All Answers 352610 Answers

you got the first one right

for (b)

3x = 15

divide both sides by 3 to get

x=5

for (c)

x = 12-8 = 4

(a) 

2cm was the radius => the area of the semicircle is 2^2 pi/2 = 2pi

the trapezoid had base 4 and 7, height 2, so (4+7)2/2 = 11

3.5 was the radius, so 12.25/2 pi = 6.125 pi

2pi+6.125pi+11 = 8.125pi+11

8.125 pi + 11 is the answer for (a)

(b)

4.1 is the radius, so area = 16.81 pi /2 = 8.405 pi

Area of triangle is 4.1 * 7.6 = 31.16

8.405 pi + 31.16 is your answer for (b)

Thanks 

I assume you mean

if its spring now what was it 3 months ago

In Australia it is clearly defined.

Winter comes before spring.

So the answer is Winter.

Our seasons are defined by months, I think in the US it works differently.

Hint:

What multiplies to give  -8

It is negative so one is neg and one is positive.

1, -8

-8,1

-4,2

-2,4

They are the only ones.  which 2 add to give 7?

It depends which month it is in spring, unless you want the specific season, which is summer

summer is your answer

\(Px - 37 = Qx - 37 \\ Px = Qx \\ P=Q \text{ leads to infinitely many solutions} \\ P\neq Q \text{ leads to the single solution } x=0\)

\(x^4-11x^3+24x^2=x^2(x^2-11x+24) \\ (4x^2-44x+96) = 4(x^2-11x+24) \\ x^4-11x^3+24x^2 - (4x^2-44x+96) =(x^2-11x+24)(x^2-4) = \\ (x-8)(x-3)(x-2)(x+2) \\ \text{and thus the solutions are }\\ x=8,~3,~2,~-2\)

Hello kminery62. The way you worded some of the problems was confusing, but I tried my best.

For question B:

9 - 6(x - 1) = 2(x - 4) + 27

9 - 6x + 6 = 2x - 8 +27          Distribute.

-8x = 4                               Move similar terms to the same side.

x = -2                                 Divide.

For question C:

5x + 12 = 87

5x = 75                               Subtract.

x = 15                                Divide.

For question D: 

(3/4)x - 5 = 4

(3/4)x = 9                            Add.

(1/4)x = 3                            Divide.

x = 12                                Multiply.

For question E:

(-5/2)x / 6 = 1

(-5/2)x = 6                           Multiply.

(1/2)x = -6/5                        Divide.

x = -12/5                             Multiply.

x = -2 1/5                            Simplify.

For question F:

6x + 3 = -1

6x = -4                                Subtract.

x = -4/6                               Divide.

x = -2/3                               Simplify.

-n + (-3) + 3n + 5 = 

-n - 3 + 3n + 5 =

-n + 3n - 3 + 5 =

3n + (-n) + 5 + (-3) =

3n - n + 5 - 3 =

2n + 2

Therefore, -n + (-3) = 3n + 5 = 2n + 2

Thank you Guest. It's kind of tricky

(1/2)x+5=2  Why ???  Because:

(1/2)x+5=2            multiply both sides by 2

x + 10 = 4              subtract 10 from both sides

x = 4 - 10

x = -6

2.47*10^-3 = 0.00247 - This is in standard notation

Here is the third one:

Solve for t:
q = (2 - 4 t)/(t + 3)

Reverse the equality in q = (2 - 4 t)/(t + 3) in order to isolate t to the left hand side.
q = (2 - 4 t)/(t + 3) is equivalent to (2 - 4 t)/(t + 3) = q:
(2 - 4 t)/(t + 3) = q

Multiply both sides by a polynomial with respect to t to clear fractions.
Multiply both sides by t + 3:
2 - 4 t = q (t + 3)

Write the linear polynomial on the right-hand side in standard form.
Expand out terms of the right hand side:
2 - 4 t = 3 q + qt

Isolate t to the left-hand side.
Subtract q t + 2 from both sides:
t (-q - 4) = 3 q - 2

Solve for t.
Divide both sides by -q - 4:
t = (2 - 3q) /( q + 4)

Here is the second one:

Solve for w:
y - aw = 2w - 1

Isolate w to the left-hand side.
Subtract 2 w + y from both sides:
w (-a - 2) = -y - 1

Solve for w.
Divide both sides by -a - 2:
 w = (y + 1) / (a + 2)

(x^5  - 5x^4  + 6x^3  - 5x^2  + 16x  - 3) : (x - 3)  =  x^4 - 2x^3 - 5x + 1  
 x^5  - 3x^4                            
 ———————————————————————————————————————
      - 2x^4  + 6x^3  - 5x^2  + 16x  - 3
      - 2x^4  + 6x^3                    
      ——————————————————————————————————
                      - 5x^2  + 16x  - 3
                      - 5x^2  + 15x     
                      ——————————————————
                                  x  - 3
                                  x  - 3
                                  ——————
                                       0

A

That is a very big number!! But, there is a method to your teacher's madness!! Since you are willing to do the homework, I will give you a hint and you finish the job!.

Take 10^10 and divide it by the powers of 5 as follows:

(10^10) / 5  +  (10^10) / 5^2  +  (10^10) / 5^3  +  ............+  (10^10) / 5^14.

Note: It is very important that you divide them one at a time and take the INTEGER PART ONLY. Disregard the fractional part. Once you calculate them all, then add them up and DIVIDE THE TOTAL BY 2. Again, if there is a fraction, disregard it and keep the integer part only. That should give you a lot of zeros!!. Good Luck. 

Here is the first one:

Solve for g:
w = 7 - sqrt(g)

Reverse the equality in w = 7 - sqrt(g) in order to isolate g to the left hand side.
w = 7 - sqrt(g) is equivalent to 7 - sqrt(g) = w:
7 - sqrt(g) = w

Isolate terms with g to the left hand side.
Subtract 7 from both sides:
-sqrt(g) = w - 7

Multiply both sides by a constant to simplify the equation.
Multiply both sides by -1:
sqrt(g) = 7 - w

Eliminate the square root on the left-hand side.
Raise both sides to the power of two:
 g = (7 - w)^2 = w^2 - 14 w + 49

ever think that B might be female?

First find the mean of the data numbers

2+4+7+8+9   / 5   = 6

Take the diff from the mean of each of the numbers

4   2  1  2  3      Square these and find the mean

16 4  1  4  9        Mean = 6.8    and take the square root of this     2.6 = S.D.

Ok thanks. 

Do you know why ?

If so would you care to share your knowledge?

a) 

Since these are composite figures, it is probably best to think about each composite figure as each individual shape that it is comprised of. Let's first discern the perimeter of the trapezoid.

We know three of the four sides, and the fourth side is of equal length to its opposite side. The fourth side is not exposed, so we can ignore it from the calculation. Just add these lengths together to obtain the perimeter. Multiply this perimeter by two because there are two trapezoids in this diagram.

\(P_{\text{trpzd}}=(13+9+7)\text{mm}\\ P_{\text{trpzd}}=29\text{mm}\\ 2P_{\text{trpzd}}=58\text{mm}\)

The only length left to determine is the arc length of the circular arc. The circumference of the circle is \(C=2\pi r\), but we only want 44 of the 360 degrees of the circle. Add these perimeters together to find the final perimeter. 

\(\text{Arclength}=2\pi r*\frac{44}{360}; r=9\text{mm}\\ \text{Arclength}=\frac{99\pi}{45}\text{mm}\approx6.9115\text{mm}\)

Add these perimeters together to find the final perimeter:

\(2P_{\text{trpzd}}+\text{Arclength}=58\text{mm}+6.9115\text{mm}\\ 2P_{\text{trpzd}}+\text{Arclength}=64.9115\text{mm}\approx64.91\text{mm}\)

And that's that!

b)

The base of the equilateral triangle is 7cm. The collinear radii of the circular arc are also 7cm.  This means that 

\(P_{\triangle\text{ & radii}}=(7+7+7)\text{cm}\\ P_{\triangle\text{ & radii}}=21\text{cm}\)

The only thing left to do is find the arc length, just like the previous problem. There are two arcs this time. Make sure to take that into account.

\(\text{Arclength}=2\pi r*\frac{120}{360}; r=7\text{cm}\\ \text{Arclength}=\frac{14\pi}{3}\text{cm}\\ 2\text{Arclength}=\frac{28\pi}{3}\text{cm}\approx29.3215\text{cm}\)

Find the sum of these perimeters again:

\(P_{\triangle\text{ & radii}}+\text{Arclength}=21\text{cm}+29.3215\text{cm}\\ P_{\triangle\text{ & radii}}+\text{Arclength}=50.3215\text{cm}\approx50.32\text{cm}\)

a)

In order to determine the vertical asymptote with a given rational function, you must set the denominator equal to zero and solve. Let's do that. 

\(f(x)=\frac{ax+b}{x-3}\)

The horizontal asymptote has three conditions that you must always examine in any rational function. Every condition compares the degree of the numerator to the degree of the denominator. They are the following:

• If the degree of the numerator is less than the degree of the denominator, then a horizontal asymptote occurs at y=0.
• If the degree of the numerator is equal to the degree of the denominator, then a horizontal asymptote occurs at the ratio of the leading coefficients of the numerator and denominator.
• If the degree of the numerator is greater than the degree of the denominator, then no horizontal asymptotes exist.

The degree of the numerator, in this case, is 1, and the degree of the denominator is also 1. According to the conditions above, we must find the "ratio of the leading coefficients of the numerator and denominator."

\(f(x)=\frac{-4x+b}{x-3}\)

Finally, we will use the last tidbit of information regarding this particular rational function, the location of the x-intercept. We only have one variable remaining, so we can just substitute this coordinate into the function and solve for b:

The final rational function, then, is \(f(x)=\frac{-4x+b}{x-3}\). You're done. 

b) 

This problem is almost exactly the same as the one I showcased above. Use the same techniques as I have showcased above, and you will be fine. 

For 1, join P to A, B and C forming three triangles. Calculate the area of each and say that their sum is equal to the area of ABC. 

The area of APC, for example, will be its base, AC (=1) multiplied by its height (= PE.cos30) divided by 2, etc. .

The wording for 2 seems a bit loose, what, exactly, is meant by equiangular ?

Which angles are equal ?

If its just the angles between successive sides, then its obvious.