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3 years ago

I need help. Please

Mathematics

2 answers:

zubka84 [21]3 years ago

5 0

Answer:

Answer is A.

Step-by-step explanation:

If you take the current cords of 'C' and add the cords it gives you: (4.5,-6.5). You end out with (3.5,-9,2). Hope this helps!

zlopas [31]3 years ago

3 0

I think the answer is A.

Hope this helps.

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This isn't really a school-related question.It's more of a conflict I have with my self. So: I have a friend that constantly cop

ycow [4]

Well to be honest, you need to straight forward to her and tell her that you do not appreciate her copying you and tell her you dont want to be rude or anything, but this needs to stop.

8 0

3 years ago

Please help me I have been stuck on this question for ages

Finger [1]

Answer:28 green and 35 red

Step-by-step explanation:

Given

If there are r red counter and g green counter then

Probability of drawing a green counter is $P(g)=\frac{4}{9}$

and $P(g)=\frac{\text{No of g counter}}{\text{Total no of counter}}$

Thus $\frac{\text{No of g counter}}{\text{Total no of counter}}=\frac{4}{9}$

$\frac{g}{g+r}=\frac{4}{9}$

$\Rightarrow 9g=4g+4r$

$\Rightarrow 5g=4r\quad \ldots(i)$

Also if 4 red and 2 green counter is added the probability of drawing a green counter is

$P(g)=\frac{10}{23}=\frac{\text{No of g counter}}{\text{Total no of counter}}$

$\Rightarrow \frac{10}{23}=\frac{g+2}{g+2+r+4}$

$\Rightarrow \frac{10}{23}=\frac{g+2}{g+r+6}$

$\Rightarrow 10g+10r+60=23g+46$

$\Rightarrow 10r+14=13g\ quad \ldots(ii)$

Substitute the value of g in equation (ii)[/tex]

$\Rightarrow 10\times \frac{5}{4}g+14=13g$

$\Rightarrow \frac{25}{2}g+14=13g$

$\Rightarrow g=28$

Therefore $r=35$

Thus there 28 green counter and 35 red counter

6 0

3 years ago

Ben makes $39,600 a year. What is the maximum amount he can afford for a'

Olin [163]

Answer:

B:825

Step-by-step explanation:

Hopes this helps:)

3 0

3 years ago

Here are some prices customers paid for different items at a farmer’s market. Find the cost for 1 pound of each item.

posledela

Answer:

1 pounds of apples $=$ $\$ 1·25

1 pounds of cheese $=$ $\$ 7

1 pounds of coffee beans $=$ $\$ 5·5

1 pounds of fudge $=$ $\$ 9

1 pounds of pumpkin $=$ $\$ 0·88

Step-by-step explanation:

Unit Rate is the ratio between two different quantities with different units

The ratio of cost per items is

$=$ $\frac{dollars}{pounds}$

4 pounds of apples $= \$ 5

4 pounds $= \$ 5

Divided 4 both the side

1 pounds of apples $=$ $\$ 1·25

$\frac{1}{2}$ pounds of cheese $= \$ 3·50

$\frac{1}{2}$ pounds $= \$ 3·50

Divided $\frac{1}{2}$ both the side

1 pounds of cheese $=$ $\$ 7

$\frac{3}{2}$ pounds of coffe beans $= \$ 8·25

$\frac{3}{2}$ pounds $= \$ 8·25

Divided $\frac{3}{2}$ both the side

1 pounds of coffee beans $=$ $\$ 5·5

$\frac{3}{4}$ pounds of fudge $= \$ 6·75

$\frac{3}{4}$ pounds $= \$ 6·75

Divided $\frac{3}{4}$ both the side

1 pounds of fudge $=$ $\$ 9

$\frac{25}{4}$ pounds of pumpkin is $= \$ 5·50

$\frac{25}{4}$ pounds $= \$ 5·50

Divided $\frac{25}{4}$ both the side

1 pounds of pumpkin $=$ $\$ 0·88

8 0

3 years ago

In 1981, the Australian humpback whale population was 350 and has increased at a rate of 14% each year since then.)

dimulka [17.4K]

Answer:

In 1981, the Australian humpback whale population was 350

Po = Initial population = 350

rate of increase = 14% annually

P(t) = Po*(1.14)^t

P(t) = 350*(1.14)^t

Where

t = number of years that have passed since 1981

Year 2000

2000 - 1981 = 19 years

P(19) = 350*(1.14)^19

P(19) = 350*12.055

P(19) = 4219.49

P(19) ≈ 4219

Year 2018

2018 - 1981 = 37 years

P(37) = 350*(1.14)^37

P(37) = 350*127.4909

P(37) = 44621.84

P(37) ≈ 44622

There would be about 44622 humpback whales in the year 2018

5 0

3 years ago