How to round 47,397 to the nearest ten

VARVARA [1.3K]

Answer:

$\begin{gathered} A=\text{ 12}\degree \\ B=\text{ 114}\degree \\ C=54\degree \end{gathered}$

Step-by-step explanation:

To calculate the angles of the given triangle, we can use the law of cosines:

$\begin{gathered} \cos (C)=\frac{a^2+b^2-c^2}{2ab} \\ \cos (A)=\frac{b^2+c^2-a^2}{2bc} \\ \cos (B)=\frac{c^2+a^2-b^2}{2ca} \end{gathered}$

Then, given the sides a=2, b=9, and c=8.

$\begin{gathered} \cos (A)=\frac{9^2+8^2-2^2}{2\cdot9\cdot8} \\ \cos (A)=\frac{141}{144} \\ A=\cos ^{-1}(\frac{141}{144}) \\ A=11.7 \\ \text{ Rounding to the nearest degree:} \\ A=12º \end{gathered}$

For B:

$\begin{gathered} \cos (B)=\frac{8^2+2^2-9^2}{2\cdot8\cdot2} \\ \cos (B)=\frac{13}{32} \\ B=\cos ^{-1}(\frac{13}{32}) \\ B=113.9\degree \\ \text{Rounding:} \\ B=114\degree \end{gathered}$ $\begin{gathered} \cos (C)=\frac{2^2+9^2-8^2}{2\cdot2\cdot9} \\ \cos (C)=\frac{21}{36} \\ C=\cos ^{-1}(\frac{21}{36}) \\ C=54.3 \\ \text{Rounding:} \\ C=\text{ 54}\degree \end{gathered}$

3 0

1 year ago

Money Flow The rate of a continuous money flow starts at $1000 and increases exponentially at 5% per year for 4 years. Find the

77julia77 [94]

Answer:

Present value = $4,122.4

Accumulated amount = $4,742

Step-by-step explanation:

Data provided in the question:

Amount at the Start of money flow = $1,000

Increase in amount is exponentially at the rate of 5% per year

Time = 4 years

Interest rate = 3.5% compounded continuously

Now,

Accumulated Value of the money flow = $1000e^{0.05t}$

The present value of the money flow = $\int\limits^4_0 {1000e^{0.05t}(e^{-0.035t})} \, dt$

= $1000\int\limits^4_0 {e^{0.015t}} \, dt$

= $1000\left [\frac{e^{0.015t}}{0.015} \right ]_0^4$

= $1000\times\left [\frac{e^{0.015(4)}}{0.015} -\frac{e^{0.015(0)}}{0.015} \right]$

= 1000 × [70.7891 - 66.6667]

= $4,122.4

Accumulated interest = $e^{rt}\int\limits^4_0 {1000e^{0.05t}(e^{-0.035t}} \, dt$

= $e^{0.035\times4}\times4,122.4$

= $4,742

8 0

3 years ago

Please help with this algebra question

valina [46]

Answer:

i think its

$P(x) = i ^{3}$

Step-by-step explanation:

8 0

3 years ago

Find the surface area of the trapezoidal

Fofino [41]

Answer:

916 in.

Step-by-step explanation:

Step by Step.

5*10 = 50, and 50*2 = 100

10*4 = 40, and 40 * 2 = 80

(4*6)/2 = 12, and 12 * 2 = 24

10*10 = 100, and 100 * 2 = 200

16*16 = 256, and 256 * 2 = 512

Adding this together, we get: 916 inches.

6 0

3 years ago

Someone please help !! I don’t know what I’m doing with this !!

dimulka [17.4K]

Answer:

a) d(sinh(f(x)))/dx = cosh(f(x))·df(x)/dx

b) d(cosh(f(x))/dx = sinh(f(x))·df(x)/dx

c) d(tanh(f(x))/dx = sech(f(x))²·df(x)/dx

d) d(sech(4x+2))/dx = -4sech(4x+2)tanh(4x+2)

Step-by-step explanation:

To do these, you need to be familiar with the derivatives of hyperbolic functions and with the chain rule.

The chain rule tells you that ...

(f(g(x)))' = f'(g(x))g'(x) . . . . where the prime indicates the derivative

The attached table tells you the derivatives of the hyperbolic trig functions, so you can answer the first three easily.

__

a) sinh(u)' = sinh'(u)·u' = cosh(u)·u'

For u = f(x), this becomes ...

sinh(f(x))' = cosh(f(x))·f'(x)

__

b) After the same pattern as in (a), ...

cosh(f(x))' = sinh(f(x))·f'(x)

__

c) Similarly, ...

tanh(f(x))' = sech(f(x))²·f'(x)

__

d) For this one, we need the derivative of sech(x) = 1/cosh(x). The power rule applies, so we have ...

sech(x)' = (cosh(x)^-1)' = -1/cosh(x)²·cosh'(x) = -sinh(x)/cosh(x)²

sech(x)' = -sech(x)·tanh(x) . . . . . basic formula

Now, we will use this as above.

sech(4x+2)' = -sech(4x+2)·tanh(4x+2)·(4x+2)'

sech(4x+2)' = -4·sech(4x+2)·tanh(4x+2)

_____

Here we have used the "prime" notation rather than d( )/dx to indicate the derivative with respect to x. You need to use the notation expected by your grader.

__

<em>Additional comment on notation</em>

Some places we have used fun(x)' and others we have used fun'(x). These are essentially interchangeable when the argument is x. When the argument is some function of x, we mean fun(u)' to be the derivative of the function after it has been evaluated with u as an argument. We mean fun'(u) to be the derivative of the function, which is then evaluated with u as an argument. This distinction makes it possible to write the chain rule as ...

f(u)' = f'(u)u'

without getting involved in infinite recursion.

7 0

3 years ago