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3 years ago

17.) Bonus question ! All or nothing

Mathematics

2 answers:

Slav-nsk [51]3 years ago

7 0

This is the way of solving the equation.

Vaselesa [24]3 years ago

3 0

You take the logarithm of both sides of the equation to remove the variable from the exponent.
x=-3ln(2)+ln(4)/12ln(2)-ln(4)
x≈-0.5

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Answer:

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3 years ago

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Zinaida [17]

Answer:

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Step-by-step explanation:

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3 years ago

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Select all quantities that are proportional to the diagonal length of a square.

horsena [70]

Answer:

B. Perimeter of a square and

C. Side length of a square

Step-by-step explanation:

if n= side length of square then

Area of square is $n^{2}$
Perimeter of a square is 4×n
diagonal length of a square is $\sqrt{2}$ × n

Thus,

Perimeter of square can be expressed as

$2\sqrt{2}$ ×diagonal length of a square

Side length of a square can be expressed as

$\frac{1}{\sqrt{2} }$ ×diagonal length of a square

but Area of square is

$\frac{1}{\sqrt{2} }$ ×n×diagonal length of a square

As a Result, Area of square is also dependent of the value n, wheras in other cases it is a proportion of diagonal length of a square

5 0

3 years ago

SOLVE THIS EXTREMELY IMPORTANT EQUATION

finlep [7]

Answer:

Option B; EF = ( About ) 4.47 units, Perimeter of Δ EFG = ( About ) 12.94 units

Step-by-step explanation:

If we were to consider the height of this triangle EFG, it would be 4 units of length, supposedly splitting base GF into two ≅ parts, each 2 units of length. First let us name the point drawn to base GF ⇒ point H, so that the height of Δ EFG ⇒ EH. Now as EH splits GF into two ≅ parts, by Converse to Coincidence Theorem, Δ EFG ⇒ Isosceles Δ;

EH and FH are legs of a right triangle EFH, so that Pythagorean Theorem can be applied to solve for the length of EF and EG, knowing that as Δ EFG ⇒ Isosceles Δ, EF ≅ EG;

( EH )^2 + ( FH )^2 = ( EF )^2,

( 4 )^2 + ( 2 )^2 = ( EF )^2 ⇒ take square of 4 & 2,

16 + 4 = ( EF )^2 ⇒ combine like terms,

( EF )^2 = 20 ⇒ take square root on either side to solve for EF,

EF = √ 20 = ( About ) 4.47 units = EG,

Perimeter of Δ EFG = EF + GF + EG = 4.47 + 4 + 4.47 = ( About ) 12.94 units,

Solution; Option B

6 0

3 years ago

3. What is the radius of a circle whose area is 64π cm squared?

nataly862011 [7]

Answer:

Area = 64π

Area \: = \pi {r}^{2}Area=πr

64 \: \pi \: = \pi {r}^{2}64π=πr

{r}^{2} = \dfrac{64\pi}{\pi}r

64π

{r}^{2} = 64r

=64

r \: = \: \sqrt{64} = 8r=

Radius = 8 units

Finding the Diameter -

Diameter = Radius x 2 = 8 x 2 = 16 .

\bold{Diameter \: is \: 16 \: units}Diameteris16units

8 0

3 years ago