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Andru [333]
3 years ago
12

The factored form of x2 + 3x + 4 is (x + 3)(x + 1) TRUE OR FALSE

Mathematics
2 answers:
FromTheMoon [43]3 years ago
7 0
False

because
(a+b)(c+d)=ab+bc+ad+bc
we look at bd

(x+3)(x+1)
bd=3*1=3

the last number is 4
not 3
if we were to expand it we would get
x²+4x+3



false
Effectus [21]3 years ago
3 0
False, the first one shows multiplication while the other just shows addition

Hope I helped ;)
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In a different dimension, cows have $x+3$ legs, with 3 toes on each leg. Chickens have $2x-4$ legs, with 7 toes on each leg. If
Juli2301 [7.4K]

Answer:

20x-10

Step-by-step explanation:

Cows have x+3 legs and 3 toes on each leg, if we want to find the number of toes of ONE cow we have to multiply the legs by the toes giving the following:

3(x+3) \quad (1)

But we have two cows so (1) we have to multiply it by 2 giving a total of:

2(3(x+3))\\6(x+3)\\6x+18 \quad (2)

By the other hand chickens have 2x-4 legs and 7 toes on each leg, if we want to find the number of toes we have to multiply the legs by the toes giving the following:

7(2x-4)\\14x-28 \quad (3)

To find the total amount of toes we have to add (2) and (3)

(6x+18)+(14x-28)\\20x-10

6 0
3 years ago
A plant grows 10 cm each week. The plant is 60 cm now.how many weeks will it take to reach 1 meter
Whitepunk [10]

Answer:

4

Step-by-step explanation:

We know 100 centimeters = 1 meter.

We know the plant grows 10 cm in 1 week.

100 - 60 = 40.

40 cm remaining / 10 cm per week = 4 weeks.

4 0
3 years ago
Suppose the number of insect fragments in a chocolate bar follows a Poisson process with the expected number of fragments in a 2
leonid [27]

Answer:

a)The expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b)0.6004

c)19.607

Step-by-step explanation:

Let X denotes the number of fragments in 200 gm chocolate bar with expected number of fragments 10.2

X ~ Poisson(A) where \lambda = \frac{10.2}{200} = 0.051

a)We are supposed to find the expected number of insect fragments in 1/4 of a 200-gram chocolate bar

\frac{1}{4} \times 200 = 50

50 grams of bar contains expected fragments = \lambda x = 0.051 \times 50=2.55

So, the expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b) Now we are supposed to find the probability that you have to eat more than 10 grams of chocolate bar before ending your first fragment

Let X denotes the number of grams to be eaten before another fragment is detected.

P(X>10)= e^{-\lambda \times x}= e^{-0.051 \times 10}= e^{-0.51}=0.6004

c)The expected number of grams to be eaten before encountering the first fragments :

E(X)=\frac{1}{\lambda}=\frac{1}{0.051}=19.607 grams

7 0
3 years ago
Help with 15 thanks
igor_vitrenko [27]
I think the answer would be C because variability means average.
8 0
3 years ago
Principal, $2000; Annual interest rate, 5.4%; time, 2 years; The balance after 2 years is?
Sloan [31]
P = $2000
R = 5.4%
T = 2 yrs

Interest = PTR/100 = 2000*2*5.4/100 = $216
Balance = $2000 + $216 = $2216
3 0
3 years ago
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