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2 years ago

Select the equivalent expression:

Mathematics

1 answer:

guapka [62]2 years ago

7 0

Answer:

On opening the bracket, we have (x^-28)/(x^56)

This is the same as (x^-28) . 1 / (x^56)

Following the law of indices 1/(x^a) = x^-a, we have 1 / (x^56) = (x^-56)

Hence (x^-28) . (x^-56)

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Which is the greatest rational number? A. 3 2/7 B. 3 1/4 C. 3.31 D. 3.36

lana [24]

Answer:

Step-by-step explanation:

3 1/4 = 3.25

3 2/7 = 3.29

3.29>3.25

3.31>3.29

3.36>3.31

so the answer is D.

4 0

3 years ago

What function is the inverse of the exponential function y = (1.5)x?

vagabundo [1.1K]

Correct question is;

What function is the inverse of the exponential function y = 1.5^(x)?

Answer:

y = log_1.5_x

Step-by-step explanation:

The inverse of exponential functions is usually written in form of logarithm.

For example inverse of y = p^(x) will be written as; y = log_p_(x)

Similarly applying this same pattern to our exponential function y = 1.5^(x), we have the inverse as;

y = log_1.5_x

6 0

2 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:

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