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2 years ago

8/3T=4/1T Could you help mi solve it pls

Mathematics

1 answer:

katrin [286]2 years ago

4 0

Answer:

For this problem t=0

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I would like for some to help me because I really need help

torisob [31]

Answer:

whats the question

Step-by-step explanation:

........

7 0

3 years ago

A point P(x,y) moves along the graph of the equation y = x3 + x2 + 2. The x-values are changing at the rate of 2 units per secon

CaHeK987 [17]

Using implicit differentiation, it is found that y is changing at a rate of 10 units per second.

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The equation is:

$y = x^3 + x^2 + 2$

Applying implicit differentiation in function of t, we have that:

$\frac{dy}{dt} = 3x^2\frac{dx}{dt} + 2x\frac{dx}{dt}$

x-values changing at a rate of 2 units per second means that $\frac{dx}{dt} = 2$
Point Q(1,4) means that $x = 1, y = 4$ .

We want to find $\frac{dy}{dt}$ , thus:

$\frac{dy}{dt} = 3x^2\frac{dx}{dt} + 2x\frac{dx}{dt}$

$\frac{dy}{dt} = 3(1)^2(2) + 2(1)(2) = 6 + 4 = 10$

y is changing at a rate of 10 units per second.

A similar problem is given at brainly.com/question/9543179

5 0

2 years ago

Find an nth-degree polynomial function with real coefficients satisfying the given conditions.

lesya692 [45]

Answer:

$\displaystyle f(x) = -2(x-2)(x^2+4)$

Step-by-step explanation:

We want to find a third degree polynomial with zeros x = 2 and x = 2i and f(-1) = 30.

First, note that by the Complex Root Theorem, since x = 2i is a root, x = -2i must also be a root.

Hence, we will have the three factors:

$\displaystyle f(x) = a(x-(2))(x-(2i))(x-(-2i))$

Where a is the leading coefficient.

Expand and simplify the second and third factors:

$\displaystyle \begin{aligned} (x-(2i))(x-(-2i)) &= (x-2i)(x+2i) \\ \\ &= x(x-2i)+2i(x-2i) \\ \\ &= (x^2 - 2ix) + (2ix - 4i^2) \\ \\ &=x^2 + 4\end{aligned}$

Hence:

$\displaystyle f(x) = a(x-2)(x^2+4)$

Since f(-1) = 30:

$\displaystyle \begin{aligned} f(x) &= a(x-2)(x^2+4) \\ \\ (30) &= a((-1)-2)((-1)^2+4) \\ \\ 30 &= -15a \\ \\ a&= -2\end{aligned}$

In conclusion, third degree polynomial function is:

$\displaystyle f(x) = -2(x-2)(x^2+4)$

5 0

3 years ago

How many real solutions does the equation y2− 8y + 16 = 0 have?

saveliy_v [14]

We can solve this equation by completing the square. Notice that $4*2=8$ and $4^{2}=16$ . Therefore, we can turn $y^{2}-8y+16=0$ into $(y-4)^{2}=0$ . y, in this equation, only has one real solution: 4. So your answer would be (B).

4 0

3 years ago

Solve for x: x/3 = -48

oksano4ka [1.4K]

Answer:

-144

Step-by-step explanation:

x=-48 times 3

8 0

2 years ago

Read 2 more answers