What is the solution of the equation 16-17+ Vb=1?<br> b = 3<br> O b = 9<br> O b = 81<br> no solution

Help me. Please due today

Schach [20]

Answer:

2). As x-> -∞, f(x)->∞

As x-> ∞, f(x)-> -∞

5). As x-> -∞, f(x)-> -∞

As x-> ∞, f(x)-> ∞

3). As x-> -∞, f(x)-> -∞

As x-> ∞, f(x)-> ∞

6). As x-> -∞, f(x)-> ∞

As x-> ∞, f(x)-> ∞

Step-by-step explanation:

I just watched a quick video so you can't completely trust me, but i tried my best. Hopefully someone more trustworthy for this comes in.

6 0

2 years ago

PLEASE HELP ASAP!!!!!!!!!!!!!!!!!!!!!!!!

expeople1 [14]

Answer:

Step-by-step explanation:

2nd one

6 0

3 years ago

How to find the area of a polygon

ipn [44]

The rectangle is 7 x 4 so the area is 28 sq cm
The triangle has a base of 4 and a height of 3.
The formula is A = 1/2 * b*h
You have 2 triangles so:
2(1/2)*4*3= 12 sq cm
Add together:
28 + 12= 40 sq cm

6 0

3 years ago

Carolina y su mamá compraron tres tortas de sabores diferentes para la reunión con sus amigos del colegio. De la torta de naranj

Karo-lina-s [1.5K]

Answer: 2 1/5 partes de torta

Step-by-step explanation:

Para contestar esta pregunta debemos sumar cada parte de torta:

0.7+9/12+6/8 =

Convertimos las fracciones en decimales, para tener todos los valores expresados de la misma forma. Dividiendo:

9/12=0.75

6/8=0.75

Sumamos :

0.7+0.75+0.75 = 2.2 partes = 2 1/5 partes

En conclusión, Carolina y sus amigos comieron 2 1/5 partes de torta.

5 0

3 years ago

The circular stream of water from a faucet is observed to taper from a diameter of 20 mm to 10 mm in a distance of 50 cm. Determ

steposvetlana [31]

Solution:

From Bernoulli equation

$\frac{1}{2}$ ρ $V_{1}^{2}$ + ρgh = $\frac{1}{2}$ ρ $V_{2}^{2}$ , where ρ is density of water, h – height difference and $V_{1}$ and $V_{2}$ are the velocities in upper and lower cross sections correspondingly. We take into account that the pressure in both cross-sections is the same and equal to the atmospheric one.

From the continuity and assuming water incompressible:

$\pi r_{1}^{2}$ $V_{1}$ = $\pi r_{2}^{2}$ $V_{2}$ , where $A_{1}$ and $A_{2}$ are the corresponding cross-sections. As the lower diameter is twice as low as the upper one, we can conclude that $V_{1} = \frac{V_{2}}{4}$

Inserting it back into Bernoulli equations produces:

$V_{1} = \sqrt{\frac{2}{15}gh} = 0.26m/s$ and the flow rate is

$\pi r_{1}^{2}V_{1} = \pi r_{2}^{2}V_{2} = 8.10^{-5} \frac{m^{3} }{s}$

6 0

3 years ago

What is the solution of the equation 16-17+ Vb=1? b = 3 O b = 9 O b = 81 no solution