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1 year ago

Hello can you please help me with one or two please

Mathematics

1 answer:

Zolol [24]1 year ago

6 0

For question number 2, the answer is A 6x+9

Explanation:

$\begin{gathered} 3(2x+3) \\ \text{Clearing the brackets, we get} \\ 6x+9 \end{gathered}$

Question Number 19:

$\begin{gathered} 8x-2x+4 \\ 6x+4 \\ \text{Factorizing, we get} \\ 2(3x+2) \end{gathered}$

So, the correct answer for question number 19 is 6x + 4.

You might be interested in

A children's pony ride at a zoo has ponies attached to a carousel pole in the center of a circle that the ponies walk around as

stepladder [879]

We simply need to solve for the circumference of the imaginary circle.

Given:
Diameter = 25 feet.

Circumference = 2 * radius * π OR Circumference = diameter * π

C = 25feet * 3.14
C = 78.50 feet

The pony will walk 78.50 feet to complete one trip around the circle.

3 0

3 years ago

Evaluate [(51 + 3) − 32] ÷ 9 ⋅ 2.

marin [14]

51+3 = 54

54 - 32 = 22

22/9 = 2.444

2.444 times 2 = 4.888

4 0

3 years ago

Read 2 more answers

Several co-workers carpool to work each day. the daily cost of gas and tolls, $12, is split evenly among the people who carpool

Verizon [17]

Answer:

c(x) = ($12) / x

Step-by-step explanation:

The general formula for inverse proportion is y = k/x.

In this specific problem, y represents the cost per person, k the dailyl cost ($12), and x the number of perople riding.

Then

c(x) = cost per person as a function of the number of people riding

= ($12) / x

Note that if only one person rides, that person pays $12/1, or $12; if two people, $12/2, or $6 per person, and so on.

5 0

3 years ago

A farmer has $1500 available to build an E-shaped fence along a straight river so as to create two identical pastures. The mater

Dahasolnce [82]

Answer: The largest dimensions that are possible would be 50 foot by 125 foot.

Step-by-step explanation:

Since we have given that

Amount a farmer has available = $1500

Let x be the side perpendicular to the river.

Let y be the side parallel to the river.

Number of section perpendicular to the river = 3

cost of material for the side parallel to the river = $6 per foot

Cost of material for the side perpendicular to the river = $5 per foot

So, total cost becomes

$Cost=6y+5(3x)=1500\\\\Cost=6y+15x=1500\\\\y=\dfrac{1500-15x}{6}}$

Area would be

$A=x\times y\\\\A=x(\dfrac{1500-15x}{6})\\\\A=\dfrac{1}{6}(1500x-15x^2)\\\\A'=\dfrac{1}{6}(1500-30x)$

Now, put A' = 0 to get the critical points.

So, it becomes,

$\dfrac{1}{6}(1500-30x)=0\\\\1500-30x=0\\\\1500=30x\\\\x=\dfrac{1500}{30}\\\\x=50$

$A''=\dfrac{-30}{6}=-5$

so, at x= 50 it will give maximum dimensions.

$y=\dfrac{1500-15x}{6}=\dfrac{1500-15\times 50}{6}=\dfrac{750}{6}=125$

So, the largest dimensions that are possible would be 50 foot by 125 foot.

7 0

3 years ago

A certain portfolio consisted of 5 stocks, priced at $20, $35, $40, $45, and $70, respectively. On a given day, the price of one

saveliy_v [14]

Answer:

Option E: 35, 40, 45

Step-by-step explanation:

To solve this problem we can test the options, checking if the calculations will match the result.

20, 35 and 70 remained constant.

The average price of a stoc in the portfolio before the changings were:

(20+35+40+45+70)/5 = 42

With an increase of approximately 2%, the average value will be approximately 42*1.02 = 42.84

If the average price of a stock in the portfolio rose, and the rise of one stock in percentage was bigger than the fall of the other stock, we need to choose the stock that has the bigger price as the one that increased its value.

An increase of 15% is calculated multiplying the base value by 1.15, and a decrease of 35% is calculated multiplying the base value by (1-0.35)=0.65.

So, if the stocks affected were the $40 and 45$, we have that:

(20+35+40*0.65+45*1.15+70)/5 = 40.55

That's not the correct answer.

20, 45 and 70 remained constant.

Affected: 35 and 40

(20+35*0.65+40+45*1.15+70)/5 = 40.9

That's not the correct answer.

20, 35 and 40 remained constant.

Affected: 45 and 70

(20+35+40+45*0.65+70*1.15)/5 = 40.95

That's not the correct answer.

35, 40 and 70 remained constant.

Affected: 20 and 45

(20*0.65+35+40+45*1.15+70)/5 = 41.95

That's not the correct answer.

35, 40 and 45 remained constant.

Affected: 20 and 70

(20*0.65+35+40+45+70*1.15)/5 = 42.7

That's the correct answer, with an increase in the average value of 0.7/42 = 0.0167 = 1.67%

6 0

3 years ago