Write the expression as a product of polynomials:

a(p–q)+q–p

p^2q+r^2–pqr–pr

PLZ ANSWER ASAP

Answer:

x ≥ 0

Step-by-step explanation:

The function is graphed starting at 0 and going up the positive x-axis.

These are numbers that are greater than 0; thus the domain is greater than or equal to 0, or

x ≥ 0

The domain of the function is (d) x >= 0

From the graph, we have the following highlight:

Since 0 is inclusive, we make use of the greater than or equal to inequality

Hence, the domain of the function is (d) x >= 0

Read more about domain at:

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Answer:

√74

Step-by-step explanation:

a^2 + b^2 = c^2

7^2 + 5^2 = c^2

49 + 25 = c^2

74 = c^2

√74 = √c^2

The lengths of the two pieces of wire, when one is formed into a square and the other into a circle with equal areas, are approximately 186.3 inches and 133.7 inches respectively.

To solve this problem, we first need to understand that the total perimeter of both the square and the circle made from the wire lengths equal to the original wire length, which is 320 inches. Suppose the length of the wire used for the square is 'a' and the length for the circle is 'b'. So, 'a + b = 320' inches.

Given that, the areas of both the square and the circle are equal. The area of a square is side^2 and the side of our square will be 'a/4' (since a square has four equal sides) and the area of a circle is πr^2, with the circumference being 2πr or 'b' in this case, so the radius is 'b / (2π)'.

Setting these two areas equal gives us the equation '(a/4)^2 = π * (b/ (2π))^2'. Solving this system of equations, we find that 'a' is approximately 186.3 inches and 'b' is approximately 133.7 inches.

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Answer:

The area will increase by a factor of 4

Step-by-step explanation:

4 * 4 = 16

IT'S UP THERE STEVE (Like on family feud)

Using the Angle Bisector Theorem solve for x.

The "Angle Bisector" Theorem says that an angle bisector of a triangle will divide the opposite side into two segments that are proportional to the other two sides of the triangle.

As, we see the figure, AD bisects angle A

So, Applying Angle Bisector theorem, we get,

[tex] \frac{AC}{CD} =\frac{AB}{BD} [/tex]

[tex] \frac{AC=15}{CD=3} =\frac{AB=(4x+1)}{BD=5} [/tex]

[tex] \frac{15}{3} =\frac{4x+1}{5} [/tex]

15 divide by 3 gives 5

So, we have

5=[tex] \frac{4x+1}{5} [/tex]

To get rid of fractions let us multiply by 5 on both sides

5*5=[tex] \frac{5*(4x+1)}{5} [/tex]

25=[tex] \frac{1*(4x+1)}{1} [/tex]

25=4x+1

To solve for x, let us subtract 1 from both sides

25-1=4x+1-1

24=4x+0

Or, 4x=24

To solve for x, let us divide by 4 on both sides

[tex] \frac{4}{4}x=\frac{24}{4} [/tex]

x=6

Answer: x=6

Given:

one U.S dollar equals 3.70 rands. To find how many rands does 450 U.S dollars equal to, we need to multiply 450 with 3.70

Solution:

[tex] 1 \; U.S \; dollars \; = \; 3.70 \; rands\\ \\ 450\; U.S\; dollars= 450 \times 3.7 \; rands=1665 \; rands [/tex]

Conclusion:

1665 rands equals $450.

Answer:      x<−2

(just had a test)

hope this helps

Answer:

the estimate is an over estimate because 2.19 time 6 equals 13.14 and you  sya 15 dollors so so it is an over estimate

Step-by-step explanation

i hope this helps i hate math

Final answer:

To estimate the circumference of a circle with a diameter of 13 yards, use the formula C = π * d where π is approximated as 3.14. The estimated circumference is 40.82 yards.

Explanation:

To estimate the circumference of a circle that has a diameter of 13 yards, you can use the formula for the circumference of a circle, which is π times the diameter (C = π * d). The value of π (pi) can be rounded to 3.14 for convenience in calculation. Using this rounded value of π:

Circumference = 3.14 * 13 yards= 40.82 yards

Therefore, the estimated circumference of the circle is 40.82 yards.

Answer:

y²-2y+3

Step-by-step explanation:

We write the dividend, y³-y²+y+3, under the box and the divisor, y+1, to the left of the box.

We first divide y³ by y; this is y².  We write this above the box, over -y².  We multiply the divisor by y²:  

y²(y+1) = y³+y²

This goes under the divisor.  We then subtract:

(y³-y²)-(y³+y²) = -2y².  We then bring down the next term, y; this gives us -2y²+y.

We then divide -2y² by y; this is -2y.  This goes above the box beside the y² in the quotient.  We then multiply the divisor by -2y:

-2y(y+1) = -2y²-2y

We now subtract:

(-2y²+y)-(-2y²-2y) = 3y.  We bring down the last term, 3; this gives us 3y+3.  We divide 3y by y; this is 3.  This goes beside the -2y in the quotient.  We then multiply this by the divisor:

3(y+1) = 3y+3.  We then subtract:  (3y+3)-(3y+3) = 0

This makes the quotient y²-2y+3.

Answer:

Step-by-step explanation:

If the object starts with a value [tex]V_0[/tex], and a annual percent of decay d, after a year the value will be

[tex]V_1 = V_0 * d[/tex]

after 2 years, taking now  [tex]V_1[/tex] as the starting value

[tex]V_2 = V_1 * d = V_0 * d * d [/tex]

[tex]V_2 = V_0 * d^2 [/tex]

and so on, after n years the value will be:

[tex]V_n = V_0 * d^n[/tex]

Now, in 1997 the value was $9500, in 2012 the value was $5000. Between 1997 and 2012 there are 15 years, so, our equation will be:

[tex] \$5000 = \$ 9500 * d^{15}[/tex]

Working it a little

[tex] \frac{\$5000}{\$ 9500} = d^{15}[/tex]

[tex] (\frac{\$5000}{\$ 9500})^{1/15} = d[/tex]

[tex] (0.5263})^{1/15} = d[/tex]

[tex] 0.9581 = d[/tex]

This mean that, after a year, the value will be at 95.81 %, this is, a decay rate of 4.19%.

Answer:

[tex]x=6.11  feet[/tex]

Step-by-step explanation:

Given that in a fulcrum weights are perfectly balanced.

One side 40 lb weight is there and another side 50 lb weight is given

Let x be the length of 40 lb weight from fulcrum.  Then 50 lbs is at a distance of 11-x.

Then we have since weights are perfectly balanced

[tex]40x = 50(11-x)\\90x=550\\x=6.111[/tex]

Thus we get [tex]x=6.11[/tex]feet

This is not enough information, sorry

For the first question please find the attached diagram.

As per the diagram, P is the upstream point and Q is the downstream point. The distance between P and Q is 22.5 miles.

Let the speed of the boat in the still waters of the lake be represented by S.

Then, when the boat travels upstream, the net speed of the boat will be (S-6) miles per hour because the river flows downstream and thus the speed of the boat will have to be subtracted from the speed of the river.

Now, we know that the relationship between the net speed, distance and time of travel is give as:

Distance = Net Speed x Time of travel

For the upstream ride of the board we know that Distance is 22.5 miles and Net Speed is (S-6). Therefore, the above equation will become:

[tex] 22.5=(S-6)\times T_{1} [/tex] where [tex] T_{1} [/tex] represents the time taken to travel upstream.

We can rearrange the above equation to be:

[tex] T_{1}=\frac{22.5 }{S-6} [/tex]......................(Equation 1)

By similar arguments we know that the downstream speed of the boat is S+6 and the distance travelled is the same and so the time taken to travel downstream (represented by [tex] T_{2} [/tex]) will be:

[tex] T_{2}=\frac{22.5}{S+6} [/tex]................(Equation 2)

Now, we know that the total time of travel should be 9 hours.

This means that: [tex] T_{1}+T_{2}=9 [/tex]............(Equation 3)

Plugging in the values of [tex] T_{1} [/tex] and [tex] T_{2} [/tex] from (Equation 1) and (Equation 2) into (Equation 3), we get:

[tex] \frac{22.5 }{S-6} +\frac{22.5 }{S+6}=9 [/tex]

Simplifying the above we will get a quadratic equation:

[tex] 9S^2-45S-54=0 [/tex]

The roots of this quadratic equation are:

[tex] S=-1 [/tex] and [tex] S=6 [/tex]

Since, speed cannot be negative, [tex] S=-1 [/tex] is out of consideration.

The speed of the boat in the lake is thus [tex] S=6 [/tex] miles per hour.

But we have a problem with S=6 too. The problem is that if S=6, then the boat will not be able to move upstream.

Let us solve problem 2

We are given that: [tex] \frac{x-2}{x+3}+\frac{10x}{x^2-9} [/tex]

We can rewrite it as:

[tex] \frac{x-2}{x+3}+\frac{10x}{(x-3)(x+3)} [/tex]

[tex] \frac{(x-3)(x-2)+10x}{(x-3)(x+3)} =\frac{x^2-5x+6+10x}{(x-3)(x+3)} [/tex]

Now, the numerator can be simplified as:

[tex] \frac{x^2+10x+6}{(x-3)(x+3)} =\frac{(x+3)(x+2)}{(x-3)(x+3)} =\frac{x+2}{x-3} [/tex]

Thus, our final simplified answer is:

[tex] \frac{x+2}{x-3} [/tex]

The restriction on the variable x is that it cannot be equal to either +3 or -3 as that would make the denominator of the original question equal to zero.

Thus, the restriction is [tex] x\neq \pm 3 [/tex]

Answer:

c is correct

Step-by-step explanation:

The scenario described in the question involves using a rope to whirl a bucket in a vertical circle. This relates to the physics concepts of centripetal force, tension in the rope, and gravity. The tension in the rope provides the centripetal force needed for the curved path of the bucket, and the scenario implies a constant speed and a counterbalance between tension and gravity at the highest point of the circle.

The question involves a principle of physics known as centripetal force, which is the force that makes a body follow a curved path. Its direction is always orthogonal to the motion of the body, towards the fixed point of the instantaneous center of curvature of the path.

In this case, the 60 cm rope tied to the bucket and whirled in a vertical circle becomes a system where the bucket, the force of gravity, and the tension in the rope all play a part. When the bucket is whirled, it experiences a centripetal force that keeps it moving in a circle. This force comes from the tension in the rope, which always pulls the bucket towards the center of the circle (the hand holding the rope). Hence, there are two key forces in this scenario - the force of gravity, pulling the bucket downward, and the tension in the rope providing the centripetal force.

Since the question does not state otherwise, we can reasonably assume that the movement is steady, implying the speed of the bucket is constant. In such a case, the swinging bucket will rise to a certain point, where the tension in the rope and the gravity cancel each other, causing the bucket to start falling back to the other side, therefore, forming the vertical circular path.

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