An airplane traveled 5.7 x 10 to the 2nd miles per hour for 1.4 x 10 to the 1st power hours. How far did the airplane travel

Mathematics

1 answer:

kvv77 [185]2 years ago

5 0

S = v x t
= 5.7 x 10^2 x 1.4 x 10 = 7980 Miles

You might be interested in

Dashna paid a $20 fee for a booth at an art fair so she could sell her ceramic bowls. She will earn $10 for every ceramic bowl s

katovenus [111]

Answer:

The correct answer is D

Step-by-step explanation:

-20 is your y-intercept and your graph will be increasing

6 0

3 years ago

What is EG? Round to the nearest tenth if necessary.

slava [35]

Answer:

26.

Step-by-step explanation:

As the vertex angles are equal:

24/54 = x / x+10

54x = 24x + 240

30x = 240

x = 8

So EG = x + x + 10

= 8 + 8 + 10

= 26..

7 0

3 years ago

What value of b will cause the system to have an infinite number of solutions?

irga5000 [103]

b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

<u>Solution: </u>

Need to calculate value of b so that given system of equations have an infinite number of solutions

$\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}$

Let us bring the equations in same form for sake of simplicity in comparison

$\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}$

Now we have two equations

$\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}$

Let us first see what is requirement for system of equations have an infinite number of solutions

If $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ are two equation

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ then the given system of equation has no infinitely many solutions.

In our case,

$\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}$