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hello person is there another way you could have figured that out?

Line 1 = \(x + 3y = -5\)

Line 2 = \(Ax + By = 3\)

Rearrange Line 1 into slope intercept form,

Line 1 = \(y = -x/3 - 5\)

Because Line 2 has the same slope,

Line 2 = \(y = -x/3 + b\)

Multiply Line 2 by 3 on both sides and convert to standard form,

Line 2 = \(3y = -x + 3b\)

Line 2 = \(x + 3y = 3b\)

Line 2 must pass through (-7, 2), so we can plug in these values,

Line 2 = \(-1 = 3b\)

\(b = -1/3\)

Line 2 = \(x + 3y = -1\)

Multiply both sides by -3,

Line 2 = \(-3x + -9y = 3\)

B - A = -9 - (-3) = -6

Because the box is indistinguishable, we can use the reccurence relation. \(B(n,k)=B(n−1,k−1)+kB(n−1,k)\)with \(B(n,1)=1B(n,1)=1\) and \(B(n,n)=1B(n,n)=1\). For your case, you need the sum \(B(4,1)+B(4,2)+B(4,3)\). I trust you can use the formula to complete the problem.

$2^{10} \cdot 4^{20} \cdot 8^{40} = 2^n$

$2^{10} \cdot \left(2^2 \right)^{20} \cdot \left(2^3 \right)^{40} = 2^n$

$2^{10} \cdot 2^{40} \cdot 2^{120} = 2^n$

$2^{10 + 40 + 120} = 2^n$

$n = 10 + 40 + 120$

$n = \boxed{170}$

Putting all with a base of 2,

2^10 * 2^40 * 2^120 = 2^n

Add up exponents, n = 170

let the resulting integer equal x. reword it another way, \(29x\equiv x(\text{mod }10^n)\rightarrow28x\equiv 0(\text{mod }10^n)\)where n is just some positive integer.

the prime factorization of 28 is 2*2*7, meaning that there must be at least a 5 in the prime factorization of x to multiply with one of the 2's in 28(for its last digit to equal zero and for n to be 1). If we were to plug 5 in, we see that it doesn't work, meaning that x must be at least 2 digits. That means that x must have at least 2 5's in its prime factorization for it to multiply with the 2 2's in 28(for its last 2 digits to equal zero and for n to be 2). Testing out 5*5=25, we can see that it works, and we are sure it's the smallest possible value, so the answer is \(25*29=\boxed{725}\)

Monday = 1 (mod 7)

21 (mod 7) = 0 mod 7 is obviously Sunday given that information.

The simplest polynomial is $ax^2 + bx + c$

Let $x = k \cdot c$ for $k \in \mathbb{Z}$

We have \(\begin{align*} ax^2 + bx + c &= a(kc)^2 + b(kc) + c\\ &= ak^2c^2 + bkc + c\\ &= c(ak^2c + bk + 1) \end{align*}\)

Because there is are at least two factors for $ax^2 + bx + c,$ $P(x)$ is composite.

January 8th, 15th, and 22nd would also be a Monday as we can add 7 to get the same day of the week. 1 day before Monday is Sunday, the 21st.

Let $\angle{A} = x.$

Let the change in angles be $n.$

We have: $x+(x+n)+(x+2n)+(x+3n) = 360$

$x+n = 72$

We want to find $x+3n.$

Rewrite first equation: $3x+5n + 72 = 360$

$3x + 5n = 288$

Rewrite third equation: $3x + 3n = 216$

$2n = 72$

$3n = \frac{3}{2} \cdot 72 = 108$

$n = 36$

$x + 36 = 72$

$x = 36$

$x + 3n = 36 + 108 = \boxed{144}$

Rearranging line L_1 into intercept form, 
y = -5/8x - 9/8

If line L_2 is perpendicular, it would have slope 8/5,
y = 8/5x + b
 

Plugging in (10, -10), 

-10 = 16 + b 

b = -26

Therefore m+b is 8/5 + (-26) = -24.4 or -24 2/5

The Pascal's Triangle for (m+n)^7 has coefficients, 1, 7, 21, 35, 35, 21, 7, 1. If both m and n are odd, then either of them to the power of anything would result in an odd number. Because odd * odd is odd, and all of the coefficients are odd, I think all 8 terms would be odd

To be a multiple of 2, it would have to have a one's digit of 2, 4, 6, 8. (0 is not included as you cannot have a hundredths digit with a 0.) This makes all the possible palindromes xyx, where x is one of these digits, and y is any digit. This gives 4 possibilities for x, and 10 possibilities for y, creating 40 even palindromes. To find all of the 3 digit palindromes, it is again in format xyx, but in this case x can be any number giving 100 possibilities (10*10). 40/100 = 2/5 probability.

Because a quadrilateral angles sum up to 360 degrees, A, B, C, and D must sum to 360.

Since it's a arithmetic sequence with B being 72, we can write them as 

A, B, C, D
72-x, 72, 72+x, 72+2x

Add them all up to get, 288+2x = 360. Simplify to get x = 36, and angle D was equal to 72 + 2x, so angle D is 144 degrees.

is there any other way to prove????????

( k+1)^3 - k ^3  = 1801

k^3  + 3k^2 +3k+1    - k^3 = 1801

3k^2 + 3k -1800 = 0

k^2 + k -600 = 0                   Qudratic formula shows k= 24   (the positive answer)     then k+1 = 25

Depth can be 2.05 m  width can then be 4.10 m and length can be 6.15 m    volume = 51.691 m³   max

this is a dependant probability thing

so what we have is  A A A A 2 2 2 2 3 3 3 3 -- what is the chance of us getting all distinct letters?

eachtime we pull, we take one right card, and then the probability to find the other ones is by eliminating the left three of that particular card, so:

$ \frac{12}{12} \frac{8}{11} \frac{4}{10}   $

therefore, the prob. of us pulling an ace, a two and a three is $ \frac{16}{55}  $

Using Desmos, I found that tan a + tan b = -11/8.

Explaination for b)?

(a) See the graph below.

(b) g(x) = 1/3*f(2x + 2).

(c) We stretch the graph horiztonally by a factor of 2, then stretch the graph vertically by a factor of 3, then shift down 2 units.

There are 6 + C(6,2) = 21 possible numbers.

Slope,m, between the points = -6/3 = -2

point slope form    (using point (-3,-5) <====== you could use the other point instead....same result)

(y+5)= -2(x+3)    re-arrange

2x + y = -11          you can finish, I hope !