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2 years ago

10

The code for puzzle one escape the rooms

1 answer:

Anna35 [415]2 years ago

4 0

Answer:

there Is no question to that cluld you explain. sorry I don't know how to comment.

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Plzzzzzzzzzzzzzzzzzz help quick

Bezzdna [24]

Answer:

The correct way to set up the slope formula for the line that passes through points (5 , 0) and (6 , -6) is $\frac{-6-0}{6-5}$ ⇒ C

Step-by-step explanation:

The formula of the slope of a line passes through points $(x_{1},y_{1})$ and $(x_{2},y_{2})$

is $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

∵ The line passes through points (5 , 0) and (6 , -6)

∴ $x_{1}$ = 5 and $x_{2}$ = 6

∴ $y_{1}$ = 0 and $y_{2}$ = -6

Substitute these values in the formula of the slope

∵ $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

∴ $m=\frac{-6-0}{6-5}$

Let us look to the answer and find the same formula

The answer is:

The correct way to set up the slope formula for the line that passes through points (5 , 0) and (6 , -6) is $\frac{-6-0}{6-5}$

8 0

3 years ago

In a survey of 200 residents, 75 said they support the proposed town budget. If there are 10,000 total residents, about how many

Archy [21]

If 75 out of 200 people support the town budget than 3,750 out of 10,000 support it

7 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:

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

or

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

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

or

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

To the nearest hundredth, what is the length of line segment AB?

Vesnalui [34]

Answer:

5.83 units

Step-by-step explanation:

See attached picture. You can use the Pythagorean Theorem. I added some lines to show the legs of a right triangle. AB is the hypotenuse.

$3^2+5^2=(AB)^2\\\\9+25 = (AB)^2\\\\34=(AB)^2\\\\AB=\sqrt{34} \approx 5.83$

5 0

2 years ago

What is 0.181 rounded to 2 decimal places?

AveGali [126]

0.18 because the third decimal place is less than 5

7 0

3 years ago