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3 years ago

Which is the simplified form of r^-7 + s^-12? arrows to the left of -7 and -12 represent exponents,

Mathematics

2 answers:

polet [3.4K]3 years ago

7 0

$a^{-b}=\dfrac{1}{a^b}\\\\ r^{-7}+s^{-12}=\dfrac{1}{r^7}+\dfrac{1}{s^{12}}\implies \text{D}$

Marrrta [24]3 years ago

3 0

Answer is D, 1/r^7 + 1/s^12. Im positive 100%

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At a local grocery store, the cost of apples varies directly with the weight of the apples. If a 5-pound bag

son4ous [18]

Answer:

$4.72

Step-by-step explanation:

7.87÷5=1.574

1.574×3=4.722

so, $4.72 is your answer.

6 0

3 years ago

How do I solve this ?

Nat2105 [25]

Answer:

Step-by-step explanation:

I'm just going to go out on a limb here and assume that those are a set of poorly made parenthesis. This, then, would be your equation:

$x+\frac{x}{7}+\frac{1}{11}(x+\frac{x}{7})=60$

The first thing to do is distribute through the parenthesis to get:

$x+\frac{x}{7}+\frac{x}{11}+\frac{x}{77}=60$

Next is to get rid of the denominators by multiplying by the LCM of 77:

$77(x+\frac{x}{7}+\frac{x}{11}+\frac{x}{77}=60)$ which gives you:

77x + 11x + 7x + x = 4620 and

96x = 4620 so

$x=\frac{385}{8}$

6 0

3 years ago

Need help #3. The answer is shown, but I don’t know how to get to the answer. Please teach and show steps.

Sunny_sXe [5.5K]

Answer:

Step-by-step explanation:

We are given a right triangle with a base of x feet and a height of h feet, where x is constant and h changes with respect to time t.

The angle in radians is defined by:

$\displaystyle \tan(\theta)=\frac{h}{x}$

And we want to find the relationship that describes dθ/dt and dh/dt.

So, we will differentiate both sides with respect to t where x is a constant:

$\displaystyle \frac{d}{dt}[\tan(\theta)]=\frac{d}{dt}\Big[\frac{h}{x}\Big]$

Differentiate. Apply the chain rule on the left. Again, remember that x is just a constant, so we can move it outside the derivative operator. Therefore:

$\displaystyle \sec^2(\theta)\frac{d\theta}{dt}=\frac{1}{x}\frac{dh}{dt}$

Since we know that tan(θ)=h/x, h is the opposite side of our triangle and x is the adjacent. Therefore, by the Pythagorean Theorem, our hypotenuse will be:

$\text{Hypotenuse}=\sqrt{h^2+x^2}$

Since secant is the ratio of the hypotenuse to adjacent:

$\displaystyle \sec(\theta)=\frac{\sqrt{h^2+x^2}}{x}$

So:

$\displaystyle \sec^2(\theta)=\frac{x^2+h^2}{x^2}$

By substitution, we have:

$\displaystyle \Big(\frac{x^2+h^2}{x^2}\Big)\frac{d\theta}{dt}=\frac{1}{x}\frac{dh}{dt}$

By multiplying both sides by the reciprocal of the term on the left:

$\displaystyle \frac{d\theta}{dt}=\frac{1}{x}\Big(\frac{x^2}{x^2+h^2}\Big)\frac{dh}{dt}$

Therefore:

$\displaystyle \frac{d\theta}{dt}=\frac{x}{x^2+h^2}\frac{dh}{dt}$

Our answer is A.

3 0

3 years ago

a construction company charges $15 per hour for debris removal , plus a one -time fee for the use of the trash dumpster.the tota

ser-zykov [4K]

Let h be the numbers of hours, f be the one-time fee and c the cost charged. The equation is

$c=15h+f$

Since you pay $15 per four, plus the fee. We can solve this equation for the fee:

$f=c-15h$

If we plug c=195 and h=9, we have

$f=195-15\cdot 9=195-135=60$

6 0

3 years ago

Change this radical to an algebraic expression with fractional exponents. √y^2 The exponent on y is:

Wewaii [24]

The power rule (or law of indices) related to root is given
$x^{ \frac{1}{2} }= \sqrt[2]{x}$

We have $\sqrt{ y^{2} }$
Rewrite according to the rule we have $y^{ \frac{2}{2} } = y^{1}$

The exponent on y is '1'

8 0

3 years ago

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