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shusha [124]
3 years ago
7

Write an equation. Let x be the unknown number.

Mathematics
2 answers:
Assoli18 [71]3 years ago
4 0
3-5x = -5
X =8/5
I hope this helps
cluponka [151]3 years ago
4 0

Answer:

3 - 5x = - 5

Step-by-step explanation:

Subtracting five times a number from 3 is negative 5 in an equation is :

3 - 5x = - 5

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Anthony bought a new computer. He made an initial payment of $100 to the store, and he will pay $25 each month until the compute
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Step-by-step explanation:

the first payment is 100 so that is added on. $25 per month so 25m

t=25m+100

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What metric unit would best measure the length of a school hallway
Ne4ueva [31]
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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
2 years ago
The length of a rectangle is three inches more than the width. the area of the rectangle is 180 inches. find the width of the re
ch4aika [34]

We have a rectangle with length L that is 3 inches more than the width W. Then we can write this as:

L=W+3

The area of the rectangle is 180 square inches.

We have to find the width W.

As the area is equal to the product of the length and the width, we can write this equation and solve for W as:

\begin{gathered} A=180 \\ L\cdot W=180 \\ (W+3)\cdot W=180 \\ W^2+3W=180 \\ W^2+3W-180=0 \end{gathered}

We have a quadratic equation. The roots of this equation will be the mathematical solutions.

We can find the roots using the quadratic formula:

\begin{gathered} W=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ W=\frac{-3\pm\sqrt[]{3^2-4\cdot1\cdot(-180)}}{2\cdot1} \\ W=\frac{-3\pm\sqrt[]{9+720}}{2} \\ W=\frac{-3\pm\sqrt[]{729}}{2} \\ W=\frac{-3\pm27}{2} \\ W_1=\frac{-3-27}{2}=-\frac{30}{2}=-15 \\ W_2=\frac{-3+27}{2}=\frac{24}{2}=12 \end{gathered}

The solutions are W = -15 and W = 12.

The first one is not valid, as W has to be greater than 0.

Then, the solution to our problem is W = 12 in.

Answer: the width is W = 12 inches.

6 0
1 year ago
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