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3 years ago

Lim as x goes to 0: (2sinxcosx)/(2x^2+x)=

1 answer:

8 0

$\displaystyle\lim_{x\to0}\frac{2\sin x\cos x}{2x^2+x}=2\left(\lim_{x\to0}\frac{\sin x }x\right)\left(\lim_{x\to0}\frac{\cos x}{x+1}\right)=2(1)\left(\frac1{0+1}\right)=2$

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Help me out please with domain and range and the last question please (algebra 2) no links please I will report

baherus [9]

Answer:

domain = ( -infinity, infinity) = all real numbers

range = ( -1, infinity)

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2 years ago

How to do this question plz answer me step by step plzz

mojhsa [17]

Answer:

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2 years ago

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Liz got home from school at 4:33. She decided to bake muffins as an after-school snack. It took Liz 55 minutes to prepare and ba

IgorC [24]

Answer:

The muffins were done at 5:28

Step-by-step explanation:

you add 4:33 to 55 minutes knowing well that 60mins makes one hour

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2 years ago

Evaluate 10 8 ÷ 2 − 4. (5 points) 10 5 6 14

vodka [1.7K]

10+8/2-4

PEMDAS
parenthasese
exxonents
multiplication or division
addiotn or subtration next

10+8/2-4
8/2 first
4
10+4-4
add
10+4=14
minus
14-4=10

answer is 10

7 0

3 years ago

The volume V of an ice cream cone is given by V = 2 3 πR3 + 1 3 πR2h where R is the common radius of the spherical cap and the c

Nuetrik [128]

Answer:

The change in volume is estimated to be 17.20 $\rm{in^3}$

Step-by-step explanation:

The linearization or linear approximation of a function $f(x)$ is given by:

$f(x_0+dx) \approx f(x_0) + df(x)|_{x_0}$ where $df$ is the total differential of the function evaluated in the given point.

For the given function, the linearization is:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh$

Taking $R_0=1.5$ inches and $h=3$ inches and evaluating the partial derivatives we obtain:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh\\V(R, h) = V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh$

substituting the values and taking $dx=0.1$ and $dh=0.3$ inches we have:

$V(R_0+dR, h_0+dh) =V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh\\V(1.5+0.1, 3+0.3) =V(1.5, 3) + (\frac{2 \cdot 3 \pi \cdot 1.5}{3} + 2 \pi 1.5^2)\cdot 0.1 + (\frac{\pi 1.5^2}{3} )\cdot 0.3\\V(1.5+0.1, 3+0.3) = 17.2002\\\boxed{V(1.5+0.1, 3+0.3) \approx 17.20}$

Therefore the change in volume is estimated to be 17.20 $\rm{in^3}$

4 0

3 years ago