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2 years ago

Determine an equivalent algebraic monomial expression using the laws of exponents. Show your work.

Mathematics

1 answer:

Serga [27]2 years ago

4 0

The equivalent algebraic monomial expression of the expression given as (-8a^5b)(3ab^4) is -24a^6b^5

<h3>How to determine an equivalent algebraic monomial expression?</h3>

The expression is given as:

(-8a^5b)(3ab^4)

Multiply -8 and 3

So, we have:

(-8a^5b)(3ab^4) = (-24a^5b)(ab^4)

Multiply a^5 and a (a^5 * a = a^6)

So, we have:

(-8a^5b)(3ab^4) = (-24a^6b)(b^4)

Multiply b and b^4

So, we have:

(-8a^5b)(3ab^4) = -24a^6b^5

Hence, the equivalent algebraic monomial expression of the expression given as (-8a^5b)(3ab^4) is -24a^6b^5

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What is an equation of the line that passes through the points (7,-6)(3,-6)

Agata [3.3K]

Answer:

Step-by-step explanation:

Since both points have their y coordinate as -6, and this is a line,

Then your answer would be y = -6

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3 years ago

How many miles are equivalent to 3.1 kilometers? 1 kilometer is about 0.62 mile

Anton [14]

Answer:

1.922 miles is equivalent to 3.1 kilometers.

Step-by-step explanation:

Given:

1 km = 0.62 miles.

3.1 km = ? miles

Now, in order to find the number of miles equivalent to 3.1 km, we use unitary method with the help of the conversion factor.

From the conversion factor, we can conclude that kilometer is a smaller unit compared to that of mile.

So, conversion factor = $\frac{0.62\ miles}{1\ km}$

The formula to transform 'n' kilometers to miles is given as:

$=Conversion\ factor\times n$

Therefore, 3.1 km = $\frac{0.62\ miles}{1\ km}\times 3.1\ km$

3.1 km = $1.922$ miles

Therefore, 1.922 miles is equivalent to 3.1 kilometers.

4 0

3 years ago

Please help like right now

liubo4ka [24]

Answer:

y-intercept
Slope
rise over run
negative
positive
Undefined
Zero
Slope
y-intercept

Step-by-step explanation:

I'm not sure about what the answer is for the last two questions. If somebody else does know please right it in the comment section. :)

5 0

3 years ago

WILL MARK BRAINLIEST <br><br> what is the domain of (f/g) (x)?

tatyana61 [14]

Given that $f(x) = \sqrt{7-x}$ and $g(x) = \sqrt{x + 2}$ , we can say the following:

$\Bigg(\dfrac{f}{g}\Bigg)(x) = \dfrac{f (x)}{g(x)} = \dfrac{\sqrt{7 - x}}{\sqrt{x+2}}$

Now, remember what happens if we have a negative square root: it becomes an imaginary number. We don't want this, so we want to make sure whatever is under a square root is greater than 0 (given we are talking about real numbers only).

Thus, let's set what is under both square roots to be greater than 0:

$\sqrt{7 - x} \Rightarrow 7 - x \geq 0 \Rightarrow x \leq 7$

$\sqrt{x + 2} \Rightarrow x + 2 \geq 0 \Rightarrow x \geq -2$

Since both of the square roots are in the same function, we want to take the union of the domains of the individual square roots to find the domain of the overall function.

$x \leq 7 \,\,\cup x \geq -2 = \boxed{-2 \leq x \leq 7}$

Now, let's look back at the function entirely, which is:

$\Bigg( \dfrac{f}{g} \Bigg)(x) = \dfrac{\sqrt{7 - x}}{\sqrt{x+2}}$

Since $\sqrt{x + 2}$ is on the bottom of the fraction, we must say that $\sqrt{x + 2} \neq 0$ , since the denominator can't equal 0. Thus, we must exclude $\sqrt{x + 2} = 0 \Rightarrow x + 2 = 0 \Rightarrow x = -2$ from the domain.

Thus, our answer is Choice C, or $\boxed{ \{ x | -2 < x \leq 7 \}}$ .

<em>If you are wondering why the choices begin with the $x |$ symbol, it is because this is a way of representing that $x$ lies within a particular set.</em>

6 0

3 years ago

The equation of a circle is (x + 6)^2 + (y - 4)^2 = 16. The point (-6, 8) is on the circle.

pantera1 [17]

Answer:

y = 8 is the equation of tangent.

Step-by-step explanation:

The equation of the tangent to the circle at (-6,8) is of the form:

y = mx + c

where m is the slope of the tangent and c is the y-intercept.

The point (-6,8) lies on the circle and the tangent line as well.

Hence (-6,8) satisfies the line equation:

8 = m(-6) + c ⇒ c-6m = 8 -------------1

We know that slope of two perpendicular lines are related as:

$m_{1}\times m_{2}=-1$

At any point on the circle, the normal line at a point is always perpendicular to the tangent line at that point.

Hence :

$m_{normal} \times m_{tangent}=-1$

We can find the slope of the normal at point (-6,8) as it passes through the centre of the circle (-6,4) by using the two-points formula for slope.

$m=\frac{y_2-y_1}{x_2-x_1}$

$=\frac{8-4}{-6+6}$

= ∞

Slope of the normal is infinity and hence slope of tangent is -1/∞ = 0

Hence m=0

Putting m=0 in equation 1 we get:

c = 8

The equation of tangent line at (-6,8) is:

y = 8

6 0

3 years ago