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3 years ago

Graph each equation Y=-2|-3X+3|+4 NEED HELP ASAP!!

Mathematics

1 answer:

HACTEHA [7]3 years ago

3 0

Graph the absolute value by using the vertex and a few selected points

Point One: X= -1, Y= -8

Point Two: X= 0, Y= -2

Point Three: X= 1, Y= 4

Point Four: X= 2, Y= -2

Point Five: X= 3, Y= -8

The graph should look something like a mountain or an upside down V

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determine if the ordered pair is a solution of the equation is (2 4) a solution of y = -3× - 2? true or false

vesna_86 [32]

Answer:

False

Step-by-step explanation:

y = -3× - 2

4 = -3(2) - 2

4 = -6 - 2

4 = -8

4 0

3 years ago

Approximately how many times greater os 6x10-5 then 5x10-9?

lions [1.4K]

Answer:

Step-by-step explanation:

6 x 10 = 60

60 - 5 = 55

5 x 10 = 50

50 - 9 = 41

55 - 41 = 14

6 0

3 years ago

Read 2 more answers

Whats the equation to find the total cost ?

Bezzdna [24]

Y = 8x
Is the equation

7 0

3 years ago

I NEED HELP PLEASE! :)<br> thanks!

BigorU [14]

Add all the chips to find the total amount.

7 + 9 + 3 + 6 = 25 chips

Since there are 6 blue chips (6/25), that's the probability of just getting once.

When you pick a blue chip and it doesn't get replaced, then that means there is one fewer blue chip and one fewer from the total amount.

5/24

Multiply both probabilities.

6/25 * 5/24 = 30/600

Simplify.

30/600 → 1/20

Therefore, the answer is B

Best of Luck!

7 0

3 years ago

The points A BC :(2, 2,1), :(1,1,3), :(2,0,5) − are the vertices of a right triangle. The radius of the sphere with center at th

ElenaW [278]

Answer:

r=1

Step-by-step explanation:

First we need to know the length of each side of the triangle, so we use the formula of the vector modulus:

$|AB|= \sqrt{(b_{1}-a_{1})^{2}+(b_{2}-a_{2})^{2}+(b_{3}-a_{3})^{2}$

By doing so, we find:

$|AB|=\sqrt {6}\\|BC|=\sqrt {6}\\|AC|=2\sqrt {5}$

With this we know that the triangle is not right, but, we assume the longest side as the hypotenuse of the problem.

As we have two equal sides, we know that the line between point |AB| and the center of the hypotenuse is perpendicular, therefore, we can calculate it using Pythagoras theorem:

$|BC|^{2}=r^{2}+(\frac{|AC|}{2})^{2}\\\\r^{2}=|BC|^{2}-(\frac{|AC|}{2})^{2}\\\\r^{2}=(\sqrt{6})^{2}- (\frac{2\sqrt{5}}{2})^{2}\\\\r^{2}=6-5=1\\r=1$

4 0

3 years ago