All categories

2 years ago

Multiply the given polynomials. Answer should be simplified and written in standard form.

Mathematics

1 answer:

vampirchik [111]2 years ago

8 0

Answer: 14k²-99k+52.

Step-by-step explanation:

$(2k-13)*(7k-4)=\\2k*(7k-4)-13*(7k-4)=\\14k^2-8k-91k+52=\\14k^2-99k+52.$

You might be interested in

Help please need to rewrite the table if you know how write the equation for the table

andrew11 [14]

Answer:

a. linear: y = 42-2x
b. non-linear: y = x(x +3)/2

Step-by-step explanation:

A function is linear if x-values are evenly spaced (all have the same difference) and y-values are evenly spaced (all have the same difference).

a) x-values have a difference of 6 -3 = 9 -6 = 12 - 9 = 3. y-values have a difference of 30 -36 = 24 -30 = 18 -24 = -6. Both these differences are constant, so the function is linear.

The ratio of y-differences to x-differences is -6/3 = -2, so that is the slope of the line. We can use the point-slope form to discover an equation in slope-intercept form.

Point-slope form of the equation for a line with slope m through point (h, k) can be written as ...

y = m(x -h) +k

Here, we have m = -2, and the first point is (h, k) = (3, 36). Then our line's equation can be written as ...

y = -2(x -3) +36

y = -2x +42 . . . . . . . eliminating parentheses

b) The x-values in this table have a constant difference of ...

3 -1 = 5 -3 = 7 -5 = 2

The y-values have differences of 9 -2 = 7, 20 -9 = 11, 35 -20 = 15. These are not constant, so the relation is non-linear. These first differences have differences of ...

11 -7 = 15 -11 = 4 . . . . . second differences are constant

When second differences are constant, the relation can be described by a second degree polynomial. We can write some equations to discover what that polynomial is.

Generic form:

ax² +bx +c = y

Filling in three of the given points, we have three equations in a, b, c:

a·1² + b·1 +c = 2

a·3² +b·3 +c = 9

a·5² +b·5 +c = 20

Subtracting the first equation from the other two eliminates c and gives two equations in a and b:

a·(9 -1) +b(3 -1) = 9 -2 . . . . . . . . . 8a +2b = 7

a·(25 -1) + b·(5 -1) = 20 -2 . . . . . 24a +4b = 18

Subtracting twice the first of these equations from the second, we can eliminate b:

a(24 -2·8) = 18 -2·7

8a = 4

a = 1/2 . . . . . divide by 8

Substituting this into the first of the equations in a and b, we get:

8·(1/2) +2b = 7

2b = 3 . . . . . subtract 4

b = 3/2 . . . . divide by 2

Substituting for a and b in the first of our original equations, we find ...

(1/2)·1 + (3/2)·(1) +c = 2

2 +c = 2 . . . simplify

c = 0 . . . . . . subtract 2

So, the table in part b can be described by the quadratic equation ...

y = (1/2)x² + (3/2)x

5 0

4 years ago

Use the given transformation to evaluate the integral. (15x + 15y) dA R , where R is the parallelogram with vertices (−1, 4), (1

MA_775_DIABLO [31]

Answer:

$\int_R 15x+15y dA = \frac{8}{16875}$

Step-by-step explanation:

Recall the following: x = 15u+15v, y = -60u+15v. So, x-y = 75u. Then u = (x-y)/75. 4x+y = 75v. Then v = (4x+y)/75.

We will see how this transformation maps the region R to a new region in the u-v domain. To do so, we will see where the transformation maps the vertices of the region.

(-1,4) -> ((-1-4)/75,(4(-1)+4)/75) = (-1/15, 0)

(1,-4)->(1/15,0)

(3,-2)->(1/15,2/15)

(1,6)->(-1/15,2/15)

That is, the new region in the u-v domain is a rectangle where $\frac{-1}{15}\leq u \leq \frac{1}{15}, 0\leq v \leq \frac{2}{15}$ .

We will calculate the jacobian of the change variables. That is

$\left |\begin{matrix} \frac{du}{dx}& \frac{du}{dy}\\ \frac{dv}{dx}& \frac{dv}{dy}\end{matrix}\right|$ (we are calculating the determinant of this matrix). The matrix is

$\left |\begin{matrix} \frac{1}{75}& \frac{-1}{75}\\ \frac{4}{75}& \frac{1}{75}\end{matrix}\right|=(\frac{1}{75^2})(1+4) = \frac{1}{15\cdot 75}$ (the in-between calculations are omitted).

We will, finally, do the calculations.

Recall that

$15x+15y = 15(15u+15v) + 15(-60u+15v) = (15^2-15\cdot 60 )u+2\cdot 15^2v = 15^2(-3)u+2\cdot 15^2 v$

We will use the change of variables theorem. So,

$\int_R 15x+15y dA = \int_{\frac{-1}{15}}^{\frac{1}{15}}\int_{0}^{\frac{2}{15}} 15^2(-3)u+2\cdot 15^2 v \cdot (\frac{1}{15^2\cdot 5}) dv du = \int_{\frac{-1}{15}}^{\frac{1}{15}}\int_{0}^{\frac{2}{15}}\frac{-3}{5}u+\frac{2}{5}v dvdu$

This si because we are expressing the original integral in the new variables. We must multiply by the jacobian to guarantee that the change of variables doesn't affect the value of the integral. Then,

$\int_{\frac{-1}{15}}^{\frac{1}{15}}\int_{0}^{\frac{2}{15}}\frac{-3}{5}u+\frac{2}{5}v dvdu = \int_{\frac{-1}{15}}^{\frac{1}{15}}\frac{-3}{5}u\cdot \frac{2}{15} + \frac{2}{5}\cdot \left.\frac{v^2}{2}\right|_{0}^{\frac{2}{15}}du = \frac{-3}{5}\left.\frac{u^2}{2}\right|_{\frac{-1}{15}}^{\frac{1}{15}}\cdot \frac{2}{15} + \frac{2}{5}\cdot \left.\frac{v^2}{2}\right|_{0}^{\frac{2}{15}} = \frac{8}{16875}$

5 0

4 years ago

Which conclusions can be drawn from the results of the survey? Check all that apply. There is no association between using socia

den301095 [7]

The sample space of an experiment is changed when some additional information pertaining to the outcome of the experiment is received. The effect of such information is to reduce the sample space by excluding some outcomes as being impossible which before receiving the information were believed possible . The probabilities associated with such a reduced sample space are called conditional probabilities.

Knowing a student’s grade level does not help determine if he or she uses social media.

The given survey explains the conditional probability.

The students of the reduced sample space i.e. of 9th grade have the probability of 81% who use the social media.

The students of 8th grade that use social media can be greater or less than 81%.

Therefore this experiment does not help determine if he or she uses social media on the basis of his/her grade.

Only the last option is correct.

The conditional probability can be understood by the following.

brainly.com/question/10739947

brainly.com/question/4403090

7 0

3 years ago

Which pair of expressions below