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2 years ago

Find the surface area of this prism please (no links please)

Mathematics

1 answer:

horsena [70]2 years ago

4 0

Answer:

192 ft^2

Step-by-step explanation:

to find the surface area of a shape, you must find the area of each side and add them together. the base of this shape is a rectangle with a length of 11ft and a height of 3ft; to calculate the area of a rectangle, you multiply the length by the height. 3*11 is 33, so the base of this prism has an area of 33ft^2. since this is a prism, we know that the side parallel to the base are the same shape, so they both have an area of 33ft^2. 33*2= 66. the two other large sides have a height of 4.5 and a length of 11; 4.5*11 is 49.5, there are two shapes with these dimensions, so 49.5*2=99. finally, the two shapes on either end of the prism have a height of 4.5 and a length of 3. 4.5*3= an area of 13.5, two of them, so 13.5*2=27. now you just add all of these numbers together: 66+99+27=192.

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Im bad at money :/ do please help me

Semenov [28]

The third answer

1 quarter, 6 dimes, 1 nickel, 7 pennies

6 0

2 years ago

Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and

Lemur [1.5K]

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates $(x_A,y_A)$

Vectors AB and AB' are perpendicular, then

$\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0$

Vectors AC and AC' are perpendicular, then

$\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0$

Now, solve the system of two equations:

$\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.$

Subtract these two equations:

$5x_A-y_A-8=0\Rightarrow y_A=5x_A-8$

Substitute it into the first equation:

$x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2$

Then

$y_{A_{1,2}}=5\cdot \dfrac{34}{13}-8 \text{ or } 5\cdot 2-8\\ \\=\dfrac{66}{13}\text{ or } 2$

Rotation by 90° counterclockwise about A(2,2) gives image points B' and C' (see attached diagram)

8 0

2 years ago

B= b= \,\,(\sqrt[3]{36}) ^3 ( 3 36 ) 3

Tems11 [23]

Uhm what’s your question

6 0

2 years ago

Assume ABC Company deposits $90,000 with First National Bank in an account earning interest at 6% per annum, compounded semi-ann

maw [93]

Answer:

Step-by-step explanation:

I'm goig to assume that the formula we need here is the following:

$A(t)=P(1+\frac{r}{n})^{(n)(t)$

where A(t) is the amount in the account after the compounding is done, n is the number of times per year the compounding occurs, r is the rate in decimal form, and t is the time in years. Filling in accordingly,

$A(t)=90000(1+\frac{.06}{2})^{(2)(5)}$ and simplifying a bit,