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Igoryamba
3 years ago
14

CORRECT ANSWER GETS BRAINIEST! THIS IS URGENT!!

Mathematics
2 answers:
shusha [124]3 years ago
8 0

Answer:

11.08

Step-by-step explanation:

dexar [7]3 years ago
6 0

Answer: x=11.075

Step-by-step explanation:

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Again stuck why because our teacher got fired so we are stuck with substitute teacher
Anna11 [10]

Answer:

27 > b

Step-by-step explanation:

7 > -20 + b

7 + 20 > -20 + 20 + b

27 > b

4 0
4 years ago
Please help I'm so confused! I need the surface area. Thank you
kondaur [170]

Answer:

The answer is : 23.25 in ²

Step-by-step explanation:

There are 3 triangles with the same lengths: height = 5in and base = 5 in

There is one triangle: height = 4.3 in base = 5 in

Formula

Area = height x base / 2

For the triangles

Area = 5 in x 5 in /2 = 25 /2 = 12.5 in²

For the triangle

Area = 4.3 x 5 /2 = 21.5 /2 = 10.75 in²

Total surface area = 12.5 + 10.75 = 23.25 in²

3 0
4 years ago
Some rectangles are not parallelograms
VARVARA [1.3K]

Answer:

I'm not sure what your asking, but, no, all rectangles are parallelograms.

I found this over the internet, and I hope it helps you understand why a rectangle is always a parallelogram, but a parallelogram is not always a rectangle:

It is true that every rectangle is a parallelogram, but it is not true that every parallelogram is not a rectangle. For instance, take a square. It's a parallelogram — it is a quadrilateral with two pairs of parallel faces. But it is also a rectangle — it is a quadrilateral with four right angles.

5 0
4 years ago
Anybody help me to solve this question. ​
Mumz [18]

Answer:

\dfrac{1}{(b-c)},\dfrac{1}{(c-a)} ,\dfrac{1}{(a-b)} are\ in\ AP

Step-by-step explanation:

Given that (b-c)^2, (c-a)^2 , (a-b)^2 are in AP

To prove: \dfrac{1}{(b-c)},\dfrac{1}{(c-a)} ,\dfrac{1}{(a-b)} are in AP

From given as we know if p , q, r are in AP then 2q= p+r.

2(c-a)^2= (b-c)^2+(a-b)^2\\\\\Rightarrow 2(c^2+a^2-2ac)=b^2+c^2-2bc+a^2+b^2-2ab\\\\\Rightarrow 2c^2+2a^2-4ac= 2b^2+c^2+a^2 -2bc-2ab\\\\\Rightarrow a^2+c^2-2b^2-4ac= -2bc-2ab\\\\\Rightarrow a^2-2b^2+c^2= 4ac-2bc-2ab

Now

\dfrac{1}{(b-c)},\dfrac{1}{(c-a)} ,\dfrac{1}{(a-b)}2\dfrac{1}{(c-a)} =\dfrac{1}{(b-c)}+\dfrac{1}{(a-b)}\\\\\Rightarrow \dfrac{2}{(c-a)}= \dfrac{a-b+b-c}{(b-c)(a-b)} \\\\\Rightarrow \dfrac{2}{(c-a)}= \dfrac{a-c}{(b-c)(a-b)} \\\\\Rightarrow2(b-c)(a-b) = (c-a)(a-c) \\\\\Rightarrow 2(ab-b^2-ac+bc)= -(a-c)^2\\\\\Rightarrow 2ab- 2b^2-2ac+2bc = -a^2-c^2+2ac\\\\\Rightarrow a^2-2b^2+c^2=4ac-2ab-2bc

Which is the result of AP

.

Hence proved

6 0
3 years ago
Hellpppppppppppppppppppppppppppppppppppppppp
natulia [17]

Answer:

._.     -_-     -.-    .-.   --__--

Step-by-step explanation:

8 0
3 years ago
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