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3 years ago

Abcd is a square and eghd is a rectangle .Prove that area (abcd)=area (eghd)

2 answers:

8 0

It is not possible to prove this. As D is common between both figures, in the rectangle, we can only infer that two sides are the same as that of the square. We need to prove that the other two sides are the same as the side of the square for both areas to be equal.

Ivahew [28]3 years ago

6 0

They have in common only one summit. you miss something because you can't prove that with only one summit in common

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How many rational numbers are between 1 and 6?

gregori [183]

one ans six are infinite

6 0

3 years ago

(The photo is too big to fit so don't mind that.)

mafiozo [28]

Answer:

Step-by-step explanation:

1'st step: $a^{2}$ + $b^{2}$ = $c^{2}$ is the formula for the pythagorean theorem

2'nd step : $18^{2}$ + $80^{2}$ = $c^{2}$

3'rd step: 324+6400 = $c^{2}$

4'rd step: 6724= $c^{2}$

5'th step: $\sqrt{6724}$ =82

6'th step : therefore the value of c is 82 cm

8 0

3 years ago

Is 690,269 divisible by 9? Yes or No?

jonny [76]

Answer:

False.

Step-by-step explanation:

It is false because when dividing an odd number by an odd number, it usually isn't divisible. In this case, 690,26<u>9</u>. Is an odd number, because the last number, 9, is odd, which makes the whole thing odd. So, odd divided by odd never turns out well. (By the way the answer is 76,696.5555556).

8 0

2 years ago

Find the surface area of the sphere

Leokris [45]

Answer:7857.14cm square

Step-by-step explanation:

surface area of a sphere =4*22/7*25*25=7857.14cm square

8 0

3 years ago

In engineering, the modulus of elasticity is a way to measure how much an object deforms along an axis when opposing forces are

Allushta [10]

Step-by-step explanation:

Consider an engineering material of initial length Lo, Area (A), Modulus of elasticity (E) and applied a force P due to which change in the length of the material is δ2 from it’s original length (Lo)

Initial length of the material is Lo. Hence, at time t = 0 when no force applied on the material the length of the material will not change (i.e., at time t=0, δ1 = 0)

Modulus of elasticity of the material:

$E=\frac{P \cdot L_{o}}{A\left[\delta_{2}-\delta_{1}\right]}$

Area of the material:

$E=\frac{P \cdot L_{o}}{A\left[\delta_{2}-\delta_{1}\right]}$

$A=\frac{P \cdot L_{o}}{E\left[\delta_{2}-\delta_{1}\right]}$

Length of the material:

$E=\frac{P \cdot L_{0}}{A\left[\delta_{2}-\delta_{1}\right]}$

$L_{0}=\frac{E \cdot A\left[\delta_{2}-\delta_{1}\right]}{P}$

5 0

3 years ago