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2 years ago

Please select the best answer from the choices providedABCD

Mathematics

1 answer:

Shalnov [3]2 years ago

3 0

Answer:

Option A

Step-by-step explanation:

$A=\begin{bmatrix}-3 & -9 & -3\\ -4 & -3 & 9\\ 7 & 9 & -10\end{bmatrix}$

$-7A=\begin{bmatrix}21 & 63 & 21\\ 28 & 21 & -63\\ -49 & -63 & 70\end{bmatrix}$

$B=\begin{bmatrix}9 & 1 & 6\\ 3 & -10 & 5\\ -1 & 9 & -9\end{bmatrix}$

$-4B=\begin{bmatrix}-36 & -4 & -24\\ -12 & 40 & -20\\ 4 & -36 & 36\end{bmatrix}$

$-7A-4B=\begin{bmatrix}-15 & 59 & -3\\ 16 & 61 & -83\\ -45 & -99 & 106\end{bmatrix}$

Option A is the answer.

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A florist gathered data about the weekly number of flower deliveries he made to homes and to businesses for several weeks. He us

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The number of deliveries that are predicted to be made to homes during a week with 50 deliveries to business is 87 deliveries

Step-by-step explanation:

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The number of home deliveries = x

The number of delivery to businesses = y

The line of best fit is y = 0.555·x + 1.629

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We substitute y = 50 in the line of best fit to get;

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2 years ago

Two cross sections of a right hexagonal pyramid are obtained by cutting the pyramid with planes parallel to the hexagonal base.

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Answer:

The larger cross section is 24 meters away from the apex.

Step-by-step explanation:

The cross section of a right hexagonal pyramid is a hexagon; therefore, let us first get some things clear about a hexagon.

The length of the side of the hexagon is equal to the radius of the circle that inscribes it.

The area is

$A=\frac{3\sqrt{3} }{2} r^2$

Where $r$ is the radius of the inscribing circle (or the length of side of the hexagon).

Now we are given the areas of the two cross sections of the right hexagonal pyramid: $A_1=216\:ft^2\: \:\:\:A_2=486\:ft^2$

From these areas we find the radius of the hexagons:

$r_1=\sqrt{\frac{2A_1}{3\sqrt{3} } } =\sqrt{\frac{2*216}{3\sqrt{3} } }=\boxed{9.12ft}$

$r_2=\sqrt{\frac{2A_2}{3\sqrt{3} } } =\sqrt{\frac{2*486}{3\sqrt{3} } }=\boxed{13.68ft}$

Now when we look at the right hexagonal pyramid from the sides ( as shown in the figure attached ), we see that $r_1$ $r_2$ form similar triangles with length $H$

Therefore we have:

$\frac{H-8}{r_1} =\frac{H}{r_2}$

We put in the numerical values of $r_1$ , $r_2$ and solve for $H$ :

$\boxed{H=\frac{8r_2}{r_2-r_1} =\frac{8*13.677}{13.68-9.12} =24\:feet.}$

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2 years ago

You deposit $300 in a savings account that pays 6% interest compounded semiannually. How much will you have at the middle of the

Makovka662 [10]

Answer:

The total amount accrued, principal plus interest, from compound interest on an original principal of $ 300.00 at a rate of 6% per year compounded 2 times per year over 0.5 years is $ 309.00.

The total amount accrued, principal plus interest, from compound interest on an original principal of $ 300.00 at a rate of 6% per year compounded 2 times per year over 1 year is $ 318.27.

Step-by-step explanation:

a) How much will you have at the middle of the first year?

Using the formula

$A\:=\:P\left(1+\frac{r}{n}\right)^{nt}$

where

Principle = P

Annual rate = r

Compound = n

Time = (t in years)

A = Total amount

Given:

Principle P = $300

Annual rate r = 6% = 0.06 per year

Compound n = Semi-Annually = 2

Time (t in years) = 0.5 years

To determine:

Total amount = A = ?

Using the formula

$A\:=\:P\left(1+\frac{r}{n}\right)^{nt}$

substituting the values

$A=300\left(1+\frac{0.06}{2}\right)^{\left(2\right)\left(0.5\right)}$

$A=300\cdot \frac{2.06}{2}$

$A=\frac{618}{2}$

$A=309$ $

Therefore, the total amount accrued, principal plus interest, from compound interest on an original principal of $ 300.00 at a rate of 6% per year compounded 2 times per year over 0.5 years is $ 309.00.

Part b) How much at the end of one year?

Using the formula

$A\:=\:P\left(1+\frac{r}{n}\right)^{nt}$

where

Principle = P

Annual rate = r

Compound = n

Time = (t in years)

A = Total amount

Given:

Principle P = $300

Annual rate r = 6% = 0.06 per year

Compound n = Semi-Annually = 2

Time (t in years) = 1 years

To determine:

Total amount = A = ?

so using the formula

$A\:=\:P\left(1+\frac{r}{n}\right)^{nt}$

so substituting the values

$A\:=\:300\left(1+\frac{0.06}{2}\right)^{\left(2\right)\left(1\right)}$

$A=300\cdot \frac{2.06^2}{2^2}$

$A=318.27$ $

Therefore, the total amount accrued, principal plus interest, from compound interest on an original principal of $ 300.00 at a rate of 6% per year compounded 2 times per year over 1 year is $ 318.27.

3 0

2 years ago

Please select the best answer from the choices providedABCD​

Please select the best answer from the choices providedABCD