The global maxima is at what point? A. c B. b C. a and b D. a

The quantity of four minus one all raised to the five-minus- two power

leva [86]

Answer:

(4-1)^5-2

Step-by-step explanation:

The quantity means parenthesis, around four minus one, (closed quantity), all raised to the power (exponent) five minus 2

3 0

3 years ago

Read 2 more answers

Plz help me with Q:13 And Q:14

PolarNik [594]

For question 13, you need to look at all the numbers that are less than 1/2 and then count the number of marks in total .
Answer: 7 B
For question 14, you need to first make an equation that suits the circumstances and the situations in this problem.
$25+$24x=$193
$24x=$168
x=7 weeks
x stands for the number of weeks that she has to wait for in order to save enough money to purchase the camera.
Answer: 7 weeks A

5 0

3 years ago

Read 2 more answers

How many pounds of $2.4/lb trail mix should a grocer combine with 5 lb of $1.5/lb trail mix to get $1.9/lb trail mix?

user100 [1]

4
1.5x5=7.5 + (2.4 x 4)
/9
=1.9

8 0

2 years ago

What is the best estimate of the sum of the fraction? 31/4 +17/3 +8/18

Bogdan [553]

The best estimate of the sum of the expression given is $\rm 13 \dfrac {31}{26}$

What is a Fraction ?

A fraction is written as p/q , where q $\rm \ne$ 0

It is asked to determine the sum of the fraction

31/4 + 17/3 + 8/18

Taking the LCM of 4 , 3 , 18

LCM of 4 , 3 ,18 is 36

Therefore

$\rm \dfrac{ 31 *9 + 17 *12 + 8 *2}{36 }$

(279+204+16) /36

499 / 36

$\rm 13 \dfrac {31}{26}$

Therefore the best estimate of the sum of the expression given is $\rm 13 \dfrac {31}{26}$ .

To know more about Fraction

brainly.com/question/10354322

#SPJ1

3 0

2 years ago

(x +y)^5 Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=x,\:\:b=y$

$=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$

so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

$=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5$

$=1\cdot \:1\cdot \:x^5$

$=x^5$

also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5x^4y}{1}$

$=5x^4y$

similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

$=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

Therefore,

$\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

6 0

2 years ago