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3 years ago

Adding and subtracting functionsg(t) = −3t^3 − 4t f (t) = 4t − 4 Find g(t) + f (t)

Mathematics

1 answer:

ElenaW [278]3 years ago

7 0

I believe the answer is -27t - 4

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PLEASE HELP!! I’LL GIVE BRAINLIEST!!

lakkis [162]

Answer:

Step-by-step explanation:

[z^4/6²]^-3 = z^-12/6^-6 = 6^6/z^12

6 0

3 years ago

Determine which of the sets of vectors is linearly independent. A: The set where p1(t) = 1, p2(t) = t2, p3(t) = 3 + 3t B: The se

defon

Answer:

The set of vectors A and C are linearly independent.

Step-by-step explanation:

A set of vector is linearly independent if and only if the linear combination of these vector can only be equalised to zero only if all coefficients are zeroes. Let is evaluate each set algraically:

$p_{1}(t) = 1$ , $p_{2}(t)= t^{2}$ and $p_{3}(t) = 3 + 3\cdot t$ :

$\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0$

$\alpha_{1}\cdot 1 + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot (3 +3\cdot t) = 0$

$(\alpha_{1}+3\cdot \alpha_{3})\cdot 1 + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot t = 0$

The following system of linear equations is obtained:

$\alpha_{1} + 3\cdot \alpha_{3} = 0$

$\alpha_{2} = 0$

$\alpha_{3} = 0$

Whose solution is $\alpha_{1} = \alpha_{2} = \alpha_{3} = 0$ , which means that the set of vectors is linearly independent.

$p_{1}(t) = t$ , $p_{2}(t) = t^{2}$ and $p_{3}(t) = 2\cdot t + 3\cdot t^{2}$

$\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0$

$\alpha_{1}\cdot t + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot (2\cdot t + 3\cdot t^{2})=0$

$(\alpha_{1}+2\cdot \alpha_{3})\cdot t + (\alpha_{2}+3\cdot \alpha_{3})\cdot t^{2} = 0$

The following system of linear equations is obtained:

$\alpha_{1}+2\cdot \alpha_{3} = 0$

$\alpha_{2}+3\cdot \alpha_{3} = 0$

Since the number of variables is greater than the number of equations, let suppose that $\alpha_{3} = k$ , where $k\in\mathbb{R}$ . Then, the following relationships are consequently found:

$\alpha_{1} = -2\cdot \alpha_{3}$

$\alpha_{1} = -2\cdot k$

$\alpha_{2}= -2\cdot \alpha_{3}$

$\alpha_{2} = -3\cdot k$

It is evident that $\alpha_{1}$ and $\alpha_{2}$ are multiples of $\alpha_{3}$ , which means that the set of vector are linearly dependent.

$p_{1}(t) = 1$ , $p_{2}(t)=t^{2}$ and $p_{3}(t) = 3+3\cdot t +t^{2}$

$\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0$

$\alpha_{1}\cdot 1 + \alpha_{2}\cdot t^{2}+ \alpha_{3}\cdot (3+3\cdot t+t^{2}) = 0$

$(\alpha_{1}+3\cdot \alpha_{3})\cdot 1+(\alpha_{2}+\alpha_{3})\cdot t^{2}+3\cdot \alpha_{3}\cdot t = 0$

The following system of linear equations is obtained:

$\alpha_{1}+3\cdot \alpha_{3} = 0$

$\alpha_{2} + \alpha_{3} = 0$

$3\cdot \alpha_{3} = 0$

Whose solution is $\alpha_{1} = \alpha_{2} = \alpha_{3} = 0$ , which means that the set of vectors is linearly independent.

The set of vectors A and C are linearly independent.

4 0

3 years ago

Find the Volume! ^^<br> (Giving brainiest)

Darya [45]

Answer:

The total volume of the composite figure is 420 ft³.

Step-by-step explanation:

I'm not sure but I solved it as fast as I could

I apologize if it's incorrect :(

7 0

2 years ago

What is the slope of the line joining (8, 8) and (12, 6)?<br> answer choices attached

Margaret [11]

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 8}}\quad ,&{{ 8}})\quad 
% (c,d)
&({{ 12}}\quad ,&{{ 6}})
\end{array}
\\\quad \\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{6-8}{12-8}$

3 0

3 years ago

Read 2 more answers

Someone please help me

Karo-lina-s [1.5K]

Area = 2 · (h · d + h · w + d · w)
area = 2 · (2 · 6 + 2 · 2 + 6 · 2)
area = 2 · (12 + 4 + 12)
area = 2 · (28)
area = 56

answer
s = 56 cm^2 (second choice)

3 0

3 years ago

Adding and subtracting functionsg(t) = −3t^3 − 4t f (t) = 4t − 4 Find g(t) + f (t)​

Adding and subtracting functionsg(t) = −3t^3 − 4t f (t) = 4t − 4 Find g(t) + f (t)