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vazorg [7]
3 years ago
10

Please help i’m sorry!!! 15 points again!

Mathematics
1 answer:
morpeh [17]3 years ago
7 0

Answer:

Guess I'll agree with the other person.

Step-by-step explanation:

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The cost D in dollars for n bars of candy at 75 cents per bar<br><br> Answer
skelet666 [1.2K]

Answer:

60 candy bars and 110 drinks will sell for $265. 120 candy bars and 90 drinks will sell for $270. How much does each candy bar sell for?Step-by-step explanation:

thats why

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2 years ago
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If –3 + i is a root of the polynomial function f(x), which of the following must also be a root of f(x)? –3 – i –3i 3 – i 3i
Troyanec [42]

Answer:

-3-i is the correct answer.

Step-by-step explanation:

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3 years ago
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Find the area of the parallelogram 67ft by 60ft by 52ft
Nikitich [7]
I’m sorry I don’t know wish I could help
4 0
4 years ago
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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
If the circumference of a circle measures 3/2pi cm, what is the area of the circle in terms of pi?
bagirrra123 [75]

Answer:

D - Area = 9/16pi cm²

Step-by-step explanation:

in order to evaluate this question , we will first of all find the radius using the formula for finding the circumference of a circle , we will then find the area of a circle.

<u>solution</u>

given that circumference of a circle = 3/2picm

recall that the formula for finding the circumference of a circle = 2πr

3/2picm = 2πr

multiply through by 2

3picm = 4pir

divide both side by 4pi

3pi cm/4pi = r

r = 3/4 picm

Recall that the Area of  a circle = πr²

Area = pi ( 3/4)² cm

Area = 9/16pi cm²

7 0
3 years ago
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