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3 years ago

Sz – [32 – (10z + 2)]

Mathematics

1 answer:

Dmitry_Shevchenko [17]3 years ago

5 0

Sz－[32－(10z＋2)]
＝Sz－(32－10z－2)
＝Sz－(30－10z)
＝Sz－30＋10z

Hope my answer helped u :)

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Artyom0805 [142]

Answer:

The total cost of producing 6 widgets is $231.

Step-by-step explanation:

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3 years ago

Mr. Juarez opened a savings account with an initial deposit of $560 and will not make any additional deposits or withdrawals. Th

olga_2 [115]

Answer:

$576.80

Step-by-step explanation:

We have been given that Mr. Juárez opened a savings account with an initial deposit of $560 and will not make any additional deposits or withdrawals. The account earns 1% simple interest.

We are asked to find the total amount that Mr. Juárez will have in his account at the end of 3 years.

We will use simple interest formula to solve our given problem.

$A=P (1+rt)$

A = Final amount after t years,

P = Principal amount,

r = Annual interest rate in decimal form,

t = Time in years.

Let us convert 1% into decimal form,

1%=1/100=0.01

P=$560 and t=3

A=$560 (1+0.01(3))

A=$560 (1+0.03)

A= $560 (1.03)

A= $576.80

Therefore, Mr. Juárez will have $576. 80 in his account at the end of 3 years. Hope this helps!

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3 years ago

Can someone please explain this to me? Thanks!

Makovka662 [10]

Answer: Choice D

$\displaystyle F\ '(x) = 2x\sqrt{1+x^6}\\\\$

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Explanation:

Let g(t) be the antiderivative of $g'(t) = \sqrt{1+t^3}$ . We don't need to find out what g(t) is exactly.

Recall by the fundamental theorem of calculus, we can say the following:

$\displaystyle \int_{a}^{b} g'(t)dt = g(b)-g(a)$

This theorem ties together the concepts of integrals and derivatives to show that they are basically inverse operations (more or less).

So,

$\displaystyle F(x) = \int_{\pi}^{x^2}\sqrt{1+t^3}dt\\\\ \displaystyle F(x) = \int_{\pi}^{x^2}g'(t)dt\\\\ \displaystyle F(x) = g(x^2) - g(\pi)\\\\$

From here, we apply the derivative with respect to x to both sides. Note that the $g(\pi)$ portion is a constant, so $g'(\pi) = 0$

$\displaystyle F(x) = g(x^2) - g(\pi)\\\\ \displaystyle F \ '(x) = \frac{d}{dx}[g(x^2)-g(\pi)]\\\\\displaystyle F\ '(x) = \frac{d}{dx}[g(x^2)] - \frac{d}{dx}[g(\pi)]\\\\ \displaystyle F\ '(x) = \frac{d}{dx}[x^2]*g'(x^2) - g'(\pi) \ \text{ .... chain rule}\\\\$

$\displaystyle F\ '(x) = 2x*g'(x^2) - 0\\\\ \displaystyle F\ '(x) = 2x*g'(x^2)\\\\ \displaystyle F\ '(x) = 2x\sqrt{1+(x^2)^3}\\\\ \displaystyle F\ '(x) = \boldsymbol{2x\sqrt{1+x^6}}\\\\$