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sukhopar [10]
2 years ago
10

The design for a mural is 16 in. Wide and 9 in. High. What are the dimensions of the largest possible complete mural that can be

painted on a wall 24 ft wide by 14 ft high?
Mathematics
1 answer:
Alexandra [31]2 years ago
6 0

Answer:

Largest possible mural that can be painted on the wall is 24 feet wide and 13.5 feet high.

Step-by-step explanation:

Design of a mural is 16 in. wide and 9 in. high.

Dimensions of the wall are 24 ft by 14 ft.

If we enlarge the size of the design along the width the wall, scale factor to be used,

Scale factor = \frac{\text{Width of the wall}}{\text{Width of the design}}

                    = \frac{24}{16}

                    = \frac{3}{2}

With the same scale factor, height of the design will be,

\frac{3}{2}=\frac{\text{Height of the wall used}}{\text{Height of the design}}

\frac{3}{2}=\frac{\text{Height of the wall used}}{9}

Width of the wall used = \frac{3\times 9}{2}

                                     = 13.5 ft

Therefore, largest possible mural that can be painted on the wall is 24 feet wide and 13.5 feet high.  

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The distance of points P and B is 5
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Need help with me question
Irina-Kira [14]

Answer:

x = -6       x=5

Step-by-step explanation:

x^2 +x -30 =0

Factor

What two numbers multiply to -30 and add to 1

6 * -5 = -30

6-5 =1

(x+6) (x-5) =0

Using the zero product property

x+6 =0    x-5=0

x+6-6=0-6    x-5+5=0+5

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8 0
3 years ago
PLS HELP ME!!! The figure is made up of a cone and a hemisphere. To the nearest whole number, what is the approximate volume of
Pachacha [2.7K]

Hello!

The figure is made up of a cone and a hemisphere. To the nearest whole number, what is the approximate volume of this figure? Use 3.14 to approximate π . Enter your answer in the box. cm³

Data: (Cone)

h (height) = 12 cm

r (radius) = 4 cm (The diameter is 8 being twice the radius)

Adopting: \pi \approx 3.14

V (volume) = ?

Solving: (Cone volume)

V = \dfrac{ \pi *r^2*h}{3}

V = \dfrac{ 3.14 *4^2*\diagup\!\!\!\!\!12^4}{\diagup\!\!\!\!3}

V = 3.14*16*4

\boxed{V = 200.96\:cm^3}

Note: Now, let's find the volume of a hemisphere.

Data: (hemisphere volume)

V (volume) = ?

r (radius) = 4 cm

Adopting: \pi \approx 3.14

If: We know that the volume of a sphere is V = 4* \pi * \dfrac{r^3}{3} , but we have a hemisphere, so the formula will be half the volume of the hemisphere V = \dfrac{1}{2}* 4* \pi * \dfrac{r^3}{3} \to \boxed{V = 2* \pi * \dfrac{r^3}{3}}

Formula: (Volume of the hemisphere)

V = 2* \pi * \dfrac{r^3}{3}

Solving:

V = 2* \pi * \dfrac{r^3}{3}

V = 2*3.14 * \dfrac{4^3}{3}

V = 2*3.14 * \dfrac{64}{3}

V = \dfrac{401.92}{3}

\boxed{ V_{hemisphere} \approx 133.97\:cm^3}

What is the approximate volume of this figure?

Now, to find the total volume of the figure, add the values: (cone volume + hemisphere volume)

Volume of the figure = cone volume + hemisphere volume

Volume of the figure = 200.96 cm³ + 133.97 cm³

\boxed{\boxed{\boxed{V = 334.93\:cm^3 \to Volume\:of\:the\:figure \approx 335\:cm^3 }}}\end{array}}\qquad\quad\checkmark

Answer:

The volume of the figure is approximately 335 cm³

_______________________

I Hope this helps, greetings ... Dexteright02! =)

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Step-by-step explanation:

Complementary Angles add up to 90°

Supplementary Angles add up to 180°

60 + 30 = 90

8 0
3 years ago
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