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4 years ago

HELP FAST PLEASE Find u*v, if u=13j and v=i+3j

1 answer:

6 0

$\bf \begin{cases}
u=13j\implies 0i+13j\implies \ \textless \ 0,13\ \textgreater \ \\
v=i+3j\implies 1i+3j\implies \ \textless \ 1,3\ \textgreater \ 
\end{cases}
\\\\\\
\ \textless \ 0,13\ \textgreater \ \cdot \ \textless \ 1,3\ \textgreater \ \implies (0\cdot 1)+(13\cdot 3)\implies 39$

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A scientist is cooling a liquid at a steady rate. The temperature of the liquid changed by −82 1/2 °C over 2 1/5 minutes. What w

dangina [55]

Answer: $37.5^{\circ}$ or $37\dfrac{1}{2}^{\circ}$

Step-by-step explanation:

Given : The temperature of the liquid changed by $-82\dfrac{1}{2}^{\circ}C$ over $2\dfrac{1}{5}$ minutes.

We can write $[tex]82\dfrac{1}{2}=\dfrac{82\times2+1}{2}=\dfrac{165}{2}$

$2\dfrac{1}{5}=\dfrac{11}{5}$

Now, the statement becomes ,

The temperature of the liquid changed by $-\dfrac{165}{2}^{\circ}C$ over $\dfrac{11}{5}$ minutes.

i.e. temperature changed in $\dfrac{11}{5}$ minutes = $-\dfrac{165}{2}^{\circ}C$

Now, by Unitary method,

The temperature changed in 1 minute = $\dfrac{165}{2}\times\dfrac{5}{11}^{\circ}C$

⇒ The temperature changed in 1 minute = $37.5^{\circ}$ or $37\dfrac{1}{2}^{\circ}$

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3 years ago

arthur wants to buy an item that costs p dollars before tax. using a 6% sales tax rate, write two different expressions that rep

Mkey [24]

For this case we have the following variable:

p: cost of the item that Arthur wants to buy before tax

The expression for the 6% tax is given by:

$\frac{6}{100} p$

Or equivalently:

$0.06p$

Therefore, two different expressions for the total cost are:

Expression 1:

$p + \frac{6}{100} p$

Expression 2:

$p + 0.06p$

To prove that they are equal, suppose that the item costs $ 100:

Expression 1:

$p + \frac{6}{100} p$

$100+ \frac{6}{100} 100$

$100 + 6$

$106$

Expression 2:

$p + 0.06p$

$100 + 0.06 (100)$

$100 + 6$

$106$

Since the cost is the same, then the expressions are the same.

Answer:

Two different expressions that model the problem are:

$p + \frac{6}{100} p$

$p + 0.06p$

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3 years ago

6 in.

tiny-mole [99]

Answer:

Step-by-step explanation:

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3 years ago

Y=-|x+3| domain and range

sergejj [24]

Answer:

the domain is that x is a set of real numbers.

the range is y ≤ -3

Step-by-step explanation:

y = -|x + 3|

y is either positive or equal to zero.

Now, x is all real numbers because any value of x used will yield a valid value of y.

Thus, we can say that the domain is that x is a set of real numbers.

Now, for the range:

The minus sign in front of the absolute value indicates that the function has a maximum value.

Thus, the range is y ≤ -3

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3 years ago

t^2+4t+4 t^2-2t+1 Consider the parametric curve a. Find points on the parametric curve where the tangent lines are vertical. b.

kondaur [170]

Answer:

a) $t = 1$ , b) $t = -2$ , c) Concave upward: $(-\infty, 1)$ . No intervals being concave downward.

Step-by-step explanation:

a) Tangent line is vertical if slope is undefined, whose points are associated with discontinuities of the function. Rational functions have discontinuities associated with the denominator:

$f(t) = \frac{t^{2}+4\cdot t +4}{t^{2}-2\cdot t + 1}$

By factorizing each component, the function is re-arranged as:

$f(t) = \frac{(t+2)^{2}}{(t-1)^{2}}$

There is no avoidable discontinuities. The only point where tangent line is vertical is $t = 1$ .

b) Tangent line is horizontal if slope is equal to zero, whose point is associated to the points that makes function equal to zero. Rational functions have horizontal tangent lines if numerator is equal to zero and denominator is different to zero. Hence, the only point where tangent line is horizontal is $t = -2$ .

c) An interval is concave upward if exist an absolute minimum inside, which can be found by the help of the First and Second Derivative Tests.

$f'(t) = \frac{2\cdot (t+2)\cdot (t-1)^{2}-2\cdot (t-1)\cdot (t+2)^{2}}{(t-1)^{4}}$

$f'(t) = 2\cdot (t+2)\cdot (t-1)\cdot \left[\frac{t-1-t-2}{(t-1)^{4}}\right]$

$f'(t) = -\frac{6\cdot (t+2)\cdot (t-1)}{(t-1)^{4}}$

$f'(t) = -\frac{6\cdot (t+2)}{(t-1)^{3}}$

The only critical point is:

$-6\cdot (t+2) = 0$

$t = -2$ (which coincides with the result of point b)

$f''(t) = -6\cdot \left[\frac{(t-1)^{3}-3\cdot (t-1)^{2}\cdot (t+2)}{(t-1)^{6}}\right]$

$f''(t) = -6\cdot \left[\frac{1}{(t-1)^{3}}-\frac{3\cdot (t+2)}{(t-1)^{4}} \right]$

The value associated with the critical point is:

$f''(-2) = \frac{2}{9}$

Which means that critical point is an absolute minimum, and, consequently, the interval that is concave upward is $(-\infty, 1)$ . There is no absolute maximums and, therefore, there is no interval that is concave downward.

7 0

4 years ago