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3 years ago

A student claims that 8^3 x 8^-5 is greater than 1. Explain whether the student is correct or not.

1 answer:

7 0

Remember:
a^m * a^n=a^(m+n)
a^-m=1/a^m

8³ x 8⁻⁵=8³⁻⁵=8⁻²=1/8²=1/64<1

Answer: the student is not correct, because 1/64<1, and 8³ x 8⁻⁵=1/64.

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Answer:

8 times larger

Step-by-step explanation:

The radius of the large sphere is double the radius of the small sphere.

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2 years ago

Combine like terms 30x + 6y - 24x - y =

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Answer:

6x+5y

Step-by-step explanation:

Combining like terms (x and y)

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3 years ago

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How do you simplify this and write as a polynomial in standar form? (x-9)3x-7)+(3x^2-5x+2)

nasty-shy [4]

Answer:

Standard form of $(x-9)(3x-7) + (3x^{2} - 5x+2)$ $=6x^{2} - 39x + 65$

Step-by-step explanation:

Here, the given expression is $(x-9)(3x-7) + (3x^{2} - 5x+2)$

Now, simplifying the above expression in parts, we get

$(x-9)(3x-7) = 3x^{2} - 7x -27x + 63 = 3x^{2} - 34x + 63$

hence, combining both parts:

$(x-9)(3x-7) + (3x^{2} - 5x+2)$ $=(3x^{2} -34x +63) + (3x^{2} - 5x+2)$

= $6x^{2} - 39x + 65$

The above expression is of the STANDARD FORM: $ax^{2} +bx + c$

Hence, the standard form of $(x-9)(3x-7) + (3x^{2} - 5x+2)$ $=6x^{2} - 39x + 65$

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3 years ago

At what values of x does f(x) = x^3 - 2x^2 -4x+1 satisfy the mean value theorwm on [0,1]

denis-greek [22]

Answer:

x=1/3

Step-by-step explanation:

A function f is given as

$f(x) = x^3-2x^2-4x+1$ in the interval [0,1]

This function f being an algebraic polynomial is continuous in the interval [0,1] and also f is differntiable in the open interval (0,1)

Hence mean value theorem applies for f in the given interval

$f(1) = 1-2-4+1 = -4\\f(0) = 1$

The value

$\frac{f(1)-f(0)}{1-0} =\frac{-4-1}{1} =-5$

Find derivative for f

$f'(x) = 3x^2-4x-4$

Equate this to -5 to check mean value theorem

$3x^2-4x-4=-5\\3x^2-4x+1=0\\\\(x-1)(3x-1) =0\\x= 1/3 : x = 1$

We find that 1/3 lies inside the interval (0,1)

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