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UkoKoshka [18]
3 years ago
9

(2x-6)(x +11) What is the answer?

Mathematics
1 answer:
ycow [4]3 years ago
5 0

Answer:

2x^2+16x-66

Step-by-step explanation:

Use FOIL (First, Outside, Inside, Last)

(2x-6)(x+11)

F- 2x*x=2x^2

O-2x*11=22x

I--6*x=-6x

L--6*11=-66

2x^2+22x-6x-66

Combine like terms

2x^2+16x-66

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a circular native american dwlelling has a diameter of 24 feet. how much are inside the home? use 3.14 for { } (pi)
Marizza181 [45]

Answer :

A = 452.16  \: {ft}^{2}

step-by-step explanation:

From the question,

the diameter:

d = 24 \: ft

The radius :

r =  \frac{diameter}{2}=  \frac{24}{2}=12 \: ft

\pi = 3.14

Area of a circle is given as

A = \pi {r}^{2}

By substitution,

A = 3.14( {12}^{2} )

\implies \: A = 3.14 \times 144

\implies \: A = 452.16  \: {ft}^{2}

Hence the area of the home is

A = 452.16  \: {ft}^{2}

8 0
3 years ago
Which description best fits the graph
soldier1979 [14.2K]

Answer:

b

Step-by-step explanation:

4 0
2 years ago
Joe is taking the train from his hometown to Georgetown and back to his hometown. The distance to Georgetown is 350 miles. How m
ivanzaharov [21]

Answer:

it would be A.

Step-by-step explanation:

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4 0
3 years ago
A formula is given that states 6x-k=wk. Solve this formula for k in terms of the other variables in the equation. ​
nadya68 [22]

Answer:

k = 6x/(w + 1)

Step-by-step explanation:

6x - k = wk

6x = wk + k

k(w+1) = 6x

k = 6x/(w + 1)

3 0
3 years ago
Suppose that each child born is equally likely to be a boy or a girl. Consider a family with exactly three children. Let BBG ind
Gemiola [76]

Answer:

(a)

S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}

(b)

i.

1\ girl = \{GBB, BBG, BGB\}

P(1\ girl) = 0.375

ii.

Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}

P(Atleast\ 2 \ girls) = 0.5

iii.

No\ girl = \{BBB\}

P(No\ girl) = 0.125

Step-by-step explanation:

Given

Children = 3

B = Boys

G = Girls

Solving (a): List all possible elements using set-roster notation.

The possible elements are:

S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}

And the number of elements are:

n(S) = 8

Solving (bi) Exactly 1 girl

From the list of possible elements, we have:

1\ girl = \{GBB, BBG, BGB\}

And the number of the list is;

n(1\ girl) = 3

The probability is calculated as;

P(1\ girl) = \frac{n(1\ girl)}{n(S)}

P(1\ girl) = \frac{3}{8}

P(1\ girl) = 0.375

Solving (bi) At least 2 are girls

From the list of possible elements, we have:

Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}

And the number of the list is;

n(Atleast\ 2 \ girls) = 4

The probability is calculated as;

P(Atleast\ 2 \ girls) = \frac{n(Atleast\ 2 \ girls)}{n(S)}

P(Atleast\ 2 \ girls) = \frac{4}{8}

P(Atleast\ 2 \ girls) = 0.5

Solving (biii) No girl

From the list of possible elements, we have:

No\ girl = \{BBB\}

And the number of the list is;

n(No\ girl) = 1

The probability is calculated as;

P(No\ girl) = \frac{n(No\ girl)}{n(S)}

P(No\ girl) = \frac{1}{8}

P(No\ girl) = 0.125

7 0
3 years ago
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