I NEED HELP!! NO LINKS OR I WILL REPORT

Mathematics

1 answer:

Rufina [12.5K]3 years ago

7 0

49degrees
60meters
24meters

You might be interested in

marias mother told her to go to the health food store and buy 10 pounds of apples each bag weighed 5 pounds and contained 15 app

Yanka [14]

Answer:

30 apples

Step-by-step explanation:

10 pounds of apples is 2 bags

5 + 5 = 10 lbs

15 apples per bag x 2 bags = 30 apples

Hope this helps :)

8 0

3 years ago

Use the graph to calculate the following

sweet-ann [11.9K]

Answer:

f(x)= { 1 if x≤-2

x2 if -2<x≤0

ln(x) if x>0

Step-by-step explanation:

The first part of the graph (from left to right) is a constant funcion. Since there isn't any pendant, the y value is constant at y=1. This remains this way until x≤-2 (because the value at x=-2 is filled in on the line and blanked out on the curve above)

The second part of the graph, between x>-2 & x=0, is a cuadratic formula.

Finally, the last part of the graph corresponds to the formula ln(x), which is only valid for x>0.

4 0

4 years ago

Let f(x) = (1 − x2)2/3. Which type of non-differentiable point exists at x = 1?.

slava [35]

$f(x)=(1-x^2)^{\frac{2}{3}}\implies \cfrac{df}{dx}=\cfrac{2}{3}(1-x^2)^{-\frac{1}{3}}\implies \cfrac{df}{dx}=\cfrac{2}{3\sqrt[3]{1-x^2}}$

when it comes to a rational expression, we can get critical points from, zeroing the derivative "and" from zeroing the denominator alone, however the denominator provides critical valid points that are either "asymptotic" or "cuspics", namely that the function is not differentiable or not a "smooth line" at such spot.

if we get the critical points from the denominator on this one, we get x = ±1, both of which are cuspics. Check the picture below.

3 0

2 years ago

Determine the gravitational force of attraction between two 3.5 kg bowling balls whose

Ipatiy [6.2K]

Answer:

$F_G=2.04\ .\ 10^{-10}\ Nw$

Option J

Step-by-step explanation:

Newton's Universal Law Of Gravitation

Newton discovered that all objects attract each other with a force of gravitational attraction which is proportional to their masses and inversely proportional to the square of the distance between them.

$\displaystyle F_G=G\frac{m_1m_2}{d^2}\ ,\ G=6.673\ .\ 10^{-11} N m^2/kg^2$