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kipiarov [429]
2 years ago
8

the school building was 98 feet long and 32 feet wide the last room in the building was an office and was 12 feet long what was

the total distance around the building if you measured twice
Mathematics
1 answer:
olga55 [171]2 years ago
8 0

Answer:the height of the building is around 122 feet the distance would be about 300.2 feet

       Please mark me brainlest

Step-by-step explanation:

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In April, 287 people visited an amusement park. In May, 379 people visited the same amusement park. In June, twice as many peopl
Mariulka [41]

Answer:

1332 i think

Step-by-step explanation:

287+379=666

666*2=1332

3 0
3 years ago
The numbers on two consecutively numbered gym lockers have a sum of 141. What are the locker​ numbers?
lesantik [10]

Answer:

70 and 71

Step-by-step explanation:

Consecutive means in a row

x is the first number

x+1 is the next number

x+ x+1 = 141

Combine like terms

2x+1 =141

Subtract 1 from each side

2x+1-1=141-1

2x= 140

Divide each side by 2

2x/2 =140/2

x = 70

The two lockers are 70,71

5 0
2 years ago
Read 2 more answers
How do you find the area of the shaded region
Amiraneli [1.4K]

well, first off, let's notice that we have a trapezoid with a rectangle inside it, so the rectangle is really "using up" area that the trapezoid already has.

now, if we just get the area of the trapezoid, and then the area of the rectangle alone, and then subtract that area of the rectangle, the rectangle will in effect be making a hole inside the trapezoid's area, and what's leftover, is the shaded section, that part the hole is not touching.

\bf \textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h&=height\\ a,b&=\stackrel{parallel~sides}{bases} \\\cline{1-2} h&=8\\ a&=16\\ b&=8 \end{cases}\implies A=\cfrac{8(16+8)}{2}\implies A=96 \\\\\\ \stackrel{\textit{area of the rectangle}}{(5\cdot 3)\implies 15}~\hfill \stackrel{~\hfill \textit{shaded area}}{\stackrel{\textit{trapezoid}}{96}~~ - ~~\stackrel{\textit{rectangle}}{15}~~ = ~~81}

3 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Cfrac%7B-2x%5E%7B2%7D%20%7D%7B-4x%5E%7B2%7D%20%7D" id="TexFormula1" title="\frac{-2x^{2} }{-
kakasveta [241]
Are you looking to simplify it? If so I think the answer would be 1/2.
7 0
2 years ago
What are the dimensions of a rectangle with an area of 12 square cm and a perimeter of 14 cm?
kow [346]

Answer:

For the rectangle

Length=ℓ

Breadth=b

Area is 12 cm2

ℓb=12

Perimeter is 16 cm

2(ℓ+b)=16

ℓ+b=8

Substitute b=12ℓ from first equation

ℓ+12ℓ=8

ℓ2+12=8ℓ

ℓ2−8ℓ+12=0

Use quadratic formula (x=−b±√b2−4ac2a) to find ℓ

ℓ=−(−8)±√(−8)2−(4×1×12)2×1

ℓ=8±√162

ℓ=8±42

ℓ1=8+42=6

ℓ2=8−42

Step-by-step explanation:

For the rectangle

Length=ℓ

Breadth=b

Area is 12 cm2

ℓb=12

Perimeter is 16 cm

2(ℓ+b)=16

ℓ+b=8

Substitute b=12ℓ from first equation

ℓ+12ℓ=8

ℓ2+12=8ℓ

ℓ2−8ℓ+12=0

Use quadratic formula (x=−b±√b2−4ac2a) to find ℓ

ℓ=−(−8)±√(−8)2−(4×1×12)2×1

ℓ=8±√162

ℓ=8±42

ℓ1=8+42=6

ℓ2=8−42

5 0
3 years ago
Read 2 more answers
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