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3 years ago

How many terms are in the expression a3 + 3a2b - 4ab - 11b + 7?

Mathematics

1 answer:

Damm [24]3 years ago

3 0

5 terms

a³
3a²b
-4ab
-11b
7

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Mr. Frank has 1.02 kg of fertilizer for the plants in his nursery he wants every plant to get 96 mg of fertilizer four times eac

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0.08kg of fertilizer left

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Ramsay cuts out a piece from a circular cardboard for a school project. The radius of the cardboard is 10 inches and the measure

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34 inches

Step-by-step explanation :

54/360 X 20

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Convert 4(cos(150°) + isin(150) into rectangular form.

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Answer:

The rectangular form of $4[cos(150^{o})+i sin (150^{o})]$ is $-2\sqrt{3}+2i$ .

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3 years ago

Read 2 more answers

One urn contains one blue ball (labeled B1) and three red balls (labeled R1, R2, and R3). A second urn contains two red balls (R

marusya05 [52]

Answer:

(a) See attachment for tree diagram

(b) 24 possible outcomes

Step-by-step explanation:

Given

$Urn\ 1 = \{B_1, R_1, R_2, R_3\}$

$Urn\ 2 = \{R_4, R_5, B_2, B_3\}$

Solving (a): A possibility tree

If urn 1 is selected, the following selection exists:

$B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]$

If urn 2 is selected, the following selection exists:

$B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]$

See attachment for possibility tree

Solving (b): The total number of outcome

For urn 1

There are 4 balls in urn 1

$n = \{B_1,R_1,R_2,R_3\}$

Each of the balls has 3 subsets. i.e.

$B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]$

So, the selection is:

$Urn\ 1 = 4 * 3$

$Urn\ 1 = 12$

For urn 2

There are 4 balls in urn 2

$n = \{B_2,B_3,R_4,R_5\}$

Each of the balls has 3 subsets. i.e.

$B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]$

So, the selection is:

$Urn\ 2 = 4 * 3$

$Urn\ 2 = 12$

Total number of outcomes is:

$Total = Urn\ 1 + Urn\ 2$

$Total = 12 + 12$

$Total = 24$

5 0

3 years ago

in the given kite SZ= 10yards , WZ = 10 yards TZ=12 yards , and RZ=32 yards Determine the area of the kite

Fynjy0 [20]

If we assume the given segments are those from the vertices to the point of intersection of the diagonals, it seems one diagonal (SW) is 20 yards long and the other (TR) is 44 yards long. The area (A) of the kite is half the product of the diagonals:

... A = (1/2)·SW·TR = (1/2)·(20 yd)·(44 yd)

... A = 440 yd²

5 0

3 years ago