How many small cubes are needed to make the large figure shown below?

If $1600 earned simple interest of $56.24 in 2 months, what was the simple interest rate? The simple interest rate is % (Do not

Fiesta28 [93]

Answer:

$\$21.1$

Step-by-step explanation:

We know that for principal amount P , time period T and rate of interest $R\%$ , simple interest is given by $S.I. = \frac{P\times R\times T}{100}$ .

Here ,

$P=\$1600\\T=2\,\,months=\frac{2}{12}\,\,years=\frac{1}{6}\,\,years\\S.I=\$56.24$

To find : simple interest rate i.e., $R\%$

On putting values of $P\,,\,T\,,\,S.I$ in formula , we get $S.I. = \frac{P\times R\times T}{100}$

$56.24 = \frac{1600\times R\times 1}{600}\\R=\frac{56.24\times 600 }{1600}=\frac{703\times 3}{100}=\$21.09$

Now we need to round off the answer to the nearest tenth .

So, simple interest rate is % = $\$21.09$ = $\$21.1$

8 0

3 years ago

Y = -6x + 3 y = 5x - 8 Question 1 options: (-3, -1) (3, -1) (1, -3) (1, 3)

OleMash [197]

Answer:

-3,-1 so a

Step-by-step explanation:

4 0

3 years ago

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10 POINST!!!<br> What is the area of quadrilateral QRST if QS=18 and RT=24? Show your work.

Allisa [31]

Answer:

area= 1/2 × d1 ×d2

d = diagonal

area= 1/2 ×24×18 =216 unit^2

6 0

3 years ago

6 m

dimulka [17.4K]

Answer:

51.63m^3

Step-by-step explanation:

The figure is made up of a rectangle, triangle and a semi circle

Area of the figure = Area of triangle + rectangle + semicircle

Area of the triangle = 1/2 * base * height

Area of the triangle = 1/2 * 3 * 5

Area of the triangle = 15/2

Area of the triangle = 7.5m^3

Area of the rectangle = Length * Width

Area of the rectangle = 6 * 5

Area of the rectangle = 30m^2

Area of the semicircle = πr²/2

Area of the semicircle = π(3)²/2

Area of the semicircle = 3.14(9)/2

Area of the semicircle = 3.14 * 4.5

Area of the semicircle = 14.13m^2

The area = 30 + 14.13+7.5

Area of the figure = 51.63m^3

7 0

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