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4 years ago

13

The distance from Boston to Philadelphia by plane is 436 kilometers. A pilot flies this route 252 times. Does the pilot travel m

ore or less than 100,000 kilometers?

2 answers:

zloy xaker [14]4 years ago

3 0

Answer: More

Step-by-step explanation: 436x252 = 109872.

Charra [1.4K]4 years ago

3 0

Answer:

more than 1000,000

Step-by-step example

well 436 times 252 is 109872 and 109872 minus 1000,000 is 890128 miles therefor a pilot flies more than 1000,000 miles from Boston to Philadelphia

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Use the number line below to find the midpoint of AD

Arisa [49]

Midpoint of line AD = (-7+1) ÷ 2
= -6 ÷ 2
= -3

The midpoint of AD is -3.

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3 years ago

1/2y+3/4-5/8y-7/8 pls help worth 10 pets

Natali5045456 [20]

Answer:

$\frac{-y-1}{8}$

Step-by-step explanation:

1/2y - 5/8y = 4/8y - 5/8y = -1/8y

3/4 - 7/8 = 6/8 - 7/8 = -1/8

-1/8y - 1/8 =

-y/8 - 1/8 =

$\frac{-y-1}{8}$

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3 years ago

The length of a line segment is 13 units. One end point of the line segment is (-3,7). Find four points that could be the other

monitta

Answer:

( -16,7 ) , ( 10,7 ) , ( -3,20 ) and ( -3,-6 )

Step-by-step explanation:

We are given a line segment with one end point ( -3,7 ) and length 13 units.

So, there are many possibilities for the other end points.

Four of them are:

A. 13 units left i.e. ( -3-13,7 ) = ( -16,7 )

B. 13 units right i.e. ( -3+13,7 ) = ( 10,7 )

C. 13 units up i.e. ( -3,7+13 ) = ( -3,20 )

D. 13 units down i.e. ( -3,7-13 ) = ( -3,-6 ).

So, the other four end points could be ( -16,7 ) , ( 10,7 ) , ( -3,20 ) and ( -3,-6 ).

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3 years ago

What is the solution to ln (x2 - 16) = 0?

bogdanovich [222]

ANSWER

$x = \pm \sqrt{17}$
EXPLANATION

The given equation is

$ln( {x}^{2} - 16) = 0$

Take antilogarithm of both sides to base e.

${e}^{ ln( {x}^{2} - 16) } = {e}^{0}$

This will simplify to

${x}^{2} - 16 = 1$

Group like terms to obtain,

${x}^{2} = 1 + 16$

${x}^{2} = 17$

Take square root of both sides to get,

$x = \pm \sqrt{17}$

6 0

3 years ago

Read 2 more answers

Solve the equation after making an appropriate substitution. X^6-98x^3=3375

Leni [432]

Rearrange $x^6-98x^3=3375$ to give

$x^6-98x^3-3375=0$

Let $x^3$ be a variable 'p' and so we can write $x^6$ as $x^6=(x^3)(x^3)=(p)(p)= p^2$

Rewrite the equation in terms of 'p'

$p^2-98p-3375 = 0$

where $a=1, b=-98, c=-3375$

using the quadratic formula $\frac{-b+ \sqrt{b^2-4ac} }{2a}$ and subsitute the value of $a, b, c$

$p_1= \frac{-(-98)+ \sqrt{(-98)^2-4(1)(-3375)} }{2} =125$

$p_2= \frac{-(-98)- \sqrt{(-98)^2-4(1)(-3375)} }{2} =-27$

There are two value of p; 125 and -27

Now we find the value of x

Earlier we substitute $x^3$ for $p$ and $x^6$ for $p^2$

When $p=125$ , $x= \sqrt[3]{125}=5$
When $p = -27, x= \sqrt[3]{-27}=-3$

So the final answer is the two values of x;

x = 5 OR x = -3

3 0

3 years ago