Solve each equation. 1. 5x-4=4x 2. 27-3x=3x+27

I'LL MARK BRANLIEST PLS HELP

sergij07 [2.7K]

Answer:

It's (2, 3) and (7, 6)

Step-by-step explanation:

It's the only option where both x2 and y2 are both bigger than x1 and y1

3 0

3 years ago

Read 2 more answers

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

Ariana and tom Tabitha recorded the amount of time spent on homework on 10 school daze graph of the data are shown

Sedaia [141]

Answer:

theres no picture

Step-by-step explanation:

5 0

4 years ago

Write 2/3 as a decimal.

Mekhanik [1.2K]

Answer:

0.67

Step-by-step explanation:

2/3 = 0.6666

0.6666 rounded to the nearest hundredth = 0.67

3 0

3 years ago

Two angles in a triangle add up to 138 degrees what is the size of the 3rd angle

Semenov [28]

The measurement of the second angle is 42°. All of the angles of a triangle add up to a total of 180°. If two angles add up to 138°. Then the last angle has to be 42° (180-138).

4 0

3 years ago

Solve each equation.1. 5x-4=4x2. 27-3x=3x+27