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3 years ago

Help pls do not delte my anwser i really need help

Mathematics

1 answer:

Andre45 [30]3 years ago

8 0

Answer:

2.False

3.True

4.True

5.False

6.False

7.False

Step-by-step explanation:

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Help please!!!

katovenus [111]

Answer:

x = $\frac{7+\sqrt{47}\times i }{4}$

Step-by-step explanation:

To solve quadratic systems,we always substitute one variable in terms if the other and then solve the equation.

x + 2y = 6 ---------------(1)

y - 5 = $(x-2)^{2}$ ---------------(2)

y = $(x-2)^{2}$ + 5 ---------------(3)

Substitute (3) in (1) ,

x + 2( $(x-2)^{2}$ + 5 ) = 6

$(a + b)^{2}$ = $a^{2} + 2ab + b^{2}$

x + 2( $x^{2} - 4x + 4$ + 5 ) = 6

$2x^{2} - 7x + 12=0$ --------------(4)

The roots of the quadratic equation $ax^{2} +bx+c$ is

x = $\frac{(-b) + \sqrt{(-b)^{2}-4 \times ac } }{2 \times a}$ -----------(5)

According to equation (5),solution of (4) is

x = $\frac{7 + \sqrt{(-7)^{2}-4 \times 24 } }{2 \times 2}$

x = $\frac{7+\sqrt{49-96}}{4}$

x = $\frac{7+\sqrt{47}\times i }{4}$

4 0

3 years ago

(1 point) (a) Find the point Q that is a distance 0.1 from the point P=(6,6) in the direction of v=⟨−1,1⟩. Give five decimal pla

natima [27]

Answer:

following are the solution to the given points:

Step-by-step explanation:

In point a:

$\vec{v} = -\vec{1 i} +\vec{1j}\\\\|\vec{v}| = \sqrt{-1^2+1^2}$

$=\sqrt{1+1}\\\\=\sqrt{2}$

calculating unit vector:

$\frac{\vec{v}}{|\vec{v}|} = \frac{-1i+1j}{\sqrt{2}}$

the point Q is at a distance h from P(6,6) Here, h=0.1

$a=-6+O.1 \times \frac{-1}{\sqrt{2}}\\\\= 5.92928 \\\\b= 6+O.1 \times \frac{-1}{\sqrt{2}} \\\\= 6.07071$

the value of Q= (5.92928 ,6.07071 )

In point b:

Calculating the directional derivative of $f (x, y) = \sqrt{x+3y}$ at P in the direction of $\vec{v}$

$f_{PQ} (P) =\fracx{f(Q)-f(P)}{h}\\\\$

$=\frac{f(5.92928 ,6.07071)-f(6,6)}{0.1}\\\\=\frac{\sqrt{(5.92928+ 3 \times 6.07071)}-\sqrt{(6+ 3\times 6)}}{0.1}\\\\= \frac{0.197651557}{0.1}\\\\= 1.97651557$

$\vec{v} = 1.97651557$

In point C:

Computing the directional derivative using the partial derivatives of f.

$f_x(x,y)= \frac{1}{2 \sqrt{x+3y}}\\\\ f_x (6,6)= \frac{1}{2 \sqrt{22}}\\\\f_x(x,y)= \frac{1}{\sqrt{x+3y}}\\\\ f_x (6,6)= \frac{1}{\sqrt{22}}\\\\f_{(PQ)}(P)= (f_x \vec{i} + f_y \vec{j}) \cdot \frac{\vec{v}}{|\vec{v}|}\\\\= (\frac{1}{2 \sqrt{22}}\vec{i} + \frac{1}{\sqrt{22}} \vec{j}) \cdot \frac{-1}{\sqrt{2}}\vec{i} + \frac{1}{\sqrt{2}} \vec{j}$

4 0

3 years ago

Evaluate if x=4 3x+1 A)2 B)3 C)11/4 D)13/4

mariarad [96]

Step-by-step explanation:

Given

If x = 4 Then

3x + 1 = 3 * 4 +1 = 12 + 1 = 13

The answer should be 13.

4 0

2 years ago

Please, I need help with this math question.

kondaur [170]

Step-by-step explanation:

Indeed the brainliest is t

8 0

2 years ago

Compare the two functions in the table. Will the value of function y = 2x eventually exceed the value of function y = x10?

Andrew [12]

First we rewrite the functions:
y = 2x
y = x ^ 10
We note that the second function always has values of y greater than the first function. However, there is a value of x for which the first function is greater.
For x = 1 we have:
y = 2 (1) = 2
y = (1) ^ 10 = 1
We note that:
2> 1
Answer:
Yes, the value of function y = 2x eventually exceed the value of function y = x ^ 10.

5 0

3 years ago

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