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Kamila [148]
4 years ago
11

Math question down below need answers for a b c and d please

Mathematics
1 answer:
storchak [24]4 years ago
6 0
(a). How much the T-shirt cost.
(b). 23=12+x
(c). 23-12=x; x= 11
(d) the T-shirt cost $11. 
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Using a number line how do i find 5+(-3)?​
aleksklad [387]
5+(-3)=5-3=2
And
5+(-3)=-3+5=2
4 0
4 years ago
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Which of the following will be constructed when the two endpoints of a line segment are folded so that they line up
nataly862011 [7]
B. Midpoint of a line segment
6 0
4 years ago
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Find the value of y
Rainbow [258]
You'll apply the geometric mean formula. The idea is to multiply the two pieces of the hypotenuse: 5*11 = 55

Then apply the square root to get sqrt(55)

The exact answer in terms of radicals or square roots is sqrt(55)

Using a calculator, the approximate answer is sqrt(55) = 7.416198

If your teacher wants you to use a proportion, then one way to do it is to have
11/y = y/5
11*5 = y*y
55 = y^2
y^2 = 55
y = sqrt(55)
y = 7.416198 approximately
4 0
4 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
Four different stores posted ads on special 15-oz cans of baked beans -8 for $6 = 75 cents per can -10 for $10 = $1 per can -2 f
worty [1.4K]

Answer:

75 cents per can

Step-by-step explanation:

8 for $6 = 75 cents per can

10 for $10 = $1 per can

2 for $3 = $1.50 per can

The last store sold 28-oz for $1.40 each

To obtain the cost per 15-oz can of baked beans

28-oz / 15-oz = 1.86667of 15 - oz cans of baked beans can be obtained from 28 - oz.

The price per can will now be :

$1.40 / 1.8666666

= $0.75

= 0.75 * 100 = 75 cents

Hence, for the last store, they sold at 75 cents per can similar to the first store.

5 0
3 years ago
Read 2 more answers
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