Draw an exponential graph of y=a^x(a>0, a!=0)

2−(−4)+3+(−6)−(−2)2, minus, left parenthesis, minus, 4, right parenthesis, plus, 3, plus, left parenthesis, minus, 6, right pare

Yuki888 [10]

Answer: 7

Step-by-step explanation:

2−(−4)+3+(−6)−(−2)2

solution:

– x – = +

– x + = –

= 2−(−4)+3+(−6)−(−2)2

= 6 + 3 + ( – 6) – ( 2 ) 2

= 9 + ( –6 ) – ( –2 ) 2

= 3 – (–2)2

= 3 – ( –4 )

= 7

4 0

3 years ago

A jar of marbles contains 7 blue marbles, 12 yellow marbles, 8 green marbles, and 5 red marbles. Use this information to answer

Andreyy89

The correct answer is 9/32

6 0

4 years ago

Holding a participant variable constant has the double advantage of (1) preventing the variable from becoming a confounding vari

Elodia [21]

Answer:

The provided statement is True.

Step-by-step explanation:

Holding a variable constant implies that there are no kind of relationship between the variables and the treatments.

Holding a participant variable constant has the double advantage:

(1) preventing the variable from becoming a confounding variable

(2) reducing variance within groups.

A confounding variable effects both the dependent and the independent variable remarking a relationship between the two.

Variation within group implies the difference caused by the individual values of each group.

Thus, the provided statement is True.

5 0

3 years ago

Hi :) this lesson is about 30-60-90 triangles.

hjlf

Answer:

5 times the square root of 3

Step-by-step explanation:

To get the longest side of a 30-60-90 triangle, you take the shortest side and multiply it by the square root of 3.

5*(square root of 3)

4 0

3 years ago

In a fraternity with 40 members , 18 take mathematics, 5 take both mathematics and physics, and 8 take neither mathematics nor p

garri49 [273]

Answer: 14 members take physics but not mathematics

Step-by-step explanation:

Please see the attached for a diagrammatic representation of this situation.

Since there are 40 members but 8 did not offer any of mathematics or physics, then:-

40 - 8 = 32

So, 32 members of the fraternity offer maths, physics or both.

5 members offer both mathematics and physics .

Since 5 members offer both mathematics and physics, the number of members that offer maths alone = 18 - 5

Let us use "P" to represent the number of members that offer physics. Therefore, the number of members that offer only physics = (P - 5)

Now, if we add up the members that offered maths only, the members that offered both mathematics and physics and the members that offered only physics, it will be equal to the 32 members that offered at least a subject.

Therefore, (18 - 5) + 5 + (P - 5) = 32

18 - 5 + 5 + P - 5 = 32

18 + P - 5 = 32

P + 13 = 32

P = 32 - 13

P = 19

19 members offered physics out of the 32 members that offered maths, physics or both.

To find out the number of people or members that offered physics only, we subtract the number of members that offered the two subjects (5) from a total of 19 members that offered Physics

19 - 5 = 14

Therefore, 14 members of the fraternity offer only physics

5 0

3 years ago

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