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3 years ago

Brainlyess

Mathematics

2 answers:

miv72 [106K]3 years ago

8 0

Answer:

Present age of Jose is 3 n, how old will he be in 3 years!?

3n+5

So, the answer is A

coldgirl [10]3 years ago

6 0

Answer:

$\huge\colorbox{violet}{✏﹏ \: ᴀɴsᴡᴇʀ \: }$

Present age of José = 3n

Age of José in 5 years

$= 3n + 5$

➳ A. 3n + 5 is the correct answer.

ʰᵒᵖᵉ ⁱᵗ ʰᵉˡᵖˢ

$\huge\red{ \mid{ \underline{ \overline{ \tt ꧁❣ ʀᴀɪɴʙᴏᴡˢᵃˡᵗ2²2² ࿐ }} \mid}}$

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Solve (x + 3)2 + (x + 3) – 2 = 0. Let u = Rewrite the equation in terms of u. (u2 + 3) + u – 2 = 0 u2 + u – 2 = 0 (u2 + 9) + u –

Verdich [7]

Answer:

The solutions of the original equation are x=-5 and x=-2

Step-by-step explanation:

we have

$(x+3)^2+(x+3)-2=0$

Let

$u=(x+3)$

Rewrite the equation

$(u)^2+(u)-2=0$

Complete the square

$u^2+u=2$

$u^2+u+1/4=2+1/4$

$u^2+u+1/4=9/4$

rewrite as perfect squares

$(u+1/2)^2=9/4$

square root both sides

$(u+1/2)=\pm\frac{3}{2}$

$u=(-1/2)\pm\frac{3}{2}$

$u=(-1/2)+\frac{3}{2}=1$

$u=(-1/2)-\frac{3}{2}=-2$

the solutions are

u=-2,u=1

Alternative Method

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$(u)^2+(u)-2=0$

$a=1\\b=1\\c=-2$

substitute in the formula

$u=\frac{-1\pm\sqrt{1^{2}-4(1)(-2)}} {2(1)}$

$u=\frac{-1\pm\sqrt{9}} {2}$

$u=\frac{-1\pm3} {2}$

$u=\frac{-1+3} {2}=1$

$u=\frac{-1-3} {2}=-2$

the solutions are

u=-2,u=1

Find the solutions of the original equation

For u=-2

$-2=(x+3)$ ----> $x=-2-3=-5$

For u=1

$1=(x+3)$ ----> $x=1-3=-2$

therefore

The solutions of the original equation are

x=-5 and x=-2

4 0

3 years ago

Read 2 more answers

What is 8 - 7 1/4 i need the answer fast

Citrus2011 [14]

Answer:

3/4

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

A large electronic office product contains 2000 electronic components. Assume that the probability that each component operates

KIM [24]

Answer:

The probability is 0.971032

Step-by-step explanation:

The variable that says the number of components that fail during the useful life of the product follows a binomial distribution.

The Binomial distribution apply when we have n identical and independent events with a probability p of success and a probability 1-p of not success. Then, the probability that x of the n events are success is given by:

$P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}$

In this case, we have 2000 electronics components with a probability 0.005 of fail during the useful life of the product and a probability 0.995 that each component operates without failure during the useful life of the product. Then, the probability that x components of the 2000 fail is:

$P(x)=\frac{2000!}{x!(2000-x)!}*0.005^{x}*(0.995)^{2000-x}$ (eq. 1)

So, the probability that 5 or more of the original 2000 components fail during the useful life of the product is:

P(x ≥ 5) = P(5) + P(6) + ... + P(1999) + P(2000)

We can also calculated that as:

P(x ≥ 5) = 1 - P(x ≤ 4)

Where P(x ≤ 4) = P(0) + P(1) + P(2) + P(3) + P(4)

Then, if we calculate every probability using eq. 1, we get:

P(x ≤ 4) = 0.000044 + 0.000445 + 0.002235 + 0.007479 + 0.018765

P(x ≤ 4) = 0.028968

Finally, P(x ≥ 5) is:

P(x ≥ 5) = 1 - 0.028968

P(x ≥ 5) = 0.971032

3 0

3 years ago

Study the pattern and predict the nth term a) 2,5,10,17,....... b)0,1,4,9,.....

Rasek [7]

A) nth term is n^2 + 1

b) nth term is (n - 1)^2

8 0

3 years ago

Tim estimates that 60÷5.7 is about 10.Will the actual quotient be greater than 10 explain

Zepler [3.9K]

Well first you can do the problem. So take 60 divided by 5.7. When it is done it will come out to about 10.5263 which is a irrational number. So yes the quotient will be larger than 10.
Hope I helped!

8 0

3 years ago