an airplaine descends 1.5 miles to an elevation of 5.25 miles. find the elevation of the plane before its descent

Mathematics

1 answer:

Maksim231197 [3]3 years ago

8 0

We can then write an equation representing this problem as:

e−1.5mi=5.25mi

Now, add 1.5mi to each side of the equation to solve for e while keeping the equation balanced:

e−1.5mi+1.5mi=5.25mi+1.5mi

e−0=6.75mi

e=6.75mi

The plane's starting elevation was 6.75 miles

Hope this helps!

You might be interested in

Evaluate the line integral, where c is the given curve. (x + 9y) dx + x2 dy, c c consists of line segments from (0, 0) to (9, 1)

viktelen [127]

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=\int_C\langle x+9y,x^2\rangle\cdot\underbrace{\langle\mathrm dx,\mathrm dy\rangle}_{\mathrm d\mathbf r}$

The first line segment can be parameterized by $\mathbf r_1(t)=\langle0,0\rangle(1-t)+\langle9,1\rangle t=\langle9t,t\rangle$ with $0\le t\le1$ . Denote this first segment by $C_1$ . Then

$\displaystyle\int_{C_1}\langle x+9y,x^2\rangle\cdot\mathbf dr_1=\int_{t=0}^{t=1}\langle9t+9t,81t^2\rangle\cdot\langle9,1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(162t+81t^2)\,\mathrm dt$
$=108$

The second line segment ( $C_2$ ) can be described by $\mathbf r_2(t)=\langle9,1\rangle(1-t)+\langle10,0\rangle t=\langle9+t,1-t\rangle$ , again with $0\le t\le1$ . Then

$\displaystyle\int_{C_2}\langle x+9y,x^2\rangle\cdot\mathrm d\mathbf r_2=\int_{t=0}^{t=1}\langle9+t+9-9t,(9+t)^2\rangle\cdot\langle1,-1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(18-8t-(9+t)^2)\,\mathrm dt$
$=-\dfrac{229}3$

Finally,

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=108-\dfrac{229}3=\dfrac{95}3$

5 0

3 years ago

1.56 times 48 equals

Crank

Answer:

It equals 74.88

Step-by-step explanation:

U jus multiply them

5 0

3 years ago

Read 2 more answers

What is 12.3 -(-9.6) =12.3 + 9.6 simplified?