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Brums [2.3K]
3 years ago
9

PLEASE HELP! WILL GIVE 70 POINTS!!

Mathematics
1 answer:
Tresset [83]3 years ago
4 0

Answer:

A=189\ mm^2

Step-by-step explanation:

<u>Surface Areas </u>

Is the sum of all the lateral areas of a given solid. We need to compute the total surface area of the given prism. It has 5 sides, two of them are equal (top and bottom areas) and the rest are rectangles.

Computing the top and bottom areas. They form a right triangle whose legs are 4.5 mm and 6 mm. The area of both triangles is

\displaystyle A_t=2*\frac{b.h}{2}=b.h=(4.5)(6)=27 mm^2

The front area is a rectangle of dimensions 7.7 mm and 9 mm, thus

A_f=b.h=(7.5)(9)=67.5 \ mm^2

The back left area is another rectangle of 4.5 mm by 9 mm

A_l=b.h=(4.5)(9)=40.5  \ mm^2

Finally, the back right area is a rectangle of 6 mm by 9 mm

A_r=b.h=(6)(9)=54 \ mm^2

Thus, the total surface area of the prism is

A=A_t+A_f+A_l+A_r=27+67.5+40.5+54=189\ mm^2

\boxed{A=189\ mm^2}

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Westkost [7]

Answer:

Margot gets from A to B faster

Step-by-step explanation:

Alan takes 2 hr and 35 min = 2. 58... hr

Margot takes 147/64 hr = 2.296.. hr

7 0
3 years ago
What are the coordinates of the x-intercepts of the parabola y = x² - 8x + 15?
tamaranim1 [39]

Answer:

(3, 0) and (5, 0)

Step-by-step explanation:

we have

y=x^{2}-8x+15

we know that

The x-intercepts are the values of x when the value of y is equal to zero

so

For y=0

x^{2}-8x+15=0

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}

in this problem we have

x^{2}-8x+15=0  

so

a=1\\b=-8\\c=15

substitute in the formula

x=\frac{-(-8)\pm\sqrt{-8^{2}-4(1)(15)}} {2(1)}

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x=\frac{8\pm2} {2}

x=\frac{8+2} {2}=5

x=\frac{8-2} {2}=3

so

x=3, x=5

therefore

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4 0
3 years ago
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UkoKoshka [18]
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Paul [167]

Answer: 18

Step-by-step explanation:

2 degrees per hour

9 hours

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3hr= 6 degrees

4hr= 8 degrees

5hr=10 degrees

6hr= 12 degrees

7hr= 14 degrees

8hr= 16 degrees

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So you are basically doing 2x9 to get your answer

4 0
3 years ago
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