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3 years ago

What is the first step in solving the equation ģx+3x+2 = 57

Mathematics

1 answer:

Svet_ta [14]3 years ago

4 0

Answer:

its Combine like terms sorry the answer is kind of late

Step-by-step explanation:

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Adele is 5 years older than Timothy. In three years, Timothy will be 2/3 of Adele’s age. What is Adele’s current age?

Verdich [7]

28 she is 28 years old

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3 years ago

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Solve the equation 3m - 7 = 8<br><br> will give brainliest and be grateful!

sertanlavr [38]

Answer:

m=5

Step-by-step explanation:

3(5)=15

therefore,

15-7=8

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3 years ago

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The legs of an isosceles triangle measure ( 2 x^4 + 2 x − 1 ) units each. The perimeter of the triangle is ( 5 x^4 − 2 x^3 + x −

nirvana33 [79]

Answer:

The base is x^4 -2x^3 -3x-1

Step-by-step explanation:

We know the perimeter of a triangle is the sum of the three sides

P = s1+s2+s3

We know the perimeter is 5 x^4 − 2 x^3 + x − 3

and two of the legs are 2 x^4 + 2 x − 1 since it is an isosceles triangle

P = 2s1 + s3

Subtract 2s1 from each side

P-2s1 =2s1 +s3 -2S1

P -2s1 =s3

Substituting what we know

5 x^4 − 2 x^3 + x − 3 - 2(2 x^4 + 2 x − 1) = s3

Distribute the -2

5 x^4 − 2 x^3 + x − 3 - 4 x^4 -4 x + 2 = s3

Combine like terms

5 x^4-4x^4 − 2 x^3 + x -4 x -3+ 2 = s3

x^4 -2x^3 -3x-1 =s3

The base is x^4 -2x^3 -3x-1

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

Show me the work for 5.34 divided by 6

andreev551 [17]

Answer:

How would we show you

5 0

3 years ago