All Answers 355923 Answers

First, let's cross multiply and simplfy. We get 

\(8 (a + 2) = b ( a + 1) \\ 8a + 16 = ab + b\)

Now, in order to progress, we have to be able to factor it, 

\(ab + b - 8a = 16 \\ ab + b - 8a - 8 = 16 - 8 \\ (a + 1) ( b - 8) = 8\)

That means the number of order pairs there are is the number of factors 8 has, including negatives. 

There are 8 factors, meaning the answer is 8. 

Here are the ordered pairs that work. 

\((3, 10), (1, 12), (0, 16), (7, 9), (-5, 6), (-3, 4), (-2, 0), (-9, 7)\)

Thanks! :)

Nice answer, NotThat Smart !!!!

First off, let's combine some like terms and move all terms to one side. 

We get \(6x^{2}-21x-31=0\). Unfortunately, this can't be factored without radicals, so let's use the Qudratic Formula. 

We get \(x=\frac{21\pm \sqrt{(-21)^{2}-4\cdot 6(-31)}}{2\cdot 6}\). Simplifying, we get 

\(x=\frac{\sqrt{1185}+21}{12}\\ x=\frac{-\sqrt{1185}+21}{12}\)

We want to take the biggest value of x we could, which would be \(x=\frac{\sqrt{1185}+21}{12}\). 

Thanks! :)

Fill in the blanks to make a quadratic whose roots are $1$ and $-1.$
x^2 + ___ x + ___  

Since the roots are +1 and –1 then the factors are (x – 1) and (x + 1) respectively.    

Multiplied out, the result is x² – 1    

Therefore the first blank holds a          0    

and the second blank holds a            –1    

_.

Thank you for helping!!! I had trouble on multipling. 

Thank you so much!! I had trouble on how to multiply! You cleared that up!!
 

There are 4 children, which means there are 4! = 4 x 3 x 2 x 1 = 24 ways to seat the children.

There are also 4 adults, which means there are, again, 4! = 4 x 3 x 2 x 1 = 24 ways to seat the adults.

Now, we just have to do 24*24 = 576 since each arrangement of children can be met by 24 ways to arrange the adults.  

There are 576 ways to do this!

Thanks! :)

In order to find all of ordered pairs that work, we just have to write an equation. 

If the two graphs intersect, then we have \(x^3 + x^2 - 3x + 5 = x^3+2x^2\). 

Combing all like terms and moving all terms to one side, we get \(x^2+3x-5 = 0\). 

Using the quadratic equation, we get 

\(x=\frac{\sqrt{29}-3}{2} \approx 1.193\\ x=\frac{-\sqrt{29}-3}{2} \approx -4.193\)

Plugging these two values back into our equations from earlier, we get

\(y \approx 4.541 \\ y\approx -38.541\)

So our two orders pairs approximately are

\((1.193, 4.541), (-4.193, -38.541)\)

Thanks! :)

The area of rectangle $WXYZ$ is 47.

Problem 3:

We can solve this problem by strategically manipulating the inequalities and using our understanding of exponents. Here's how to approach it:

Break down the exponentials:

Notice that 2 raised to the power of 200 appears on the left side of the first inequality. We can rewrite it as (2^2)^100 which is equal to 4^100.

Simplify the inequalities:

With the simplification from step 1, the inequalities become: 4^100 < n^100 < (130n)^{50}

Taking the 50th root:

To compare the terms more effectively, let's take the 50th root of all three parts of the inequalities. This is a valid operation because all terms are positive. Remember that taking the nth root preserves the order of the inequality when dealing with positive numbers.

Applying the 50th root: 2 < n < √(130n)

Squaring both sides:

To get rid of the square root, we can square both sides of the inequality. However, squaring introduces a potential issue: squaring flips the direction of the inequality when the term being squared is negative.

Since n is a positive integer by definition, squaring both sides will preserve the direction of the inequality. So we have: 4 < n^2 < 130n

Analyze the terms:

The leftmost inequality (4 < n^2) tells us that n^2 must be greater than 4. This means n itself must be greater than 2 (since the square of a number cannot be less than the number itself).

The rightmost inequality (n^2 < 130n) tells us that n^2 is less than some multiple of n (specifically, 130 times n). This can only be true if n itself is less than 130.

Identify valid n:

So, to satisfy both inequalities, n must be greater than 2 and less than 130.

Now we consider all the positive integers within this range: 3, 4, 5, ..., 128, 129.

Counting valid integers:

There are a total of 129 - 3 + 1 = 127 positive integers that satisfy the conditions.

Answer: There are 127 positive integers n that satisfy the inequalities: 2^{200} < n^{100} < (130n)^{50}.

Problem 2:

The expression we're given involves squares of several terms. Here's how to find the minimum value for all real numbers x:

Completing the Square: Notice that each term in the expression is a squared term of the form (x + a)^2. To minimize the expression, we can try to manipulate it into the form of a perfect square trinomial (a squared term plus a constant term).

Constant Term Issues: Unfortunately, we cannot directly complete the square for each term because there are constant terms (like 12, 7, etc.) within the parentheses. Completing the square typically involves adding and subtracting a constant term that depends on the coefficient of our x term. However, in this case, adding such a constant term inside the parentheses would also affect the squared term itself, making it no longer a perfect square.

Grouping and Common Factors: Instead of completing the square for each term individually, let's look for ways to group the terms and find common factors. We can rewrite the expression as:

(x + 12)^2 + (x + 7)^2 + (x + 3)^2 + (x - 4)^2 + (x - 8)^2 =

(x^2 + 12x + 144) + (x^2 + 14x + 49) + (x^2 + 6x + 9) + (x^2 - 8x + 16) + (x^2 - 16x + 64)

Notice that each term except the first has a common factor of (x^2 + px + q), where p and q are constants depending on the specific term.

Factoring and Simplifying: Now we can factor out the common factors and group the remaining terms:

(x^2 + 12x + 144) + (x^2 + 14x + 49) + (x^2 + 6x + 9) + (x^2 - 8x + 16) + (x^2 - 16x + 64) =

(x^2 + 12x + 144) + (x^2 + 14x + 49) + (x^2 + 6x + 9) + (x^2 - 8x + 16) + (x^2 - 16x + 64)

= (x^2) + (x^2) + (x^2) + (x^2) + (x^2) + (12x + 14x + 6x - 8x - 16x) + (144 + 49 + 9 + 16 + 64)

= 5(x^2) - 14x + 282

Non-Negative Term: The first term, 5(x^2), is always non-negative because any real number squared is non-negative. The minimum value it can take is 0, which occurs when x = 0.

Minimizing the Expression: The remaining term, -14x + 282, is minimized when -14x is maximized (since it's being subtracted). However, -14x is maximized when x is minimized (remember x is a real number).

Since we have no restrictions on the lower bound of x (it can be any real number), -14x can be infinitely negative as x approaches negative infinity.

Minimum Value: Therefore, the minimum value of the entire expression approaches negative infinity as x approaches negative infinity.

However, the question asks for the minimum value for ALL real numbers x. Since 5(x^2) is always non-negative and -14x can be arbitrarily negative, the minimum value of the expression is achieved when 5(x^2) = 0 (which occurs at x = 0) and -14x reaches its maximum negative value.

Answer: The minimum value of the expression is 282.

Relating medians and angle bisectors:

In a triangle, when a median (such as BM) intersects an altitude (such as AH) inside the triangle, the median is also an angle bisector.

This is a useful property to remember for problems like this.

Angle relationships:

Since BM is a median and an angle bisector of angle B, we have:

∠ABM = ∠MBC (angle bisector property)

We are also given that angle ACB = 41 degrees.

Angle sum in triangle ABC:

The angles in any triangle add up to 180 degrees. Therefore, in triangle ABC:

∠ABC + ∠ACB + ∠BAC = 180°

Solving for ∠MBC:

Since BM is a median, it divides triangle ABC into two triangles with equal areas (triangle ABM and triangle MBC). Because they share the same base (BM), the ratio of their areas is equal to the ratio of their altitudes. In this case, both triangles share the same altitude (AH), so their areas must be equal.

Knowing that triangle ABM and triangle MBC have equal areas, we can express this using their angles:

1/2 * base * height (triangle ABM) = 1/2 * base * height (triangle MBC)

Since the base (BM) is the same for both triangles, we can simplify this to:

∠ABM = ∠MBC (angles are proportional to their areas in a triangle)

We established earlier that due to the angle bisector property, ∠ABM = ∠MBC.

Now, let x represent the measure of angle MBC (which is also equal to angle ABM).

From the angle sum property in triangle ABC:

∠ABC + 41° + ∠BAC = 180°

Since ∠ABM (or x) bisects angle B:

∠ABC + 41° + x = 180°

Substitute ∠ABM with x:

x + 41° + x = 180°

Combine like terms:

2x + 41° = 180°

Subtract 41° from both sides:

2x = 139°

Divide both sides by 2:

x = 69.5°

First one

Cross-multiply and  we  get

2x √(6x) + 4√x  = x√3 + √2       square both sides

4x^2 (6x) + 16x  √(6x^2) + 16x  =  3x^2 + 2x√6 + 2

24x^3 + 16x^2 √6 + 16x  = 3x^2 + 2x√6 + 2         factor the left side

8x ( 3x^2 + 2x√6 + 2 ) =  (3x^2 + 2x√6 + 2)          divide both sides by (3x^2 + 2x√6 + 2)    

8x  =  1

x =  1/8 

Let B = (0,0)   C  =(32,0)

To find the height of the triangle we can construct two lines  using the tangents 

y = (3/5)x

And

y = (-1/4) (x -32) =  (-1/4)x + 8

To find the x coordinate   of their intersection

(3/5()x =(-1/4)x + 8

(3/5 + 1/4)x =  8

[(12 + 5] / 20]  x  =  8

[17/20] x = 8

x = (8*20] / 17  =  160/17

The height of the triangle (AD)  = (3/5)(160/17)  = 480 / 85 = 96/17

Area ABC  = (1/2) BC * AD =  (1/2) 32 * (96/17)  = 1536 / 17

sinA =  1/3

So

BD / AB  =  1/3

BD / 24  = 1/3

BD = 24 (1/3)  = 8

sin C = 1/2

Angle C  = arcsin (1/2)  = 30°

Then triangle BDC  is a 30-60-90  right triangle

BD  = 8

So  BC   =  2BD  = 2 * 8  =   16

So first off, let's calculate the distance Will fovers before Grace even begins to row. 

We have \((45 min)(50 m/min) = 2250 m \). 

When Grace starts to row, the two have to cover \((2800 m) – (2250 m) = 550 m \)

Since they are rowing at eachother at the same time, if we let t be the time it takes, we have \( (50)(t) + (30)(t) = 550 \). 

Solving this, we get

 \( 80t = 550 \\ t = 550/80 = 6.875 \)

This rounds to approximately 6 minutes and 52 seconds. 

This means in addition to the 45 minutes Will takes at the beggining, they will take an additional 6 minutes and 52 seconds to meet. 

Now, we some math to find that they meet at \(2:51:52\) P.M. 

Thanks! :)

First, let's multiply g(x) by f(x). I won't show all the steps cuz I'm a bit lazy....

But make sure use the Distributive property to expand \(\left(\:-x^2+\:8x\:-\:5\right)\left(\:x^3\:-\:11x^2\:+\:2x\right)\). 

In the end, we get

\(-x^5+19x^4-95x^3+71x^2-10x\)

The leading term is the term that is first, which is ofcourse, \(-x^5\)

The leading coefficient is the coefficient of the leading term, which is \(-1\)

The degree is the highest raised power, which is \(5\)

The constant term is the term with no x values, which there is none in this case, so \(0\)

The coefficient of x^2 is clearly \(71\)

I hope I answered all of your questions!

Thanks! :)

Let the money of the bill be x. 

(a) From the problem, we know that \(\frac{x/3 + 9}{x/3 + (x/3-9)} = \frac{5}{6}\)

We just have to solve for x now. 

\(\frac{x+27}{2x-27}=\frac{5}{6}\)

\(6x+162=10x-135 \\ 4x=297\\x=74.26\)

The total bill was $74.25

(b) We just have to work backwards now. 

First off, we know at first, Calvin had to pay \(74.25/3 = 24.75\) dollars. 

However, he then had 9 dollars paid for him, so we have \(24.75-9=15.75\). 

Calvin had to pay Bala $15.75 

Thanks! :)

The number of solutions is 3218.