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4 years ago

If f(x)=3x+2, what is f(5)

Mathematics

1 answer:

patriot [66]4 years ago

5 0

Answer:

Step-by-step explanation:

f(x)=3x+2

To find f (5) replace every x in f(x) by 5

That's

f(5) = 3(5) + 2

= 15 + 2

Hope this helps you

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Find the percentage of decrease from 95 to 60

balu736 [363]

The answer is about 36.9%
I just guessed and checked. 95-60= 35
So .3685×95 is the closest to equaling 35

6 0

4 years ago

Complete each equation so that it has the indicated number of solutions

Akimi4 [234]

Answer:

No solutions: 3x + 1 = 3x + b, where b ≠ 1

Infinite solutions: 2x - 4 = 2x - 4

Step-by-step explanation:

For an equation to have <em>no </em>solutions, it has to reduce to an equation that's never true. For our given equation, this means we'll need to fill in the blank with some value other than 1. If we pick 2, for instance, we'll get the equation 3x + 1 = 3x + 2, which reduces to 1 = 2, a false statement. No value of x will make the equation true.

To have <em>infinite </em>solutions, on the other hand, we need an equation that is <em>always</em> true. Filling in the blank with 4 here will give us one: 2x - 4 = 2x - 4 reduces to -4 = -4, a statement that is always true.

5 0

3 years ago

Mr Jenkins bought pavers for some landscaping projects. He used 120 for a small patio, 86 for the border of a flower bed, and 70

Nastasia [14]

All you need to do is add all of the numbers up.

120+86+70+24 =300

This means Mr. Jenkins bought 300 pavers.

6 0

3 years ago

Read 2 more answers

Estimate the sum of 24 and 36

Natasha_Volkova [10]

Round 24 to 20 and then 36 to 40 add 20 plus 40 and u get 60

7 0

3 years ago

Compute the matrix of partial derivatives of the following functions.

s344n2d4d5 [400]

For a vector-valued function

$\mathbf f(\mathbf x)=\mathbf f(x_1,x_2,\ldots,x_n)=(f_1(x_1,x_2,\ldots,x_n),\ldots,f_m(x_1,x_2,\ldots,x_n))$

the matrix of partial derivatives (a.k.a. the Jacobian) is the $m\times n$ matrix in which the $(i,j)$ -th entry is the derivative of $f_i$ with respect to $x_j$ :

$D\mathbf f(\mathbf x)=\begin{bmatrix}\dfrac{\partial f_1}{\partial x_1}&\dfrac{\partial f_1}{\partial x_2}&\cdots&\dfrac{\partial f_1}{\partial x_n}\\\dfrac{\partial f_2}{\partial x_1}&\dfrac{\partial f_2}{\partial x_2}&\cdots&\dfrac{\partial f_2}{\partial x_n}\\\vdots&\vdots&\ddots&\vdots\\\dfrac{\partial f_m}{\partial x_1}&\dfrac{\partial f_m}{\partial x_2}&\cdots&\dfrac{\partial f_n}{\partial x_n}\end{bmatrix}$

So we have

(a)

$D f(x,y)=\begin{bmatrix}\dfrac{\partial(e^x)}{\partial x}&\dfrac{\partial(e^x)}{\partial y}\\\dfrac{\partial(\sin(xy))}{\partial x}&\dfrac{\partial(\sin(xy))}{\partial y}\end{bmatrix}=\begin{bmatrix}e^x&0\\y\cos(xy)&x\cos(xy)\end{bmatrix}$

(b)

$D f(x,y,z)=\begin{bmatrix}\dfrac{\partial(x-y)}{\partial x}&\dfrac{\partial(x-y)}{\partial y}&\dfrac{\partial(x-y)}{\partial z}\\\dfrac{\partial(y+z)}{\partial x}&\dfrac{\partial(y+z)}{\partial y}&\dfrac{\partial(y+z)}{\partial z}\end{bmatrix}=\begin{bmatrix}1&-1&0\\0&1&1\end{bmatrix}$

(c)

$Df(x,y)=\begin{bmatrix}y&x\\1&-1\\y&x\end{bmatrix}$

(d)

$Df(x,y,z)=\begin{bmatrix}1&0&1\\0&1&0\\1&-1&0\end{bmatrix}$

5 0

4 years ago