Find the solution to the graph EASY 30 POINTS

Calculate the area of triangle ABC with altitude BD, given A (−6, 0), B (0, 0), C (0, 6), and D (−3, 3).

nata0808 [166]

Answer:

18 unit²

Step-by-step explanation:

If BD is altitude then AC is the base.

The length of AC is:

AC = $\sqrt{6^2+6^2} = 6\sqrt{2}$

The length of BD is:

BD = $\sqrt{3^2 + 3^2} = 3\sqrt{2}$

The area is:

A = 1/2bh
A = 1/2 * $6\sqrt{2} *3\sqrt{2}$ = 18 unit²

4 0

3 years ago

Read 2 more answers

WILL MARK YOU BRAINLIEST

Naddik [55]

I think x = 78
because 21/14 = 1.5
so x = 1.5*52
x = 78

5 0

3 years ago

From first principles, find the indicated derivatives

LenaWriter [7]

By definition of the derivative,

$\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac{\left(\frac{(s + h)^3}2 + 1\right) - \left(\frac{s^3}2 + 1\right)}{h}$

$\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac{\left(\frac{s^3+3s^2h+3sh^2+h^3}2 + 1\right) - \left(\frac{s^3}2 + 1\right)}{h}$

$\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac{\frac{3s^2h+3sh^2+h^3}2}{h}$

$\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac12 \frac{3s^2h+3sh^2+h^3}{h}$

$\displaystyle\frac{dr}{ds} = \lim_{h\to0} \frac12 (3s^2+3sh+h^2)$

$\displaystyle\frac{dr}{ds} = \frac{3s^2}2$

6 0

2 years ago

Hi, can you help me with these? It's really hard :(

Shalnov [3]

Answer:

44 Rachel earn in 10 hour

8 0

2 years ago

Read 2 more answers

A biologist is studying the growth of a particular species of algae. She writes the fall inclusion to show the radius of the alg

horrorfan [7]

Answer:

Part A)

A reasonable domain to plot the growth function is:

$0\leq d \leq 12$

Part B)

The y-intercept represents that when the biologist started her study, the radius of the algae was five millimeters.

Part C)

The average rate of change from d = 4 to d = 11 was about 0.11. This means that from the 4th day to the 11th, the radius of the algae grew, on average, at a rate of 0.11 mm per day.

Step-by-step explanation:

The radius of the algae f(d) in millimeters after d days is given by the function:

$f(d)=5(1.02)^d$

Part A)

We know that the radius of the algae was approximately 6.34 mm when the biologist concluded her study. To find the reasonable domain, we can substitute 6.34 for f(d) and solve for d. Therefore:

$6.34=5(1.02)^d$

Divide both sides by five:

$(1.02)^d=1.268$

Take the log of both sides with base 1.02:

$\displaystyle d=\log_{1.02}1.268$

Using the Change of Base Property, evaluate for d:

$\displaystyle d=\frac{\log 1.268}{\log 1.02}=11.9903...\approx 12$

So, the biologist concluded her study after 12 days.

Therefore, a reasonable domain to plot the growth function is:

$0\leq d\leq 12$

Part 2)

The y-intercept of the function is when d = 0. Find the y-intercept:

$f(0)=5(1.02)^{(0)}=5(1)=5$

Since d represent the amount of days after the study had begun, the y-intercept represents the radius of the algae on the initial day.

So, when the biologist started her study, the radius of the algae was five millimeters.

Part 3)

To find the average rate of change for a nonlinear function, we find the slope between the two endpoints on the interval.

We want to find the average rate of change of f(d) from d = 4 to d = 11.

Find the endpoints:

$f(4)=5.4121...\text{ and } f(11)=6.2168...$

And find the slope between them:

$\displaystyle m=\frac{f(11)-f(4)}{11-4}=0.1149...\approx 0.11$

Since f(d) measures millimeters and d measures days, this tells us that, on average, the radius of the algae grew by about 0.11 mm per day from the 4th day to the 11th day.

5 0

2 years ago