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4 years ago

Evaluate each expression

Mathematics

1 answer:

Oliga [24]4 years ago

5 0

1. - 4^3 = -1 * 4*4*4 = - 64
2. (-4)^3 = (-4)(-4)(-4) = - 64
3. 4^3 = 4 * 4 * 4 = 64

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PLS HELP AN OGRE OUT Marco has drawn a line to represent the perpendicular cross-section of the triangular prism. Is he correct?

Rzqust [24]

Answer:

The correct option is;

Yes, the line should be perpendicular to one of the rectangular faces

Step-by-step explanation:

A perpendicular cross-section of a triangular prism resting on a rectangular face will have a triangular cross-section section an either side of the line of the cross-section, therefore we have;

Given that the triangular prism is lying on one of the rectangular faces, the perpendicular cross-section will be represented by a line drawn along the slant height of the triangular prism, therefore the correct option is yes, the line should be perpendicular to one of the rectangular faces.

5 0

4 years ago

I need help with this question

Levart [38]

Answer:

-5.5

Step-by-step explanation:

You just need to subtract the two I'm pretty sure

8 0

3 years ago

A tank's length is 3 feet longer than its width. The height is 1 foot less than the width. The volume of the tank is 160 cubic f

Ivanshal [37]

Answer:

The width of the tank is <u>5 feet</u>.

Step-by-step explanation:

Given:

A tank's length is 3 feet longer than its width.

The height is 1 foot less than the width.

The volume of the tank is 160 cubic feet.

Now, to find the width of the tank.

Let the width of the tank be $x.$

So, length is $x+3.$

And, height is $x-1.$

The volume of tank = 160 cubic feet.

Now, to get the width we put formula:

Volume = length × width × height.

$160=(x+3)\times x\times (x-1)$

$160 =x(x^2-x+3x-3)$

$160=x(x^2+2x-3)$

$160=x^3+2x^2-3x$

<em>Subtracting both sides by 160 we get</em>:

$0=x^3+2x^2-3x-160$

$x^3+2x^2-3x-160=0$

Now, on solving the equation we get:

$x=5.$

The width = 5 feet.

Therefore, the width of the tank is 5 feet.

5 0

4 years ago

Deion has the following data: 47, u, 41, 38. if the range is 10 which could "U" be ?

slavikrds [6]

deion has the following data: 47, u, 41, 38. if the range is 10 which could "U" be ?

REmember that

The range of a set of data is the difference between the highest and lowest values in the set

If U is the highest value

Range=U-38

range=10

10=U-38

U=10+38

U=48

If U is the lowest value

Range=47-U

10=47-U

U=47-10

U=37

therefore

U coulbe between 48 and 37

4 0

1 year ago

Write the equation of a hyperbola centered at the origin with x-intercepts +/- 4 and foci of +/-2(sqrt5)

Xelga [282]

Answer:

$\frac{x^2}{16}-\frac{b^2}{4}=1$

Step-by-step explanation:

A hyperbola is the locus of a point such that its distance from a point to two points (known as foci) is a positive constant.

The standard equation of a hyperbola centered at the origin with transverse on the x axis is given as:

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

The coordinates of the foci is at (±c, 0), where c² = a² + b²

Given that a hyperbola centered at the origin with x-intercepts +/- 4 and foci of +/-2√5. Since the x intercept is ±4, this means that at y = 0, x = 4. Substituting in the standard equation:

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\\frac{4^2}{a^2}-\frac{0}{b^2} =1\\\frac{4^2}{a^2}=1\\ a^2=16\\a=\sqrt{16}=4\\ a=4$

The foci c is at +/-2√5, using c² = a² + b²:

$c^2=a^2+b^2\\(2\sqrt{5} )^2=4^2+b^2\\20 = 16 + b^2\\b^2=20-16\\b^2=4\\b=\sqrt{4}=2\\ b=2$

Substituting the value of a and b to get the equation of the hyperbola:

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\\\\frac{x^2}{16}-\frac{b^2}{4}=1$

5 0

3 years ago