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3 years ago

I need help with this question to turn it into an equation. The question is

1 answer:

7 0

Sure, here is how I broke it down. The bird is flying a constant X miles per hour with a wind speed of Y miles per hour. So, there are two equations that you use to solve. Here they are:

X+Y=12
X-Y=4

Now, you have to use substitution or elimination to solve (I will use elimination). To use elimination, you simply add both equations together to get:

2X=16
Solve for X by dividing by 2.

X=8. We already established that X is the speed of the bird. Plug X back into either of the equations to find the windspeed.

(8)+Y=12

Subtract 8 from both sides.

Y=4. We already established that Y is the windspeed.

So, the speed of the bird (X) is 8mph and the speed of the wind (Y) is 4mph.

Hope this helped!!

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X^2 + y^2 - 6y + 4x = 36

zysi [14]

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3 years ago

Distance between A (2, -1) and (2,a) is 4 units. Find the value of a where a is positive.

kolezko [41]

Answer:

a = 3

Step-by-step explanation:

Calculate the distance d using the distance formula and equate to 4

d = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }$

with (x₁, y₁ ) = (2, - 1) and (x₂, y₂ ) = (2, a)

d = $\sqrt{(2-2)^2+(a+1)^2}$ = $\sqrt{0+(a+1)^2}$ = $\sqrt{(a+1)^2}$ = a + 1 then

a + 1 = 4 ( subtract 1 from both sides )

a = 3

3 0

2 years ago

How many times does 5 go in to 47

Feliz [49]

47/5=9.40
9*5=45
47/5= 9 2/5

6 0

3 years ago

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10

zvonat [6]

The approximate difference in the ages of the two cars, which depreciate to 60% of their respective original values, is 1.7 years.

<h3>What is depreciation?</h3>

Depreciation is to decrease in the value of a product in a period of time. This can be given as,

$FV=P\left(1-\dfrac{r}{100}\right)^n$

Here, (<em>P</em>) is the price of the product, (<em>r</em>) is the rate of annual depreciation and (<em>n</em>) is the number of years.

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.

Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is $n_1$ . Thus, by the above formula for the first car,

$0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}$

Take log both the sides as,

$\log 0.6=\log (0.9)^{n_1}\\\log 0.6={n_1}\log (0.9)\\n_1=\dfrac{\log 0.6}{\log 0.9}\\n_1\approx4.85$

Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.

Thus, the depreciation price of the car is 0.6y. Let the number of year is $n_2$ . Thus, by the above formula for the second car,

$0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}$

Take log both the sides as,

$\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14$

The difference in the ages of the two cars is,

$d=4.85-3.14\\d=1.71\rm years$

Thus, the approximate difference in the ages of the two cars, which depreciate to 60% of their respective original values, is 1.7 years.

Learn more about the depreciation here;

brainly.com/question/25297296

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2 years ago

You roll a number cube twice first roll is a four second roll I roll a even number is that a defendant or independent situation

Katyanochek1 [597]

Independent because the ratio of the second cube is the same (1:6), and it isn't affected by the result of the first cube

7 0

3 years ago