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3 years ago

Please answer this please

Mathematics

2 answers:

motikmotik3 years ago

6 0

Answer:

Step-by-step explanation:

lora16 [44]3 years ago

4 0

So first what you would have to do is separate the three triangles from the square by basically slicing the top from the bottom so now you have three identical triangles and one square then just add them all together

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Бу – 2 = бу +5 what does Y equal

ivanzaharov [21]

Answer: y = 7

explanation: 6y = 6y + 7
y =

5 0

3 years ago

6m+7n+5m-3n how do I solve

Wewaii [24]

combine like terms, so 6m + 5m =11m

then 7n - 3n = 4n

ANSWER: 11m+4n

6 0

4 years ago

Read 2 more answers

the number m and 5/8 are additive inverses.drag and drop m to their correct positions on the number line.drag and drop the label

monitta

The additive inverse of a number is that number with its sign changed.

The additive inverse of 5/8 is -5/8, so that is the value of m.

The sum of a number and its additive inverse is zero. (This is actually the definition of additive inverse.)

5/8 is found at 5/8 on the number line.

m is found at -5/8 on the number line.

"sum" is found at 0 on the number line.

3 0

4 years ago

5(x+5)-25=6(2-x) can someone help me with this sum please including how you worked it out

alekssr [168]

Answer:

The answer is 12/11

Step-by-step explanation:

5x+25-25=12-6x

5x=12-6x

11x=12

x=12/11

5 0

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