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3 years ago

10/14 a/10.5 solve for A with explanation

Mathematics

1 answer:

photoshop1234 [79]3 years ago

5 0

Answer:

$a = 7.5$

Step-by-step explanation:

Given

$\frac{10}{14} = \frac{a}{10.5}$

Required

The value of a

Cross multiply

$a * 14 =10 * 10.5$

Make a the subject

$a = \frac{10 * 10.5}{14}$

$a = \frac{105}{14}$

$a = 7.5$

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A surveyor is conducting measures to see how wide a road is

drek231 [11]

Answer:

D) 45 ft

Step-by-step explanation:

The two triangles are shown below.

Given:

BC = 60 ft, CD = 24 ft and DE = 18 ft.

Since, the two triangles are similar, their corresponding sides are in proportion.

So, $\frac{AB}{DE}=\frac{BC}{CD}=\frac{AC}{CE}$

Now, consider the proportion of sides,

$\frac{AB}{DE}=\frac{BC}{CD}\\AB=\frac{BC}{CD}\times DE\\AB=\frac{60}{24}\times 18\\AB=\frac{60\times 18}{24}=45\ ft$

Therefore, the distance between A and B is 45 ft.

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4 years ago

How to tell if a graph is a function?

Ghella [55]

Look at one of the graphs you have a question about. Then take a vertical line and place it on the graph. If the graph is a function, then no matter where on the graph you place the vertical line, the graph should only cross the vertical line once.

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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.

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