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3 years ago

Cyclic quadrilateral.

Mathematics

2 answers:

tresset_1 [31]3 years ago

7 0

Write the question properly buddy

ad-work [718]3 years ago

5 0

What do you mean, you need more information in your queston p,ease try to ask it again, it makes no Sence

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Select the correct answer.

Neporo4naja [7]

Answer:

c salary refers to the fixed amount paid on a weekly bi-weekly

5 0

3 years ago

Read 2 more answers

Which equation has a graph that is perpendicular to the graph of -x + 6y = -12?

serious [3.7K]

Answer:

c) 6x + y = -52 is required equation perpendicular to the given equation.

Step-by-step explanation:

If the equation is of the form : y = mx + C.

Here m = slope of the equation.

Two equations are said to be perpendicular if the product of their respective slopes is -1.

Here, equation 1 : -x + 6y = -12

or, 6y = -12 + x

or, y = (x/6) - 2

⇒Slope of line 1 = (1/6)

Now, for equation 2 to be perpendicular:

Check for each equation:

a. x + 6y = -67 ⇒ 6y = -67 - x

or, y = (-x/6) - (67/6) ⇒Slope of line 2 = (-1/6)

but $\frac{1}{6} \times \frac{-1}{6} \neq -1$

b. x - 6y = -52 ⇒ -6y = -52 - x

or, y = (x/6) + (52/6) ⇒Slope of line 2 = (1/6)

but $\frac{1}{6} \times \frac{1}{6} \neq -1$

c. 6x + y = -52

or, y =y = -52 - 6x ⇒Slope of line 2 = (-6)

$\frac{1}{6} \times (-6) = -1$

Hence, 6x + y = -52 is required equation 2.

d. 6x - y = 52 ⇒ -y = 52 - 6x

or, y = 6x - 52 ⇒Slope of line 2 = (6)

but $\frac{1}{6} \times 6 \neq -1$

Hence, 6x + y = -52 is the only required equation .

3 0

3 years ago

Read 2 more answers

How to Graph of y=1/2x-4

Advocard [28]

Answer and explanation:

We know that this is in slope-intercept form. ( y = mx + b )

1/2 would be the slope (you would use this to graph the next point)

-4 would by your y-intercept (where the line crosses the x-axis)

With this information you could graph any point using the slope given, starting at -4 (x-axis) going down or up.

3 0

3 years ago

Read 2 more answers

Find the fifth term in the sequence<br>that is defined as follows:<br>an = 2n

Rainbow [258]

Answer:

$a_{5}=10$

Step-by-step explanation:

If you want to find the 5th term of a sequence or any term you have to substitute n for the desired term in your case the result is the following

$a_{n}=2n\\a_{5}=2(5)\\a_{5}=10$

7 0

3 years ago

Suppose that we don't have a formula for g(x) but we know that g(3) = −5 and g'(x) = x2 + 7 for all x.

Leni [432]

Answer:

$g(2.9) \approx -6.6$

$g(3.1) \approx -3.4$

The values are too small since $g''$ is positive for both values of $x$ in. I'm speaking of the $x$ values, 2.9 and 3.1.

Step-by-step explanation:

The point-slope of a line is:

$y-y_1=m(x-x_1)$

where $m$ is the slope and $(x_1,y_1)$ is a point on that line.

We want to find the equation of the tangent line of the curve $g$ at the point $(3,-5)$ on $g$ .

So we know $(x_1,y_1)=(3,-5)$ .

To find $m$ , we must calculate the derivative of $g$ at $x=3$ :

$m=g'(3)=(3)^2+7=9+7=16$ .

So the equation of the tangent line to curve $g$ at $(3,-5)$ is:

$y-(-5)=16(x-3)$ .

I'm going to solve this for $y$ .

$y-(-5)=16(x-3)$

$y+5=16(x-3)$

Subtract 5 on both sides:

$y=16(x-3)-5$

What this means is for values $x$ near $x=3$ is that:

$g(x) \approx 16(x-3)-5$ .

Let's evaluate this approximation function for $g(2.9)$ .

$g(2.9) \approx 16(2.9-3)-5$

$g(2.9) \approx 16(-.1)-5$

$g(2.9) \approx -1.6-5$

$g(2.9) \approx -6.6$

Let's evaluate this approximation function for $g(3.1)$ .

$g(3.1) \approx 16(3.1-3)-5$

$g(3.1) \approx 16(.1)-5$

$g(3.1) \approx 1.6-5$

$g(3.1) \approx -3.4$

b) To determine if these are over approximations or under approximations I will require the second derivative.

If $g''$ is positive, then it leads to underestimation (since the curve is concave up at that number).

If $g''$ is negative, then it leads to overestimation (since the curve is concave down at that number).

$g'(x)=x^2+7$

$g''(x)=2x+0$

$g''(x)=2x$

$2x$ is positive for $x>0$ .

$2x$ is negative for $x$ .

That is, $g''(2.9)>0 \text{ and } g''(3.1)>0$ .

So $2x$ is positive for both values of $x$ which means that the values we found in part (a) are underestimations.

6 0

4 years ago