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3 years ago

What is a possible integers for b? in the equation x squared+13x+36=(x+a)(x+b)

Mathematics

1 answer:

Scorpion4ik [409]3 years ago

6 0

Answer:

3) b = {4 , 9}

Step-by-step explanation:

x^2 + 13x + 36

36 = 9 * 6

13 = 9 + 4

x^2 + 13x + 36 = (x + 9)(x + 4) or = (x + 4) (x + 9)

possible b = {4 , 9}

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For the members of a neighborhood meeting group, the probability of randomly selecting a member who likes to cook is 42%, the pr

Monica [59]

Answer: Probability of randomly selecting a member who likes to cook or likes to sew = 58%

Step-by-step explanation:

Since we have given that

Probability of a randomly selecting a member who likes to cook=P(C) = 42%

Probability of a randomly selecting a member who likes to sew =P(S)= 23%

Probability of a randomly selecting a member who likes to both cook and sew

$P(C\cap S)=7\%$

As we know the "Probability rules " :

$P(C\cup S)=P(C)+P(S)-P(S\cap C)\\\\P(C\cup S)=42+23-7\\\\P(C\cup S)=58$

So, Probability of randomly selecting a member who likes to cook or likes to sew = 58%

5 0

3 years ago

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Darya [45]

Answer:

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6 0

3 years ago

Read 2 more answers

A local sorority sold hot dogs and bratwursts at the spring fling picnics. The first day they sold 8 dozen hot dogs and 13 dozen

kicyunya [14]

Answer:

1 hot dog costs $0.75

1 bratwurst costs $1.35

Step-by-step explanation:

Let x and y be the price per dozen of hot dogs and bratwursts respectively.

The first day they sold 8 dozen hot dogs and 13 dozen bratwursts for $282.60

8x + 13y = 282.60

The second day they sold 10 dozen hot dogs and 15 dozen bratwursts for a total of $333.00

10x + 15y = 333

and we have the linear system

which can be written in matrix form as

$\bf \left(\begin{array}{cc}8&13\\10&15\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}282.60\\333\end{array}\right)$

The solution would be given by

$\bf \left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{cc}8&13\\10&15\end{array}\right)^{-1}\left(\begin{array}{c}282.60\\333\end{array}\right)$

We have

$\bf \left(\begin{array}{cc}8&13\\10&15\end{array}\right)^{-1}=\left(\begin{array}{cc}-3/2&13/10\\1&-4/5\end{array}\right)$

hence

$\bf \left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{cc}-3/2&13/10\\1&-4/5\end{array}\right)\left(\begin{array}{c}282.60\\333\end{array}\right)=\left(\begin{array}{c}9\\ 16.2\end{array}\right)$