Compare using >, <, or =,<br> 64 ounces<br> 4 pounds

A jet flies at an altitude of 52,800 feet. What is the jets height in miles?

WITCHER [35]

Hi, I have an answer for you, hope is O.K.

5 0

4 years ago

I need this answered in ONE minute

Helen [10]

Answer:

$x^4 + 4x^3 + 6x^2 + 7x + 2$

Step-by-step explanation:

We are asked to multiply the given polynomials.

$(x^ 2 + 3x + 1) \times (x^2 + x + 2)$

Multiply each term of the first polynomial to each term of the second polynomial.

$x^ 2 \times (x^2 + x + 2) = x^4 + x^3 + 2x^2$

$3x \times (x^2 + x + 2) = 3x^3 + 3x^2 + 6x$

$1 \times (x^2 + x + 2) = x^2 + x + 2$

Add the results

$(x^4 + x^3 + 2x^2) + (3x^3 + 3x^2 + 6x) + ( x^2 + x + 2)$

Combine the like terms

$x^4 + 4x^3 + 6x^2 + 7x + 2$

The answer is written in descending powers of x.

7 0

3 years ago

HELP ME PLZ BRAINLIST 4 U

mylen [45]

Answer:

G

Step-by-step explanation:

I think its G because i counted the numbers...and i pretty sure its not a Negative so yah

8 0

3 years ago

Evaluate Dx / ^ 9-8x - x2^

Solnce55 [7]

It depends on what you mean by the delimiting carats "^"...

Since you use parentheses appropriately in the answer choices, I'm going to go out on a limb here and assume something like "^x^" stands for $\sqrt x$ .

In that case, you want to find the antiderivative,

$\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}$

Complete the square in the denominator:

$9-8x-x^2=25-(16+8x+x^2)=5^2-(x+4)^2$

Now substitute $x+4=5\sin y$ , so that $\mathrm dx=5\cos y\,\mathrm dy$ . Then

$\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\int\frac{5\cos y}{\sqrt{5^2-(5\sin y)^2}}\,\mathrm dy$

which simplifies to

$\displaystyle\int\frac{5\cos 
y}{5\sqrt{1-\sin^2y}}\,\mathrm dy=\int\frac{\cos y}{\sqrt{\cos^2y}}\,\mathrm dy$

Now, recall that $\sqrt{x^2}=|x|$ . But we want the substitution we made to be reversible, so that

$x+4=5\sin y\iff y=\sin^{-1}\left(\dfrac{x+4}5\right)$

which implies that $-\dfrac\pi2\le y\le\dfrac\pi2$ . (This is the range of the inverse sine function.)

Under these conditions, we have $\cos y\ge0$ , which lets us reduce $\sqrt{\cos^2y}=|\cos y|=\cos y$ . Finally,

$\displaystyle\int\frac{\cos y}{\cos y}\,\mathrm dy=\int\mathrm dy=y+C$

and back-substituting to get this in terms of $x$ yields

$\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\sin^{-1}\left(\frac{x+4}5\right)+C$

4 0

3 years ago

Which undefined terms are needed to define a line segment

Finger [1]

Unsure of what you are asking!

But if the issue here is how to define a line segment, write what you do know and then reconsider "undefined terms."

A line segment is a straight line that connects a given starting point and given ending point.

If you consider a circle of radius 3 units, the radius can be thought of as the line segment connecting the center of the circle to any point on the circumference of the circle.

If the center of a given circle is at C(0,0) and a point on the circumference is given by R(3sqrt(2),3sqrt(2)), then AC is the line segment joining these two points. This line segment has length 3 and is in the first quadrant, with coordinates x=3sqrt(2) and y=3sqrt(2) describing the end point of the segment.

3 0

3 years ago

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Compare using >, <, or =, 64 ounces 4 pounds