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3 years ago

HALPPPPP PLZZZ I'll give Brainliest for the first answer CORRECT

Mathematics

1 answer:

ddd [48]3 years ago

7 0

Answer:

Since it is area, it is most defined as base times height, so the larger rectangle would be 9x14 which is 126 and the smaller rectangle is 6x8 which is 48, and since they are together in one shape, you can add both areas which would give you an answer of 174

Step-by-step explanation:

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PLSS ANSWER MY QUESTION

Taya2010 [7]

Answer:

1) Area of frame: $140cm^2$ ; Area of picture: $40cm^2$ ; Total Area: $180cm^2$ .

2) Area 1 = $48cm^2$ ; Area 2 = $12cm^2$ ; total area = $60cm^2$ [first answer line : $60cm^2$ ]

Step-by-step explanation:

1. So to answer this question, you have to find the inner photo area first (which you will then subtract from the overall rectangle area to get the photo frame area).

To find area, you multiply length x width. For this question, there are not enough measurements given to initially multiply length and width. So, assuming that the photo is centered on the frame and the frame width is consistent along all sides, you find the difference between the overall length (12) and the picture length (5).

12 - 5 = 7

The leftover area (outside of the picture) is 7 cm (half of seven on both sides), so one side can be assumed to be : 3.5 cm

this means that 15 cm - 3.5 cm of both sides lengths (difference between the two lengths given) is the height of the photo, 15 - 3.5(2) = 8

So, the dimensions/measurements of the photo are 5 cm by 8 cm.

This means that the inner photo area is 40 (5 x 8) cm

The area of the frame will be the area of the total rectangle [12 x 15] - the inner photo area [40]. [ (12)(15) - 40 ]

[180 - 40] = 140 cm

(area is measured in squared units, so the area of the frame should be written as $140cm^{2}$ ; and the area of the photo should be written as $40cm^{2}$ )

The total area is the entire area of the rectangle: (12)(15) = 180. (This should be written as $180cm^2$ )

--------------------------------------

2) Finding the area of a shape that isn't a simple rectangle is easier if you separate the shape into smaller rectangles. I will separate the two rectangles into 4 x 12 and 2 x 6 [6 cm would be the remaining side length if you remove the length of 4].

{you can separate the shape in multiple ways, so my answer is only one answer to this problem out of other splits.}

as mentioned, the area is length x width.

4 x 12 = 48.

2 x 6 = 12. Once again, you should write area in squared units.

(the area 1 should be written as $48cm^{2}$ ; the area 2 should be written as $12cm^2$ )

you can combine the separated areas (add them) to find the total area. 48 + 12 = 60.

(the total area should be written in squared units also, so the total area should be written as: $60cm^2$ )

The area of the figure is $60cm^2$

8 0

2 years ago

Which of the following is a solution of x2 − 8x = –27?

Sergeu [11.5K]

Its 4 plus i square root of eleven

4 0

2 years ago

Read 2 more answers

What happened to rectangle A to become rectangle B in the diagram below?

Bumek [7]

Answer: Dilated and then translated

Step-by-step explanation:

8 0

1 year ago

Solve 73 make sure to also define the limits in the parts a and b

Aleks04 [339]

73.

$f(x)=\frac{3x^4+3x^3-36x^2}{x^4-25x^2+144}$

$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3$ $\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3$

Since we can't divide by zero, we need to find when:

$x^4-2x^2+144=0$

But before, we can factor the numerator and the denominator:

$\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}$

Now, we can conclude that the vertical asymptotes are located at:

$\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}$

so, for x = -3:

$\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty$ $\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty$

For x = 4:

$\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$ $\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$

4 0

8 months ago

Please help ASAP!!!!!!!!!!

3241004551 [841]

Answer:

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Step-by-step explanationbfd:

5 0

2 years ago