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Area = (1/2π * 4²) + (15 * 3) + (4 * 6)

         = 25.12 + 45 + 24

         = 94.12 km²

The pie that would have the smallest slice is:

B) Peach pie

Because the higher the number of slices, the smaller a slice it would have.

Total money he had = $62

    money spent = $(28 + 4) = $32

    so, money left = $(62 - 32) = $30

    He needs total $70 at end 

    So, money he needs by selling 2 bird houses = 70 - 30 = $40

    So, he should sell both for $40

By symmetry, there are 6 solutions.

10^10= 2^10 * 5^10 ==[10 +1] x [10 + 1] ==121 divisors

18^7= 2^7 * 3^14==[7 +1] x [14 +1] ==120 divisors

18^11= 2^11 * 3^22==[11 + 1] x [22 +1]==276 divisors.

8! ==(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 28, 30, 32, 35, 36, 40, 42, 45, 48, 56, 60, 63, 64, 70, 72, 80, 84, 90, 96, 105, 112, 120, 126, 128, 140, 144, 160, 168, 180, 192, 210, 224, 240, 252, 280, 288, 315, 320, 336, 360, 384, 420, 448, 480, 504, 560, 576, 630, 640, 672, 720, 840, 896, 960, 1008, 1120, 1152, 1260, 1344, 1440, 1680, 1920, 2016, 2240, 2520, 2688, 2880, 3360, 4032, 4480, 5040, 5760, 6720, 8064, 10080, 13440, 20160, 40320)>>Total = 96 divisors.

I count 24 divisors > 1,000.

6  Giraffes 

5  Apes 

1  Zebra       

_.

Since the number who brough an umbrella is twice the number who had raincoats, the percent who brought raincoats is 33.33%, and the number who brought an umbrella is 66.67%.  Then the number who only brough an umbrella is 66.67% - 8% = 58.67%

I assume that by N=abcd, you mean that the 4 digits must be distinct (no repetitions allowed).

4015 , 4125 , 4180 , 4235 , 4290 , 4510 , 4620 , 4675 , 4730 , 4785 , 4895 , 4950 , 5170 , 5280 , 5390 , 5610 , 5720 , 5830 , 5940 , Total =  19 such 4-digit integers.

Hint,

You cannot divide by 0

help you do what?

This is not difficult.

Draw a clock face and see when it is true.  The two hands must be at right angles to each other and since it is n o'clock the big hand must point to the 12.

Two 3x4 rectangles overlap, as shown below. Find the area of the overlapping region. (Which is shaded)

\(\begin{array}{|rcll|} \hline 3^2+x^2 &=& (4-x)^2 \\ 3^2+x^2 &=& 16 - 2*4x + x^2 \\ 3^2+x^2 &=& 16 {\color{red}{-8}}x + x^2 \\ 9 &=& 16 -8x \\ 8x &=& 16-9 \\ 8x &=& 7\\\\ x &=& \mathbf{\dfrac{7}{8}} \\ \hline \end{array}\)

area

\(\begin{array}{|rcll|} \hline A &=& 3*(4-x) \\ &=& 12-3x \\\\ &=& 12 -\dfrac{3*7}{8} \\\\ &=& 12 -\dfrac{21}{8} \\\\ &=& \dfrac{8*12-21}{8} \\\\ \mathbf{A }&=& \mathbf{\dfrac{75}{8}} \qquad \text{or} \quad \dfrac{150}{16} \\ \hline \end{array}\)

3√6cis(π/8) and 2√5cis(7π/6)

assuming only the 5 sand 5 are under the square root

\([3\sqrt6 cis(π/8) ]*[ 2\sqrt5 cis(7π/6)]\\ =6\sqrt{30}*cis(π/8) ]* cis(7π/6)]\\ =6\sqrt{30}*cis(π/8) ]* cis(7π/6)]\\ =6\sqrt{30}*e^{(\pi/8)i}*e^{(7\pi/6)i}\\ =6\sqrt{30}*e^{((1/8)+(7/6))\pi i}\\ =6\sqrt{30}*e^{((3/24)+(28/24))\pi i}\\ =6\sqrt{30}*e^{(31/24)\pi i}\\ =6\sqrt{30}*cis{(31\pi/24)}\\ \)

I have not checked it.  That is your job

LaTex

[3\sqrt6 cis(π/8) ]*[ 2\sqrt5 cis(7π/6)]\\
=6\sqrt{30}*cis(π/8) ]* cis(7π/6)]\\
=6\sqrt{30}*cis(π/8) ]* cis(7π/6)]\\
=6\sqrt{30}*e^{(\pi/8)i}*e^{(7\pi/6)i}\\
=6\sqrt{30}*e^{((1/8)+(7/6))\pi i}\\
=6\sqrt{30}*e^{((3/24)+(28/24))\pi i}\\
=6\sqrt{30}*e^{(31/24)\pi i}\\
=6\sqrt{30}*cis{(31\pi/24)}\\

4!*8C4

Zep. looks like I did so i better add 6.  That makes   44  And now we agree.

Thanks guest  :))

What is tan B?

\(tan\ B=3/\sqrt{14}=0.80178\)

  !

pls help 

Divide by ab on both sides of the equation. 

\(\dfrac ab + \dfrac ba = 4\)

Squaring gives 

\(\dfrac{a^2}{b^2} + \dfrac{b^2}{a^2} + 2 = 16\\ \dfrac{a^2}{b^2} + \dfrac{b^2}{a^2} =14\)

Let x = a/b - b/a. Then x^2 = a^2/b^2 + b^2/a^2 - 2 = 12.

Therefore, \(\dfrac ab - \dfrac ba = \pm 2 \sqrt 3\). Since a > b > 0, the negative root is rejected. \(\dfrac ab - \dfrac ba = 2\sqrt 3\).

Please try to fill in the missing details.

Previously answered here: https://web2.0calc.com/questions/number-theory_68857

When x = y = 0, \((f(0))^2 = 2f(0)\).

This implies f(0) = 0 or f(0) = 2.

Case 1: f(0) = 2.

When x = 0, 2 f(y) = f(y) + 2

f(y) = 2 for all real numbers y.

The function f(x) = 2 is a solution.

Case 2: f(0) = 0.

When x = 0, 0 = f(y) for all real numbers y.

The function f(x) = 0 is a solution.

Therefore, the possible functions are the constant functions f(x) = 0 and f(x) = 2.

Do not post A o P S homework.

The solutions (x,y,z) are (3,1,2), (-4,1,-3/2), and (5/3,9/2,2).  You can figure out the possible values of x + y from there.

Can someone please help me with this? I need it by the end of tonight.

The amount that Brian spent is 1/10 of his total paycheck, so we know that 1/10 of the paycheck is 3 dollars, since he spent only half of the remaining money. Therefore, if m is the total money he has, 1/10*m=3, so m is 30 dollars.