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Hmm...I plugged it into google translate, as I am not fluent in Vietnamese, but this is what it replied:

"base 90 times 6 times 4"

So...should it be in a different base? Base 90 seems insane, as not many people use it...

This is just algebra in disguise. 
Since the 4th angle is equal to 3x + 7 degrees, and the 5th angle is equal to 9x-43 degrees, we can form a linear equation and solve.

3x + 7 + 9x - 43 = 180      <------- 180 degrees on a line

12x + 7 - 43 = 180

12x - 36 = 180

12x = 216

x = 18

To solve for UPS, we plug 18 into the 4th angle. 
 

3(18) + 7 = 61

But, we also need to add 90 degrees:

61 + 90 = 151 degrees

Maybe you actually will see this answer. Maybe not, as you are a guest, and you can't save topics. 
 

:)

If m angle ABD=79 degrees , what are m angle ABC and m angle DBC​    

Since together they equal 79^o just add them together, solve for x, and then plug it back in. 

                                              ABC    + DBC       =  ABD   

                                             (8x – 3) + (5x + 4)  =  79  

add like terms                         13x + 1  =  79  

                                                     13x  =  78         

                                                         x  =  6 

Plug 6 back in to ABC             (8 • 6) – 3  =  45      so    angle ABC = 45^o      

Plug 6 back in to DBC             (5 • 6) + 4  =  34      so    angle DBC = 34^o   

_.

the measure of one angle of a triangle is three times the measure of a second angle. the measure of the third angle is 12 less than the sum of the other two. what are the measures of the 3 angles?

Let the angles be A, B and C

We have three equations connecting them:

A+B+C = 180

A = 3B

C = A+B-12

(I'm assuming the 12 is degrees not radians).

Can you take it from here?

Write the equation of the line perpendicular to 2x-3y=12 that passes through the point (16,-7)

Rewrite the equation as  y = (2/3)x - 4  This has slope 2/3

A perpendicular line will have slope -3/2 so the equation of a perpendicuar line is y = -(3/2)x + c where c is a constant found from the fact that the line must go through point (16, -7).

-7 = -(3/2).16 + c

c = 24 - 7 

c = 17

Hence perpendicular line has equation  y = -(3/2)x + 17

Because triangle ABC is isosceles, angle ABC = angle ACB = (180 - 30)/2 = 75°

Angles on a straight line must sum to 180°, so 4x + 2x + 75 = 180  giving x = 17.5°, which makes angle ABE = 35°.  This doesn't match any of your options!

$90 \times 6 \times 4 = 540 \times 4 = \boxed{2160}$

Let z = (sqrt(6) - sqrt(2)) + (sqrt(6) + sqrt(2))*i.  Compute z^(24).

Convert to polar notation 

z = r.exp(i.θ)   where r = √(√6 - √2)² + (√6 + √2)² ) = 4  and  θ = tan^-1( (√6 + √2)/ (√6 - √2) ) = 75°

z²⁴ = r²⁴.exp(i.24.θ) = 4²⁴.exp(1800°) = 4²⁴.(cos(1800°) + i.sin(1800°)) = 4²⁴.(cos(0°) + i.sin(0°)) = 4²⁴  

Good work Juriemagic,

It is good to see you here again too  

Hi dp,

Why haven't you posted your LaTex properly?

x/25 = 63/35

     x =63/35*25

     x = 45

The line is y = 3/2*x - 1.

(r,theta) = (4,2*pi/3).

The distance is 25 cm.

|v + w| = 2, |w + z| = 2*sqrt(3), |v + z| = 1.

System of equations, seriously, it helps.

Let the 3 numbers be x, y, and z. We know:

x + y + z = 52

2z = x

x - 8 = y

What do we see? Well, we can rewrite everything in terms of x:

x = x (obviously)

z = x/2

y = x - 8

x + x - 8 + x/2 = 52

2x - 8 + x/2 = 52

5x - 16 = 104

5x = 120

x = 24

Knowing that x = 24, we see:

x = 24

z = 12

y = 32

:)

a=1;p=0; b=1;c=1;n= (a+2*b+3*c)%7;m=(2*a+3*b+c)%7;k=(3*a+b+2*c)%7;if(n==0 and m==4 and k==4, goto loop, goto next);loop:p=p+1;printp," =",a,b,c;next:a++;if(a<15, goto4,0);a=1;b++;if(b<15, goto4, 0);a=1;b=1;c++;if(c<15, goto4,0)

a=1,  b=2,  c=3,  and abc mod 7 =(1*2*3) mod 7 =6

Since are looking for a cubic polynomial, let it be ax^3 + bx^2 + cx + d = 0.  So

$a(\sqrt[3]{2} + \sqrt[3]{4})^3 + b(\sqrt[3]{2} + \sqrt[3]{4})^2 + c(\sqrt[3]{2} + \sqrt[3]{4}) + d = 0$

This expands to

$a (2 + 3 \sqrt[3]{2^2} \sqrt[3]{4} + 3 \sqrt[3]{2} \sqrt[3]{4^2} + 4) + b(\sqrt[3]{2^2} + 2 \sqrt[3]{2} \sqrt[3]{4} + \sqrt[3]{4^2}) + c(\sqrt[3]{2} + \sqrt[3]{4}) + d = 0$

Expanding everything out, and comparing the coefficients, we get

a + 3b + 3c + d = 16,
-6b + 3c + 3d = -24,
c - 3d = 22,
d = -6.

The solution to this system is a = 1, b = 3, c = 4, d = -6, so the cubic is x^3 + 3x^2 + 4x - 6.

Acute scalene triangle.

Sides: XZ = 4    XY =2sqrt(6) = 4.899    YZ = 5.464

Area: T = 9.464
Perimeter: p = 14.363
Semiperimeter: s = 7.182

Angle ∠ Y = 45° = 0.785 rad
Angle ∠ Z = 60° = 1.047 rad
Angle ∠ X = 75° = 1.309 rad

These are the products of the 10 numbers chosen:

(-8, -6, -4, -4, -3, -2, -2, -1, 1, 2, 2, 3, 3, 4, 4, 4, 6, 6, 8, 8, 9, 12, 12, 16) Total =>> 24

If you count the positive products, you get 16

So, the probability is: 16 / 24 = 2 / 3

Let the width of the rectangle = W
The length of the rectangle     = 2W + 2

2W + 2[2W + 2] = 28
2W + 4W + 4     = 28
6W = 28 - 4 =24

W = 24 / 6 = 4 m - width of the rectangle
4 x 2 + 2   = 10 m - length of the rectangle.

Let the price of an adult's ticket = A

Let the price of a child's ticket    = C

11C + 3A = 122.......................(1)

6C  + 4A =102........................(2)

Now, you have 2 simultaneous equations. Can you solve them?

Using the Binomial Theorem, expand (-2t + 7)^3.

-8 t^3 + 84 t^2 - 294 t + 343