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2 years ago

Let be an angle in standard position. Name the quadrant in which lies. cos >0, cot >0

Mathematics

1 answer:

Alexeev081 [22]2 years ago

4 0

Answer:

Quadrant I

Step-by-step explanation:

The quadrants where $cos\theta>0$ are Quadrant I and Quadrant IV. Since $cot\theta=\frac{cos\theta}{sin\theta}$ and $sin\theta$ is positive in Quadrant I, then Quadrant I will contain values of $\theta$ where $cos\theta>0$ and $cot\theta>0$

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Need geometry help please

Tatiana [17]

Answer:

I think C

Step-by-step explanation:

8 0

3 years ago

Compute the difference quotient, StartFraction f (x plus h )minus f (x )Over h EndFraction f(x+h)−f(x) h, for the function f (

lawyer [7]

Answer:

The difference quotient for $f(x)=3x^2$ is $3 h + 6 x$ .

Step-by-step explanation:

The difference quotient is a formula that computes the slope of the secant line through two points on the graph of <em>f</em>. These are the points with x-coordinates x and x + h. The difference quotient is used in the definition the derivative and it is given by

$\frac{f(x+h)-f(x)}{h}$

So, for the function $f(x)=3x^2$ the difference quotient is:

To find $f(x+h)$ , plug $x+h$ instead of $x$

$f\left(x+h\right)=3 \left(h + x\right)^{2}$

Finally,

$\frac{f\left(x+h\right)-f\left(x\right)}{h}=\frac{\left(3 \left(h + x\right)^{2}\right)-\left(3 x^{2}\right)}{h}$

$\frac{3\left(\left(h+x\right)^2-x^2\right)}{h} \\\\\frac{3(h^2+2hx+x^2-x^2)}{h} \\\\\frac{3h\left(h+2x\right)}{h}\\\\3h+6x$

The difference quotient for $f(x)=3x^2$ is $3 h + 6 x$ .

4 0

3 years ago

Fourteen more than a number is eighteen.

AVprozaik [17]

Answer: 4

Step-by-step explanation:

18-14=4

3 0

3 years ago

Help please!!!!!!!!!

In-s [12.5K]

ANSWER

EXPLANATION

For a matrix A of order n×n, the cofactor $C_{ij}$ of element $a_{ij}$ is defined to be

$C_{ij} = (-1)^{i+j} M_{ij}$

$M_{ij}$ is the minor of element $a_{ij}$ equal to the determinant of the matrix we get by taking matrix A and deleting row i and column j.

Here, we have

$C_{11} = (-1)^{1+1} M_{11} = M_{11}$

M₁₁ is the determinant of the matrix that is matrix A with row 1 and column 1 removed. The bold entries are the row and the column we delete.

$\begin{aligned} A=\begin{bmatrix} \bf 1 & \bf -6 & \bf -4\\ \bf 7 & 0 & -3 \\ \bf -9 & 8 & -8 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 0&-3 \\ 8&-8 \end{bmatrix} \right) \end{aligned}$