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3 years ago

46 and one forth percent as a fraction in simple as form

Mathematics

2 answers:

SOVA2 [1]3 years ago

4 0

46 1/4% = 46.25/100 = 23.125/50

adelina 88 [10]3 years ago

3 0

Answer:

the answer is 46 1/4% = 46.25/100 = 23.125/50

Step-by-step explanation:

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I am not sure how sigma notation works, can somebody explain how to solve this? It isn't the second choice.

Masja [62]

Answer:

first answer is correct

Step-by-step explanation:

Hello, we want to write 2 + 4 + 6 + ... + 20

for n = 1, 2n = 2*1 =2

for n = 2, 2n = 2*2=4

for n = 3, 2n=2*3=6

etc

$\displaystyle \sum_{n=1}^{10} {2n} = 2*1 + 2*2 +2*3+2*4+...+2*10=2+4+6+8+...+20$

thanks

6 0

3 years ago

Solve 5x − 4 = 7 for x using the change of base formula log base b of y equals log y over log b.

Sauron [17]

Answer:

5.209 or A

Step-by-step explanation

5^x-4=7

reserve 4

log sides

x-4 log5 and log7

x-4=log7/log5

+4 or number of your equation

x=log5/log7 +4

use calculator

get answer

5.209

EDIT: Got a 100 on the test this is correct

5 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:

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