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3 years ago

Decompose each number: 20

Mathematics

2 answers:

insens350 [35]3 years ago

7 0

Um u need to put more details

inn [45]3 years ago

7 0

I dont get it u need to explain

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Plz help, will give brainiest

Tems11 [23]

Answer:

4x^2 + 8x + 4

4(x^2 + 2x + 1) - remove GCF of 4

4(x + 1)(x + 1) - factor

4(x + 1)^2 - collect like terms

Step-by-step explanation:

Then also expand it out by distributing:

21x^3 + 35x²

Form 1:

21x^3 + 35x² - unfactored

Form 2:

7x²(3x + 5) - factored with GCF of 7x² brought to the front

Update:

You could also multiply two binomials and make a quadratic.

Example:

(7x + 2)(3x + 5)

7x(3x + 5) + 2(3x + 5)

= 21x² + 35x + 6x + 10

= 21x² + 41x + 10

7 0

3 years ago

Based on the table below, evaluate f(1).

Jobisdone [24]

Answer:

f(1) = 24

Step-by-step explanation:

f(1) is the value of f(x) when x = 1, that is from the table

f(1) = 24

3 0

3 years ago

Please help asap! Add using the number line.

sweet-ann [11.9K]

Answer:

0.6

Step-by-step explanation:

1) 1.4-0.8 (A negative and a positive equals a negative, then you just subtract. On the number line you would start at 1.4 and go back)

hope this helps!

6 0

4 years ago

Add x^3 - 4x^2 + 1 to 3x^2 + x <br> Show Your work!

Sladkaya [172]

$\qquad \qquad\huge \underline{\boxed{\sf Answer}}$

Let's solve ~

$\qquad \sf \dashrightarrow \: (x {}^{3} - 4x {}^{2} + 1) + (3 {x}^{2} + x)$

$\qquad \sf \dashrightarrow \: {x}^{3} - 4 {x}^{2} + 1 + 3 {x}^{2} + x$

$\qquad \sf \dashrightarrow \: {x}^{3} - 4 {x}^{2} + 3x {}^{2} + x + 1$

$\qquad \sf \dashrightarrow \: {x}^{3} - {x}^{2} + x + 1$

6 0

2 years ago

The probability that a professor arrives on time is 0.8 and the probability that a student arrives on time is 0.6. Assuming thes

saul85 [17]

Answer:

a)0.08 , b)0.4 , C) i)0.84 , ii)0.56

Step-by-step explanation:

Given data

P(A) = professor arrives on time

P(A) = 0.8

P(B) = Student aarive on time

P(B) = 0.6

According to the question A & B are Independent

P(A∩B) = P(A) . P(B)

Therefore

${A}'$ & ${B}'$ is also independent

${A}'$ = 1-0.8 = 0.2

${B}'$ = 1-0.6 = 0.4

part a)

Probability of both student and the professor are late

P(A'∩B') = P(A') . P(B') (only for independent cases)

= 0.2 x 0.4

= 0.08

Part b)

The probability that the student is late given that the professor is on time

$P(\frac{B'}{A})$ = $\frac{P(B'\cap A)}{P(A)}$ = $\frac{0.4\times 0.8}{0.8}$ = 0.4

Part c)

Assume the events are not independent

Given Data

P $(\frac{{A}'}{{B}'})$ = 0.4

= $\frac{P({A}'\cap {B}')}{P({B}')}$ = 0.4

$P$ $({A}'\cap {B}')$ = 0.4 x P $({B}')$

= 0.4 x 0.4 = 0.16

$P({A}'\cap {B}')$ = 0.16

The probability that at least one of them is on time

$P(A\cup B)$ = 1- $P({A}'\cap {B}')$

= 1 - 0.16 = 0.84

ii)The probability that they are both on time

P $(A\cap B)$ = 1 - $P({A}'\cup {B}')$ = 1 - $[P({A}')+P({B}') - P({A}'\cap {B}')]$

= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56

6 0

3 years ago