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Properties of the Rhombus:

Since the given rhombus has all sides equal (4 cm) and two acute angles of 90 degrees each, it's essentially a square.

Diagonals of the Square:

The diagonals of a square bisect each other at right angles and divide the square into four congruent right triangles. Each diagonal has a length equal to the square root of twice the side length squared (sqrt(2s^2)).

In this case, the side length (s) is 4 cm. Therefore, the length of each diagonal (d) is:

d = sqrt(2 * s^2)

d = sqrt(2 * 4^2 cm^2)

d = sqrt(32) cm

d = 4√2 cm (since the square root of 32 can be simplified to 4 times the square root of 2)

Radius of the Inscribed Circle:

The inscribed circle in a square touches the midpoint of each side. Since the diagonals bisect each other at the center of the square and are perpendicular, they also pass through the center of the inscribed circle.

Therefore, the radius (r) of the inscribed circle is half the length of a diagonal divided by two (since there are two diagonals and the circle touches them both at their midpoints):

r = (d / 2) / 2

r = (4√2 cm) / 2 / 2

r = √2 cm

Answer:

The radius of the inscribed circle is √2 centimeters.

Workers and Days:

We know 5 workers can complete the job in T days (initially unknown).

Impact of Additional Worker:

Hiring one more worker reduces the completion time by 12 days, meaning it takes (T-12) days with 6 workers.

Work Relationship: There's a constant amount of work to be done. We can represent this with the following equation:

Work = Rate x Time

In this case, Work is constant (the job itself).

Rate represents the combined speed of the workers (more workers = faster rate).

Time is the number of days to complete the job.

Translating to Equations:

From the work relationship, we can write equations for both scenarios (5 and 6 workers):

5 Workers: Work = (Rate of 5 Workers) * T days

6 Workers: Work = (Rate of 6 Workers) * (T-12) days

Since the Work is the same, we can equate both expressions:

(Rate of 5 Workers) * T days = (Rate of 6 Workers) * (T-12) days

Relating Rates to Workers:

We can assume the rate of each worker is constant (say, w units of work per day).

Therefore, the Rate of n Workers = n * w. Substituting this into the equation from step 5:

5w * T = 6w * (T-12)

Solving for T:

Expand and solve for T:

5wT = 6wT - 72w

T = 72 days (This is the original completion time with 5 workers)

New Completion Time Requirement:

We need to reduce the completion time by 32 days. This means the new desired completion time is (T-32) days.

Workers Needed for Faster Completion:

We can again use the work relationship:

Work = (Rate of x Workers) * (T-32) days

We know the Work is constant and the original time (T). We need to find the number of workers (x) required to achieve the new completion time (T-32).

Solving for Additional Workers:

Since the Rate of each worker is w (assumed constant), we can rewrite the equation from step 9:

5w * T = x * w * (T-32)

We know T from step 7 (72 days). Substitute and solve for x (additional workers):

5 * 72 = x * (72 - 32)

x = 12

Answer: We already have 5 workers. To achieve the 32-day reduction, we need to hire 12 additional workers. Therefore, a total of 5 + 12 = 17 workers are needed.

In a geometric sequence, the ratio between successive terms is constant. In this case, the common ratio is -1/3, which means each term is -1/3 times the term before it.

For the sequence to be greater than 1, the terms need to be positive. Since the first term (1000) is positive and the common ratio is negative, the sequence will alternate between positive and negative terms.

Since there are 400 terms, half (200) will be positive and the other half negative. However, the question asks for terms greater than 1, not just positive terms.

Finding the first term greater than 1:

We can analyze the sequence using the formula for the nth term of a geometric sequence:

tn = a * r^(n-1)

tn: nth term

a: first term (1000)

r: common ratio (-1/3)

n: term number

We want to find the value of n (n = k) where tk > 1.

Since the terms alternate in sign, we need to consider the even terms (positive terms). Let's rewrite the formula for even terms (n = 2k):

t(2k) = 1000 * (-1/3)^(2k-1)

For tk to be greater than 1, we need the exponent of -1/3 to be negative enough to make the result positive.

The smallest even term that satisfies this condition is t4 (fourth term), where k = 2.

t4 = 1000 * (-1/3)^(2(2)-1) = 1000 * (-1/9) = -100/9 < 1

t6 (sixth term) = 1000 * (-1/3)^(2(3)-1) = 1000 * (1/27) = 100/27 > 1

Therefore, the first term greater than 1 is the sixth term (n = 6).

Counting terms greater than 1:

Since the sequence alternates in sign, every two terms after the sixth term will have one term greater than 1. With 400 terms, there are 200 even terms (positive or negative).

Out of these 200 terms, the first 4 (t2, t4, t6, and t8) won't be greater than 1. The remaining 196 terms will have 98 terms greater than 1 (every other term).

So, there are 98 terms greater than 1 in the sequence.

To solve the equation, we first try to get rid of the denominators.

Multiplying both sides of the equation by (x+1)(x+2), we get [x^2 + 2x + x^2 + x = kx(x + 1)(x + 2).]

Expanding the right side, we get [2x^2 + 3x = kx^3 + kx^2 + 2kx.]

Moving all the terms to one side, we get [0 = kx^3 + (k - 1)x^2 + (2k - 3)x.]

Since we are looking for complex solutions, we can treat x as a complex number.

For the equation to have exactly two complex solutions, the discriminant of the quadratic part must be nonzero.

That is, [(k - 1)^2 - 4 \cdot k \cdot (2k - 3) > 0.]

Expanding, we get [k^2 - 2k + 1 - 8k^2 + 24k - 12 > 0.]

This simplifies to [-7k^2 + 22k - 11 > 0.]

We can factor this as [(k - 1)(7k - 11) > 0.]

Thus, k must be in one of the intervals (−∞,1/7)∪(11/7,∞).

Now, let's check that for these values of k, the equation has exactly two complex solutions.

For k=0, the equation becomes x+1x​+x+2x​=0, which has the complex solutions x=−1 and x=−2

For k=2, the equation becomes x+1x​+x+2x​=2x, which has the complex solutions x=0 and x=−23​.

Thus, the ony possible k is 11/7.

For problem 2:

Let's denote the following:

O: Center of both circles

A and B: Endpoints of the chord (chord length AB = 10)

P: Point of tangency between the chord and the smaller circle

r: Radius of the smaller circle

R: Radius of the larger circle

Since the chord is tangent to the smaller circle at point P, we know that OP is perpendicular to AB. Additionally, since O is the center of both circles, line segment OP bisects chord AB (AP = BP = 5).

Finding the Radius of the Smaller Circle (r):

Using the Pythagorean Theorem in right triangle APO:

AP^2 + OP^2 = r^2 (since AP = BP = 5)

5^2 + OP^2 = r^2

We don't have the value of OP yet, but we can find it using the information about the larger circle and the chord.

Finding the Radius of the Larger Circle (R):

In right triangle OBP:

OB^2 + OP^2 = R^2 (since OB = half the chord length = 5)

5^2 + OP^2 = R^2

Relating the Radii (R and r):

Since the chord is tangent to the smaller circle, the distance between the center (O) and the point of tangency (P) is the radius of the smaller circle (r).

Additionally, this distance (OP) is also the difference between the radii of the larger and smaller circles (R - r).

Therefore, we have: OP = r and OP = R - r

Setting these two expressions equal to each other:

r = R - r

Solving for r (the smaller radius):

2r = R r = R/2

Substitute r in terms of R back into Equation 1:

5^2 + (R/2)^2 = R^2

Expand and rearrange:

R^2 - 5R - 25 = 0

Factor the equation:

(R - 10)(R + 2.5) = 0

Since the radius cannot be negative, we have R = 10.

Therefore, the radius of the smaller circle (r) is r = R/2 = 5.

Area of the Ring-Shaped Region:

The area of the ring is the difference between the area of the larger circle and the area of the smaller circle:

Area of Ring = π * R^2 - π * r^2

Substitute the values of R and r:

Area of Ring = π * (10)^2 - π * (5)^2

Area of Ring = π * (100 - 25)

Area of Ring = 75π

Therefore, the area of the ring-shaped region is 75π square units.

The largest term in the binomial expansion of (a+b)n occurs when k = n/2 (for even n) or k = (n - 1)/2 (for odd n) in the binomial coefficient \binom{n}{k} = \frac{n!}{k!(n-k)!}.

In this case, n = 31 (odd). Therefore, the largest term occurs when k = (31 - 1)/2 = 15.

The term we want is \binom{31}{15} = \frac{31!}{15!16!}. We can re-write this as:

\binom{31}{15} = \frac{31 \times 30 \times 29 \times \cdots \times 16}{16 \times 15 \times 14 \times \cdots \times 1}

Notice that most of the terms in the numerator and denominator cancel out, leaving us with:

\binom{31}{15} = \frac{31 \times 30 \times 29}{3 \times 2 \times 1} = 15450

The denominator, b, is thus 3 \times 2 \times 1 = \boxed{6}.

Note that angle CEA = angle DCA + angle DAC and angle BDE = angle DCE + angle EDC, so angle DCA = angle CEA - angle DAC and angle DCE = angle EDC - angle CED.

Also, angle BDE = angle ABD + angle ADB = angle AEC + angle DAC, so angle DCE + angle DCA = (angle EDC - angle CED) + angle DCA.

Therefore, angle ACE = angle ACD + angle DCE = angle CEA, which implies that triangle ACE is isosceles.