2x + 3y = 0 X + 2y = -1 Is this one solution No solution Or infinite solution

2x^2y term, constant, and coefficient pls help me pls

prohojiy [21]

Answer:

16.2

Step-by-step explanation:

because 3x2=6

I AM SOOOO SORRY IF I GET IT WRONG

8 0

3 years ago

HW HELP! Will mark you BRAINLIEST!!

blsea [12.9K]

Answer:

16. DF = - 1-(-4) = 3

17. DE = -1/1/3 -(-4)

= - 5/3 + 12/3

= 7 / 3

18. FG = 2-(-1) = 3

19. FH = 3.5 - (-1) = 4.5

20. GH = 3.5 - 2 = 1.5

21. EH = 3.5 -(-1.5) = 5

22. AC = 16

3x + 7 = 16

x = 3

23. AB = x +7

= 10

24. BD = 2x + 3x - 1

=5(3) - 1

=14

25. CE = 3x - 1 + 2x + 3

=5x +2

=17

3 0

2 years ago

What is the slope of the line? PLS Help

marin [14]

Answer:

-2/5

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

At a certain time of day, the angle of elevation of the sun is 40°. To the nearest foot,find the height of a tree whose shadow i

konstantin123 [22]

We need to find a tree such that the angle of elevation from the end of the shadow to top of the tree is 40 degrees.

The length of the shadow is the adjacent side and is 35.
The height of the tree is the opposite side. Let it be x.

Tan ratio = opposite/adjacent

tan(40) = x/35
x = 35*tan(40) = 29.37

Answer: Height of the tree is 29 feet

4 0

3 years ago

A conical container, with vertex down, has a height of 6 cm and a diameter of 2 cm. It is leaking water at

user100 [1]

Answer:

$\displaystyle \frac{dh}{dt}=-\frac{4}{\pi}\approx-1.2732\text{ centimeters per minute}$

The water level is dropping by approximately 1.27 centimeters per minute.

Step-by-step explanation:

Please refer to the attached diagram.

The height of the conical container is 6 cm, and its radius is 1 cm.

The container is leaking water at a rate of 1 cubic centimeter per minute.

And we want to find the rate at which the water level h is dropping when the water height is 3 cm.

Since we are relating the water leaked to the height of the water level, we will consider the volume formula for a cone, given by:

$\displaystyle V=\frac{1}{3}\pi r^2h$

Now, we can establish the relationship between the radius r and the height h. At any given point, we will have two similar triangles as shown below. Therefore, we can write:

$\displaystyle \frac{1}{6}=\frac{r}{h}$

Solving for r yields:

$\displaystyle r=\frac{1}{6}h$

So, we will substitute this into our volume formula. This yields:

$\displaystyle \begin{aligned} V&=\frac{1}{3}\pi \Big(\frac{1}{6}h\Big)^2h\\ &=\frac{1}{108}\pi h^3\end{aligned}$

Now, we will differentiate both sides with respect to time t. Hence:

$\displaystyle \frac{d}{dt}[V]=\frac{d}{dt}\Big[\frac{1}{108}\pi h^3\Big]$

The left is simply dV/dt. We can move the coefficient from the right:

$\displaystyle \frac{dV}{dt}=\frac{1}{108}\pi\frac{d}{dt}\big[h^3\big]$

Implicitly differentiate:

$\displaystyle\begin{aligned} \frac{dV}{dt}&=\frac{1}{108}\pi(3h^2\frac{dh}{dt})\\ &=\frac{1}{36}\pi h^2\frac{dh}{dt}\end{aligned}$

Since the water is leaking at a rate of 1 cubic centimeter per minute, dV/dt=-1.

We want to find the rate at which the water level h is dropping when the height of the water is 3 cm.. So, we want to find dh/dt when h=3.

So, by substitution, we acquire:

$\displaystyle -1=\frac{1}{36}\pi(3)^2\frac{dh}{dt}$

Therefore:

$\displaystyle -1=\frac{1}{4}\pi\frac{dh}{dt}$

Hence:

$\displaystyle \frac{dh}{dt}=-\frac{4}{\pi}\approx-1.2732\text{ centimeters per minute}$

The water level is dropping at a rate of approximately 1.27 centimeters per minute.

8 0

3 years ago