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2 years ago

The number p and 2/3 are additive inverses.

Mathematics

1 answer:

Tpy6a [65]2 years ago

4 0

Answer:

The label “Sum” is located at 0

The label “p” is located between 0 and -1 (-2/3)

The label “2/3” is located between 0 and 1

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When f(x) = 7x^2 - 2x + 4, evaluate f(-1) f(-1) =

Blizzard [7]

Answer:

Step-by-step explanation:

do u want me to elaborate on how I found it>

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2 years ago

∆PQR has vertices P(5, –2), Q(1, –4), and R(3,3).Reflect ∆PQR about the y-axis, translate 4 units down, and reflect about the x-

JulsSmile [24]

Answer:

P(-5,6) Q(-1,8) R(-3,1)

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3 years ago

Mark borrow $200 at 12% compound interest for 2 years if he makes no payments how much interest will he owe at the end of the se

Arlecino [84]

B (if annual interest),
200+12%=224
224+12%=250.88

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%20%5Cint%20t%5E2%2B1%20%5C%20dt" id="TexFormula1" title="\frac{d}{dx} \

Kisachek [45]

Answer:

$\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2} t^2+1 \ \text{dt} \ = \ 2x^5-8x^2+2x-2$

Step-by-step explanation:

$\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2} t^2+1 \ \text{dt} = \ ?$

We can use Part I of the Fundamental Theorem of Calculus:

$\displaystyle\frac{d}{dx} \int\limits^x_a \text{f(t) dt = f(x)}$

Since we have two functions as the limits of integration, we can use one of the properties of integrals; the additivity rule.

The Additivity Rule for Integrals states that:

$\displaystyle\int\limits^b_a \text{f(t) dt} + \int\limits^c_b \text{f(t) dt} = \int\limits^c_a \text{f(t) dt}$

We can use this backward and break the integral into two parts. We can use any number for "b", but I will use 0 since it tends to make calculations simpler.

$\displaystyle \frac{d}{dx} \int\limits^0_{2x} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}$

We want the variable to be the top limit of integration, so we can use the Order of Integration Rule to rewrite this.

The Order of Integration Rule states that:

$\displaystyle\int\limits^b_a \text{f(t) dt}\ = -\int\limits^a_b \text{f(t) dt}$

We can use this rule to our advantage by flipping the limits of integration on the first integral and adding a negative sign.

$\displaystyle \frac{d}{dx} -\int\limits^{2x}_{0} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}$

Now we can take the derivative of the integrals by using the Fundamental Theorem of Calculus.

When taking the derivative of an integral, we can follow this notation:

$\displaystyle \frac{d}{dx} \int\limits^u_a \text{f(t) dt} = \text{f(u)} \cdot \frac{d}{dx} [u]$
where u represents any function other than a variable

For the first term, replace $\text{t}$ with $2x$ , and apply the chain rule to the function. Do the same for the second term; replace

$\displaystyle-[(2x)^2+1] \cdot (2) \ + \ [(x^2)^2 + 1] \cdot (2x)$

Simplify the expression by distributing $2$ and $2x$ inside their respective parentheses.

$[-(8x^2 +2)] + (2x^5 + 2x)$
$-8x^2 -2 + 2x^5 + 2x$

Rearrange the terms to be in order from the highest degree to the lowest degree.

$\displaystyle2x^5-8x^2+2x-2$

This is the derivative of the given integral, and thus the solution to the problem.

6 0

2 years ago

Mary looks over reports on four of her workers. jack made 30 baskets in 5 hours. rudy made 32 baskets in 8 hours. sam made 40 ba

Elena L [17]

Well let's do some maths if each made a certain amount of baskets an hour you could divide them and figure out how many baskets were made per hour.
jack = 30/5 = 6 baskets per hour
Rudy = 32/8 = 4 baskets per hour
Sam = 40/12 =3.333 baskets per hour
Walter = 22/4 = 5.5 baskets per hour
according to this Jack had the greatest productivity

6 0

3 years ago