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3 years ago

13

Factor completely. x^2−5x−36 Enter your answer in the box.

1 answer:

myrzilka [38]3 years ago

6 0

We can factor this using the AC method.

We can split the middle term using the factors 4 and -9

x^2 + 4x - 9x - 36

Factor by grouping.

x(x + 4) - 9(x + 4)

<h3><u>(x - 9)(x + 4) are the factors of this polynomial.</u></h3>

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Sylvia placed 20 marbles in a bag: 10 red, 5 blue, green, 1 white. If

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Lynn bought a bag of grapefruit, 1 5/8 pounds of apples, and 2 3/16 pounds of bananas. The total weight of her purchase was 7 1/

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3 years ago

Read 2 more answers

Each of three coins has two sides, heads and tails. Represent the heads or tails status of each coin by a logical variable (A fo

jeka94

Answer:

(a) F(A, B, C) = AB'C' + A'BC' + A'B'C

(b) F(A, B, C) = (A' + B' + C')(A' + B + C)(A + B' + C)(A + B + C')(A + B + C)

Step-by-step explanation:

(a) If F(A, B, C) = 1 iff exactly one of the coins is heads, then either

A is heads and the others are tails (AB'C')

B is heads and the others are tails (A'BC')

C is heads and the others are tails (A'B'C)

Hence, as a minterm expansion,

F(A, B, C) = AB'C' + A'BC' + A'B'C

(b) To get the corresponding maxterm expansion, we convert to binary.

$F(A, B, C) = \sum (100, 010, 001) = \sum(4,2,1)$

The maxterm is the product of the complements.

$F(A, B, C) = \prod (0, 3, 5, 6, 7) = \prod(000, 011, 101, 110, 111)$

Expanding,

F(A, B, C) = (A' + B' + C')(A' + B + C)(A + B' + C)(A + B + C')(A + B + C)

3 0

3 years ago

Hi help please and thank you

Charra [1.4K]

Answer:

looks hard

Step-by-step explanation:

6 0

3 years ago

Use the equation and type the ordered-pairs. y = log 3 x {(1/3, a0), (1, a1), (3, a2), (9, a3), (27, a4), (81, a5)

vagabundo [1.1K]

Answer:

Considering the given equation $y = log_{3}x\\$

And the ordered pairs in the format $(x, y)$

I don't know if it is log of base 3 or 10, but I will assume it is 3.

For $(\frac{1}{3}, a_{0} )$

$x=\frac{1}{3}$

$y=a_{0}$

$y = log_{3}x\\y = log_{3}(\frac{1}{3} )\\y=-\log _3\left(3\right)\\y=-1$

So the ordered pair will be $(\frac{1}{3}, -1 )$

For $(1, a_{1} )$

$x=1$

$y=a_{1}$

$y = log_{3}x\\y = log_{3}1\\y = log_{3}(1)\\Note: \log _a(1)=0\\y = 0$

So the ordered pair will be $(1, 0 )$

For $(3, a_{2} )$

$x=3$

$y=a_{2}$

$y = log_{3}x\\y = log_{3}3\\y = 1$

So the ordered pair will be $(3, 1 )$

For $(9, a_{3} )$

$x=9$

$y=a_{3}$

$y = log_{3}x\\y = log_{3}9\\y=2\log _3\left(3\right)\\y=2$

So the ordered pair will be $(9, 2 )$

For $(27, a_{4} )$

$x=27$

$y=a_{4}$

$y = log_{3}x\\y = log_{3}27\\y=3\log _3\left(3\right)\\y=3$

So the ordered pair will be $(27, 3 )$

For $(81, a_{5} )$

$x=81$

$y=a_{5}$

$y = log_{3}x\\y = log_{3}81\\y=4\log _3\left(3\right)\\y=4$

So the ordered pair will be $(81, 4 )$

4 0

4 years ago