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(a + 2)  / ( a + 1)  =  b / 9

8 (a + 2)  = b ( a + 1)

8a + 16  = ab + b

ab + b   - 8a   = 16               multiply the coefficients on a,b and  add  to  both sides

ab + b - 8a - 8  = 16 - 8

(a + 1) ( b - 8)  =  8

Factors of 8                a           b

4, 2                             3         10

2, 4                             1         12

1, 8                             0         16

8. 1                             7          9

-4 , -2                         -5         6

-2, - 4                         -3         4

-1, -8                          -2         0

-8. -1                          -9         7

(x + 1)  ( x -1)  =  0

x^2  +  0x - 1  = 0

          Area = 144 units²    

          If one of the angles is 90^o   

          then all of the angles are 90^o   

          and this rhombus is a square.  

_.

The coefficient of y⁴ in the expansion will be 81.   

Combine terms of (2y – 5 + y – 6)⁴  to  (3y – 11)⁴   

There will be only one term raised to the fourth power   

and that's the "3y" term ...  3 raised to the fourth is 81.   

_.

A = 100 units²     

Perimeter 40 and diagonal 10 • sqrt(2) indicate that this     

figure is the special case of rectangle, which is a square.    

Each side is 10, therefore the area is 10 • 10 = 100.     

Deleted.  I misread the question.

We can solve this problem by considering two cases:

Case 1: One family has siblings in each row

There are 3 families (let's call them A, B, and C). In this case, we need to place one child from each family in each row. There are 3 choices for the first seat in the first row (child from any family).

Once the first seat is filled, there are 2 choices left for the second seat (child from one of the remaining families). The third seat is automatically filled by the remaining child.

Similarly, for the second row, there are again 3 choices for the first seat, and so on. Therefore, the number of arrangements for this case is:

Arrangements (Case 1) = 3 * 2 * 3 * 2 = 36

Case 2: Two families have siblings in one row

There are again 3 choices for which family will have siblings in the same row (A, B, or C). Without loss of generality, let's say family A has siblings in the same row.

There are now 2 choices for which child of family A sits in the first seat of the first row. The other child of family A must sit in the second seat.

The remaining child from family B can sit in the third seat. Now, for the second row, there are 2 choices for the first seat (child from either family B or C). Following the same logic as before, the number of arrangements for this case is:

Arrangements (Case 2) = 3 * 2 * 1 * 2 * 2 = 24

Total Arrangements

To get the total number of arrangements, we add the number of arrangements from both cases:

Total Arrangements = Arrangements (Case 1) + Arrangements (Case 2)

Total Arrangements = 36 + 24

Total Arrangements = 60​

There are 5!=120 ways to order the numbers 1 through 5. However, this counts an arrangement as valid if it is just a rotation of another valid arrangement.

For instance, the arrangement 1<2>3<4>5 is the same arrangement as 5<1>2<3>4 if we just rotate the boxes cyclically.

There are 5 ways to rotate the boxes, so we have overcounted exactly that many times. Therefore, the number of valid arrangements is 120 ÷ 5 = 24​.

The radius of the circle is 35.

                 7!             5040  

₇P₇  =  –––––––  =  ––––––  =  5040       

             (7 – 7)!           1   

You might question zero in the denominator,   

but zero factorial is defined as 1 so it's okay.  

_.

               AB + CD  

Area  =  —–—–—–  •  height   

                     2   

That's just the formula for the area of a trapezoid.   

Without specific values, I can't take it any further.  

_.

 How many miles did I drive?

\(s=40mph\cdot t=55mph\cdot (t-\frac{25}{60}h)\\ 40mph\cdot t=55mph\cdot t-55mph\cdot \frac{25}{60}h\\ (55-40)mph\cdot t=55mph\cdot \frac{25}{60}h\\ t=\dfrac{55\cdot \frac{25}{60}h}{15}\\ t=1.52\overline 7h\)

\(s=40mph\cdot1.52\overline 7h\\ s=61\frac{1}{9}\ miles\\ \color{blue}61\frac{1}{9}\ miles\ dit\ i\ drive.\\ \)

 !

b = bagel

m = muffin

t = toast

2t + b = 3.15

b + 4m = 7.85

t + 2b + 3m = 9.75

b = 3.15 - 2t

-3t + 3m = 3.45

m - t = 1.15

m = t + 1.15

b + 4t + 4.6 = 7.85

b + 4t = 3.25

b + 2t = 3.15

t = 0.05

b = 3.05

I think the answer is 3.05

Let the roots of Babette's monic quadratic be r and s. Since the leading coefficient is 1, the quadratic can be written as:

p(x)=(x−r)(x−s)=x2−(r+s)x+rs.

We are given that squaring the roots gives another monic quadratic with those squares as its roots. In other words:

(x−r2)(x−s2)=x2−(r2+s2)x+r2s2

Expanding the left side gives:

x2−(r2+2rs+s2)x+r2s2

Equating the coefficients of the corresponding terms in both quadratics, we get the system of equations:

r+s=r2+s2

r2s2=rs (notice this simplifies to r2s2−rs=0)

From the first equation, we can rearrange to get r2+s2−r−s=0. Factoring this gives (r−1)(s−1)=0. This means either r=1 or s=1.

We can consider two cases:

Case 1: r=1

Substituting r=1 into the second equation gives s2−s=0, which factors as s(s−1)=0. This means either s=0 or s=1. However, if s=0, then the leading coefficient of the quadratic wouldn't be 1, so we reject this case.

Therefore, in this case, s=1 and the quadratic is simply x2−2x+1=(x−1)2. There is only 1​ quadratic possible in this case.

Case 2: s=1

Substituting s=1 into the second equation gives r2−r=0, which factors as r(r−1)=0. This means either r=0 or r=1. However, if r=0, then the leading coefficient of the quadratic wouldn't be 1, so we reject this case.

Therefore, in this case, r=1 and the quadratic is again simply x2−2x+1=(x−1)2. There is only 1​ quadratic possible in this case.

Since both cases lead to only one possible quadratic, Babette could have been thinking of at most 2​ different monic quadratics.

Find the intersection points:

The line x = 2 intersects the y-axis at (2, 0).

To find the intersection of y = -x - 1 and y = 2, we set the y values equal: -x - 1 = 2 --> x = -3. Substituting this x value back into y = -x - 1 gives the point (-3, -2).

Find the midpoint of each side:

Midpoint of side connecting (2, 0) and (-3, -2):

x-coordinate: ((2) + (-3)) / 2 = -1/2

y-coordinate: ((0) + (-2)) / 2 = -1

Midpoint is (-1/2, -1).

Since the line x = 2 is a vertical line, any point on this line with a y-coordinate of -1 will be the midpoint of the side connecting (2, 0) and the intersection point on this line (which doesn't matter since it's a vertical line). Therefore, the midpoint is (2, -1).

Since all three vertices lie on the line x = 2, the center of the circle that passes through all three vertices must also lie on this line. Therefore, the center of the circle has an x-coordinate of 2.

Since the midpoint of one side is also the midpoint of another side that lies on the same line, the center of the circle must also coincide with the midpoint we found: (2, -1).

Therefore, the center of the circle is at (2,−1)​.

If there are 20 marbles in the bag, and the ratio of marbles is 4:1:2:3

adding 4+1+2+3 = 10, which when multiplied by 2 results in 20, which means there are 4 blue marbles.

The radius of the circle is 22/9.

The area of triangle ABC is 50 \sqrt{3}.