Pls help me solve this pls

QveST [7]

The answerrr is 18 for thisvprblme

8 0

1 year ago

The figure à shown on ancient coin which was once used in china.the coin is in the shape of a circle of radius 3cm with a square

lianna [129]

Answer:

(1) 2π - x² = 0 (2) x = 2.5 cm (3) perimeter = 10 cm

Step-by-step explanation:

(1)The area of the circular coin without the inner square removed is πr² where r = 3 cm is the radius of the coin. So, the area of the coin without the inner square removed is πr² = π(3 cm)² = 9π cm²

The area of the square of x sides removed from its center is x².

The area A of the each face of the coin is thus A = 9π - x²

Since the area of each face of the coin A = 7π cm²,

then

7π = 9π - x²

9π - 7π - x² = 0

2π - x² = 0

(2) Solve the equation 2π - x² = 0

2π - x² = 0

x² = 2π

x = ±√(2π)

x = ± 2.51 cm

Since x cannot be negative, we take the positive answer.

So, x = 2.51 cm

≅ 2.5 cm

(3) Find the perimeter of the square

The perimeter of the square, p is given by p = 4x

p = 4 × 2.51 cm

= 10.04 cm

≅ 10 cm

8 0

2 years ago

I just this one please someone help

Step2247 [10]

Answer:

100

Step-by-step explanation:

because 4, 3, and 2 all make 180 and 4 and 2 are both 40 so you add that which makes 80 and then you take 180 and subtract 80 which gives you 100

6 0

2 years ago

Read 2 more answers

Graded Assignment Unit Test, Part 2: Basic Geometric Shapes Answer the questions below. When you are finished, submit this assig

AfilCa [17]

Answer:

In case (a) car makes 105° turn.

In case (b) car makes 75° turn.

In case (c) car makes 105° turn.

Step-by-step explanation:

Figure is redrawn To explain properly (in attachment)

Given : streets are parallel means $\overline{AB}$ ║ $\overline{CD}$ ,

AB - 4th street , CD - 3rd street and XY - King Ave.

∠XLA = 75°

To find : (a) ∠XLB

(b) ∠LMD (left onto 3rd streat means left of car)

(c) ∠YMD (right means right side of car)

∠XLB + ∠XLA = 180° (Linear Pair = 2 adjacent angles are

supplementary)

∠XLB + 75° = 180°

∠XLB = 180 - 75

∠XLB = 105°

∴ In case (a) car makes 105° turn.

∠LMD = ∠XLA = 75° (Corresponding angles of parallel lines are equal)

∠LMD = 75°

∴ In case (b) car makes 75° turn.

∠YMD + ∠LMD = 180° (Linear Pair = 2 adjacent angles are

supplementary)

∠YMD + 75° = 180°

∠YMD = 180 - 75

∠YMD = 105°

∴In case (c) car makes 105° turn.

8 0

2 years ago

Simplify the complex fraction: ((3x-7)/x^2)/(x^2/2)+(2/x)

Archy [21]

Simplify the complex fraction: ((3x-7)/x^2)/(x^2/2)+(2/x)

$\dfrac{ \dfrac{3x-7}{x^2} }{ \dfrac{x^2}{2} } + \dfrac{2}{x} =\qquad \qquad x \neq 0\qquad x^4 \neq 0 \\ \\ \\ \dfrac{(3x-7)*2}{x^2*x^2}+ \dfrac{2}{x} = \\ \\ \\ \dfrac{6x-14}{x^4}+ \dfrac{2}{x} =\\ \\ \\ \dfrac{ 6x-14 }{ x^4 }+ \dfrac{2(x^3)}{x(x^3)}=\\ \\ \\ \boxed{ \dfrac{ 6x-14 +2x^3}{ x^4 } }$

6 0

2 years ago