All categories

3 years ago

How do you multiply a whole number with a mixed number

1 answer:

6 0

You would change the whole number to a fraction by putting it over 1 and make the mixed number an improper fraction by multiplying the whole number part of the mixed number by the denominator and then and the numerator

You might be interested in

Find area of the shape

pentagon [3]

The answer is a. You find the area of each shape and add them

4 0

3 years ago

There are 15 questions in this quiz. Im sorry if im putting too much pressure on yall

nika2105 [10]

Answer:

Step-by-step explanation:

i dont really kbow why but i think it 86

4 0

3 years ago

Read 2 more answers

Consider the function, f(x) = 3x - 3. Determine the asymptote for the function f(x).

Natali [406]

Answer:

Y= 1

Step-by-step explanation:

Given the expression f(x) = 3x - 3.

To solve for the asymptote y

Firstly we need to solve for x

3x-3=0

3x=3

x= 3/3

x=1

Hence the asymptote of the function y=1

5 0

3 years ago

Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen

Jobisdone [24]

Parameterize $S$ by the vector function

$\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k$

so that the normal vector to $S$ is given by

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}$

In this case, the normal vector is

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k$

with magnitude $\sqrt{1^2+2^2+1^2}=\sqrt6$ . The integral of $f(x,y,z)=e^z$ over $S$ is then

$\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx$

where $T$ is the region in the $x,y$ plane over which $S$ is defined. In this case, it's the triangle in the plane $z=0$ which we can capture with $0\le x\le8$ and $0\le y\le\frac{8-x}2$ , so that we have

$\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}$

5 0

3 years ago

Given the function f(x) = 3|x – 2| + 6, for what values of x is f(x) = 18?

Gwar [14]

Answer:

x = - 2, x = 6

Step-by-step explanation:

Given f(x) = 18 we require to solve

3 | x - 2 | + 6 = 18 ( subtract 6 from both sides )

3 | x - 2 | = 12 ( divide both sides by 3 )

| x - 2 | = 4

The absolute value function always returns a positive value, however, the expression inside can be positive or negative, thus

x - 2 = 4 ( add 2 to both sides )

x = 6

- (x - 2) = 4

- x + 2 = 4 ( subtract 2 from both sides )

- x = 2 ( multiply both sides by - 1 )

x = - 2

As a check substitute these values into the left side of the equation and if equal to the right side then they are the solutions

x = 6 → 3|6 - 2| + 6 = 3|4| + 6 = 3(4) + 6 = 12 + 6 = 18 ← True

x = - 2 → 3|- 2 - 2| + 6 = 3|-4| + 6 = 3(4) + 6 = 12 + 6 = 18 ← True

Hence solutions are x = - 2, x = 6

7 0

3 years ago

Read 2 more answers