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3 years ago

-(12x-6)=12x 6 solve for x

Mathematics

1 answer:

postnew [5]3 years ago

5 0

6=24x
hope this helped :)
tell me if u need anymore help

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The M&M’s website says that 24% of milk chocolate M&M’s are blue. Suppose that we buy 5 small packets of milk chocolate

SVEN [57.7K]

Answer:

The correct option is A: 24%, 24%, 24%, 24%, 24%

Step-by-step explanation:

Since the overall probability of M&M's produced is stated as 24% and the packets are filled with different colored candies at random, the probability of blue colored candies in each packet cannot be ascertained individually.

Hope that answers the question, have a great day!

7 0

3 years ago

There are 20 bucks on the summer reading list how many different ways can you choose one per week for six weeks

eduard

Answer:

There would be 120 different ways

Step-by-step explanation:

20×6=120

6 0

2 years ago

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0 = -5x + 4y in standard form

Effectus [21]

Answer:

5x - 4y = 0

Step-by-step explanation:

the equation of a line in standard form is

Ax + By = C ( A is a positive integer and B, C are integers )

given 0 = - 5x + 4y ( multiply through by - 1 )

5x - 4y = 0 ← in standard form

6 0

3 years ago

Read 2 more answers

Plz help me

GenaCL600 [577]

The correct answer is b.

6 0

3 years ago

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Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. Illustra

Rainbow [258]

Answer:

$x = t$

$y = 1 - t$

$z = 2t$

Step-by-step explanation:

Given

$x=t$

$y=e^{-t}$

$z=2t-t^2$

(0, 1, 0)

The vector equation is given as:

$r(t) = (x,y,z)$

Substitute values for x, y and z

$r(t) = (t,\ e^{-t},\ 2t - t^2)$

Differentiate:

$r'(t) = (1,\ -e^{-t},\ 2 - 2t)$

The parametric value that corresponds to (0, 1, 0) is:

$t = 0$

Substitute 0 for t in r'(t)

$r'(t) = (1,\ -e^{-t},\ 2 - 2t)$

$r'(0) = (1,\ -e^{-0},\ 2 - 2*0)$

$r'(0) = (1,\ -1,\ 2 - 0)$

$r'(0) = (1,\ -1,\ 2)$

The tangent line passes through (0, 1, 0) and the tangent line is parallel to r'(0)

It should be noted that:

The equation of a line through position vector a and parallel to vector v is given as:

$r(t) = a + tv$

Such that:

$a = (0,1,0)$ and $v = r'(0) = (1,-1,2)$

The equation becomes:

$r(t) = (0,1,0) + t(1,-1,2)$

$r(t) = (0,1,0) + (t,-t,2t)$

$r(t) = (0+t,1-t,0+2t)$

$r(t) = (t,1-t,2t)$

By comparison:

$r(t) = (x,y,z)$ and $r(t) = (t,1-t,2t)$

The parametric equations for the tangent line are:

$x = t$

$y = 1 - t$

$z = 2t$

7 0

2 years ago