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2 years ago

Answer question 2 please

Mathematics

1 answer:

julia-pushkina [17]2 years ago

4 0

Answer:

163.2

Step-by-step explanation:

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HELP ME QUICK: Suppose AB has one endpoint at A(0, 0). If (5, 3) is the midpoint of AB, what are the coordinates of point B?

DiKsa [7]

Answer:

Step-by-step explanation:

If P is midpoint of AB and: $A(0,\,0)\,,\quad P(5,\,3)\,,\quad B(x_B,\,y_B)$

then:

$x_P-x_A=x_B - x_P\qquad\quad\ \wedge\qquad y_P-y_A=y_B - y_P\\\\ 5-0=x_B-5\qquad\quad\wedge\qquad 3-0=y_B -3\\\\ x_B=5+5\qquad\qquad\wedge\qquad\ \ y_B=3+3 \\\\ {}\quad x_B=10\qquad\qquad\wedge\qquad\qquad \ y_B=6$

5 0

2 years ago

The radioactive substance cesium-137 has a half-life of

statuscvo [17]

A(t)=A0*(1/2)^(t/30)
where t=elapsed years
A0=initial mass at t=0.

7 0

2 years ago

What is the average rate of change for 5/6 and 15/14

andreev551 [17]

Answer:

Average rate of change is <u>0.80.</u>

Step-by-step explanation:

Given:

The two points given are (5, 6) and (15, 14).

Average rate of change is the ratio of the overall change in 'y' and overall change in 'x'. If the overall change in 'y' is positive with 'x', then average rate of change is also positive and vice-versa.

The average rate of change for two points $(x_1,y_1)\ and\ (x_2,y_2)$ is given as:

$R=\frac{y_2-y_1}{x_2-x_1}$

Plug in $x_1=5,x_2=15,y_1=6,y_2=14$ and solve for 'R'. This gives,

$R= \frac{14-6}{15-5}\\\\R=\frac{8}{10}\\\\R=0.80$

Therefore, the average rate of change for the points (5, 6) and (15, 14) is 0.80.

5 0

3 years ago

Which of the following is a solution to 2cos2x − cos x − 1 = 0?

Pavel [41]

Answer:

Option A is correct.

Solution for the given equation is, $x = 0^{\circ}$

Step-by-step explanation:

Given that : $2\cos^2x -\cos x -1 =0$

Let $\cos x =y$

then our equation become;

$2y^2-y-1= 0$ .....[1]

A quadratic equation is of the form:

$ax^2+bx+c =0$ .....[2] where a, b and c are coefficient and the solution is given by;

$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Comparing equation [1] and [2] we get;

a = 2 b = -1 and c =-1

then;

$y = \frac{-(-1)\pm \sqrt{(-1)^2-4(2)(-1)}}{2(2)}$

Simplify:

$y = \frac{ 1 \pm \sqrt{1+8}}{4}$

$y = \frac{ 1 \pm \sqrt{9}}{4}$

$y = \frac{ 1 \pm 3}{4}$

$y = \frac{1+3}{4}$ and $y = \frac{1 -3}{4}$

Simplify:

y = 1 and $y = -\frac{1}{2}$

Substitute y = cos x we have;

$\cos x = 1$

⇒ $x = 0^{\circ}$

and

$\cos x = -\frac{1}{2}$

⇒ $x = 120^{\circ} \text{and} x = 240^{\circ}$

The solution set: $\{0^{\circ}, 120^{\circ} , 240^{\circ}\}$

Therefore, the solution for the given equation $2\cos^2x -\cos x -1 =0$ is, $0^{\circ}$

8 0

3 years ago

Read 2 more answers

Need help with all three questions...

MA_775_DIABLO [31]

7) 2-9= -7 because, if you have two cookies and promised nine friends cookies, you would still owe 7 friends cookies or -7
8) -3 - (-4)= 1 because you owe 3 friends cookies, then you get 4 cookies and have one left over
9) 11 - (-12)= 23 because a negative plus a negative is a positive

4 0

2 years ago