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2 years ago

The graph of Y = COS X+ż is the graph of the y = cos(x) shifted in which direction?

Mathematics

1 answer:

rjkz [21]2 years ago

3 0

Answer:

2 units up

Step-by-step explanation:

You're just taking the entire graph of cos(x) and then adding 2 so it moves everything up. If you added 2 to x BEFORE taking the cosine

(like y = cos (x+2) ), then you would shift it left 2. Adding and subtracting from the x moves everything left or right. Adding or subtracting at the end just moves it up and down.

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Solve the inequality and express your answer in interval notation x^2+8x+5 <0

irakobra [83]

The interval notation is $(-\sqrt{11}-4, \sqrt{11}-4)$

Explanation:

The inequality is $x^{2}+8 x+5$

Let us complete the square, we get,

$x^{2}+8 x+5+4^{2}-4^{2}$

Simplifying, we have,

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

$(x+4)^{2}-11$

Adding both sides of the equation by 11, we get,

$(x+4)^{2}$

Since, we know that for $u^{n}$ , if n is even then $-\sqrt[n]{a}$

Thus, writing the above expression in this form, we get,

$-\sqrt{11}$

Also, if $a$ then $a$ and $u$

Then ,we have,

$-\sqrt{11}$ and $x+4$

Solving the inequalities, we get,

$-\sqrt{11}\ \ \ \ \ \$ and $x+4$

Merging the intervals, we have,

$-\sqrt{11}-4$

Hence, writing the solution in interval notation, we have,

$(-\sqrt{11}-4, \sqrt{11}-4)$

Therefore, the answer is $(-\sqrt{11}-4, \sqrt{11}-4)$

7 0

3 years ago

PLS HELP :( Solve the inequality and graph the solution on the line provided.

Travka [436]

Answer:

$x\leq -1$

Step-by-step explanation:

1. $7x\leq 7-14$

2. $7x\leq -7$

3. divide both sides by 7

4. $x\leq -1$

Closed circle on -1 and the line goes to the left

4 0

3 years ago

You are to drive to an interview in another town, at a distance of 270 km on an expressway. The interview is at 11:15 a.m. You p

Yuri [45]

Answer:

the least required speed is 103.88 km/h

Step-by-step explanation:

Data provided in the question:

Total distance = 270 km

Total time to reach = Time between 8 a.m and 11:15 a.m i.e 3.25 hours

Now,

For the distance of 110 km speed was 100 km/h

therefore, the time taken to cover 110 km = $\frac{\textup{Distance}}{\textup{speed}}$

= $\frac{\textup{110 km}}{\textup{100 km/h}}$

= 1.1 hour

For another 43 km speed was 42 km/h

therefore, the time taken to cover 43 km = $\frac{\textup{Distance}}{\textup{speed}}$

= $\frac{\textup{43 km}}{\textup{42 km/h}}$

= 1.0238 hours

Now,

The distance left to be covered = 270 - 110 - 43 = 117 km

Time left = 3.25 h - 1.1 h - 1.0238 h = 1.1262 h

Thus required speed = $\frac{\textup{117 km}}{\textup{1.1262 h}}$

= 103.88 km/h

Hence, the least required speed is 103.88 km/h

6 0

2 years ago

If the slope of a line is 4/5, does the line rise (from left to right), fall (from left to right), horizontal or vertical?

slava [35]

slope: 4/5 is a positive slope

positive slopes rises from left to right

positive slopes falls from right to left

neither horizontal or vertical.

The picture below shows some examples of equation with positive slope.

4 0

2 years ago

Read 2 more answers

Please help there are two pictures because I could not screen the whole thing!!!!1 Giving Branliest answer! YEEEEEEEEEEEET

olga nikolaevna [1]

These drawings depict the steps to get the perpendicular line using a compas (and maybe a ruler)

remember that a line goes through 2 points so we need 2 different points to construct our perpendicular line

the first one is wrong because it doesn't constraint it to be perpendicular
it only has 1 point,

2nd is wrong as well, has only 1 point, the center

3rd is right

4th is correct, hold a sec
how did that 2nd upper arc have same beginig pointn for compass?

3rd on is answer

4 0

2 years ago