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3 years ago

What's 5x10+15-40(30+20)+40+10+24

Mathematics

1 answer:

Fudgin [204]3 years ago

4 0

5x10+15-40(50)+40+10+24
50+15-40(50)+40+10+24
65-40=25+50=75+40=110+10=120+24=144

144 is your answer

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balu736 [363]

Answer:219

Step-by-step explanation:

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3 years ago

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What two numbers add together to make -11

Paladinen [302]

Answer:

-6 and -5 (There are many more combinations of numbers that can equal to -11 but this is just one of the many possibilities.

Step-by-step explanation:

-6 and -5

See you can write it like this,

-6+ (-5)

= -6-5

=-11

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3 years ago

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adiocarbon dating of blackened grains from the site of ancient Jericho provides a date of 1315 BC ± 13 years for the fall of the

Zigmanuir [339]

Answer:

$\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659$ and $\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661$

Step-by-step explanation:

The equation of the isotope decay is:

$\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }$

14-Carbon has a half-life of 5568 years, the time constant of the isotope is:

$\tau = \frac{5568\,years}{\ln 2}$

$\tau \approx 8032.926\,years$

The decay time is:

$t = 1315\,years + 2007\,years \pm 13\,years$ (There is no a year 0 in chronology).

$t = 3335 \pm 13\,years$

Lastly, the relative amount is estimated by direct substitution:

$\frac{m(t)}{m_{o}} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\mp\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{min} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{-\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659$

$\left(\frac{m(t)}{m_{o}} \right)_{max} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\frac{13\,years}{8032.926\,years} }$

$\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661$

4 0

3 years ago

Jo borrowed 3,800 for 4 months from a bank. the bank discounted the loan at 5.5%.

garri49 [273]

A.
I=prt
I=(3800)(.055)(4/12)
I=$69.67

b.
Principal-interest=money received from Bank
3800-69.67=$3730.34

c.
I=prt
69.67=3730.34(r)(4/12)
69.67=1243.45r
69.67/1243.45=1243.45/1243.45r
.056≈r
Rate is 5.60%

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3 years ago

How to solve derivative of (sin3x)/x using first principle

Leona [35]

$\dfrac{d}{dx}(\dfrac{\sin(3x)}{x})$

First we must apply the Quotient rule that states,

$(\dfrac{f}{g})'=\dfrac{f'g-g'f}{g^2}$

This means that our derivative becomes,

$\dfrac{\dfrac{d}{dx}(\sin(3x))x-\dfrac{d}{dx}(x)\sin(3x)}{x^2}$

Now we need to calculate $\dfrac{d}{dx}(\sin(3x))$ and $\dfrac{d}{dx}(x)$

$\dfrac{d}{dx}(\sin(3x))=\cos(3x)\cdot3$

$\dfrac{d}{dx}(x)=1$

From here the new equation looks like,

$\dfrac{3x\cos(3x)-\sin(3x)}{x^2}$

And that is the final result.

Hope this helps.

r3t40

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3 years ago

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