frac{\sqrt[12]{6^3} }{\sqrt[12]{2^4} }*\frac{\sqrt[12]{2^8} }{\sqrt[12]{2^8} }$

This simplifies down to

$\frac{\sqrt[12]{216*256} }{\sqrt[12]{2^{12}} }$

Since the index and the power on the 2 are both 12, they cancel each other out leaving us with just a 2! Doing the multiplication of those 2 numbers in the numerator gives us, as a final answer:

$\frac{\sqrt[12]{55296} }{2}$