All Answers 358087 Answers

Can't even copy homework correctly, are you dull or something?

imagen cheating on homework lmao so bad no skill

I apologize for my mistake. You are correct that b must be less than 7. 

Using the formula for the sum of an infinite geometric series, we have:

S = b/7 + b^2/7^2 + b^3/7^3 + b^4/7^4 + ... + b^n/7^n

This is a geometric series with first term b/7 and common ratio b/7. The sum of an infinite geometric series is:

S = a / (1 - r)

where a is the first term and r is the common ratio. 

Substituting a = b/7 and r = b/7, we get:

S = (b/7) / (1 - b/7)

Simplifying the denominator, we get:

S = (b/7) / (7-b)/7

Simplifying further, we get:

S = b / (7-b)

Since the sum of the series is equal to 7, we have:

b / (7-b) = 7

Multiplying both sides by (7-b), we get:

b = 7(7-b)

Expanding, we get:

b = 49 - 7b

Solving for b, we get:

8b = 49

b = 49/8

Since b must be less than 7, the solution is:

b = 49/8 = 6.125

Therefore, the value of b that satisfies the equation is 6.125.

Sum = F / [1  - R],  where =First term,  R==constant ratio

7 = (49/8) / [1  -  R], solve for R

R= 1 /8

b =49 / 8 = 6+1/8

The value of s is 8.

The value of s is -13

The value of s is 11.

To solve this problem, we can use the fact that the line externally tangent to both circles is perpendicular to the segment connecting the centers of the circles (the common external tangent). Let $O_1$ and $O_2$ be the centers of the circles with radii of lengths $5$ and $3$, respectively [1][2]. Let $D$ be the intersection point of the common external tangent with $AB$. Then, we have $AD=5+3=8$.

Since the line is perpendicular to $O_1O_2$, we have $\triangle O_1O_2D$ is a right triangle with hypotenuse $O_1O_2$ and leg $BC$. Therefore, by the Pythagorean Theorem, we have $$ BC^2 = O_1O_2^2 - O_1D^2 = (5+3)^2 - 5^2 = 3^2 \cdot 4 = 36. $$

Taking the square root of both sides, we get $BC=\sqrt{36}=6$, which is the length we were asked to find.

Therefore, $BC=\boxed{6}$.

The figure has area 12.5 square centimeters.

The product of p(x) and q(x) is:

p(x)q(x) = (ax^2 - 5x + 6)(bx^2 + 3x - 2) = abx^4 + (3a+5b)x^3 + (2a+15+3b)x^2 + (10a+6b)x - 12

We're given that the product has a 20x^3 term and a 30x^2 term. This means that:

3a + 5b = 20 ... (1) 2a + 3b + 15 = 30 ... (2)

Solving equations (1) and (2) simultaneously, we get:

a = 5 b = 1

Therefore, a + b = 5 + 1 = 6.

P=300;   R=0.04/12;  N=2*12;    FV=P*((1 + R)^N - 1)/ R

FV = $ 7,482.87 - balance of the account after 2 years.

The area is ab/4.

The given series is a geometric series with first term a and common ratio r.

Here, a = 1/5 and r = 3/5.

Using the formula for the sum of an infinite geometric series with |r| < 1, we have:

Sum = a / (1 - r)

Sum = (1/5) / (1 - 3/5)

Sum = (1/5) / (2/5)

Sum = 1/2

Therefore, the sum of the given series is 1/2.

Here's a non calculus approach.

Begin by noticing that

\(7^{2}+24^{2}=49+576=625=25^{2},\)

so 7, 24, 25 can be used to as the sides of a right-angled triangle.

Let the angle opposite the 7 be \(\phi,\)

then

\(\sin\phi=7/25 \text{ and } \cos\phi=24/25.\)

\(\displaystyle 7\cos\theta+24\sin\theta+7\\ = 25(\frac{7}{25}\cos\theta+\frac{24}{25}\sin\theta)+7\\ =25(\sin\phi\cos\theta+\cos\phi\sin\theta)+7 \\ =25\sin(\theta+\phi)+7.\)

Maximum value will be 25 + 7 = 32 occuring when

 \(\sin(\theta+\phi)=1, \\ \theta=\pi/2-\tan^{-1}(7/24),\\ 0\leq\theta\leq\pi/2.\)

As follows:

If the diameter of each is 2" and the total length is 1' then    12/2=6

There will be 6 semicircles on the top and 6 semicircles on the bottom which makes 6 circles

A= pi* 1^2  =  pi

To find the value of m such that the triangle enclosed by the line x+y=m and the coordinate axes has an area of 200 square units, we can use the formula for the area of a triangle:

Area = 1/2 * base * height

In this case, the base of the triangle is the distance between the x-intercept and the point where the line x+y=m intersects the y-axis. The height of the triangle is the distance between the y-intercept and the point where the line x+y=m intersects the x-axis.

To find these points, we can set x=0 and y=0 in the equation x+y=m, and solve for the corresponding variables:

When x=0, y=m, so the line intersects the y-axis at the point (0,m).
When y=0, x=m, so the line intersects the x-axis at the point (m,0).

The distance between the x-intercept and the point (0,m) is m, and the distance between the y-intercept and the point (m,0) is m. Therefore, the base and height of the triangle are both equal to m.

Plugging in the values of base and height into the formula for the area, we get:

Area = 1/2 * m * m = 1/2 * m^2

We want the area to be 200 square units, so we can set up an equation:

1/2 * m^2 = 200

Solving for m, we get:

m^2 = 400

m = ±20

Since we are looking for a positive value of m, the answer is:

m = 20

Therefore, the value of m such that the triangle enclosed by the line x+y=20 and the coordinate axes has an area of 200 square units is 20. 

Using the change of base formula for logarithms, we can rewrite the given equations as:

log m / log 36 = log n / log 42 = log (3m + 4n) / log 49

Let's start by equating the first two expressions:

log m / log 36 = log n / log 42

log m log 42 = log n log 36

log (m^log 42) = log (n^log 36)

m^log 42 = n^log 36

Taking the log of both sides with base 42, we get:

log 42 (m^log 42) = log 42 (n^log 36)

log 42 m = log 36 n

Using this result, we can write:

log m / log 36 = log n / log 42 = log (3m + 4n) / log 49

log (3m + 4n) / log 49 = log m / log 36 = log n / log 42

log (3m + 4n) = log 49 (m^(log 36) ) = log 49 (n^(log 42) )

3m + 4n = 49^(log 36) * m = 49^(log 42) * n

Using the relationship between logarithms and exponents, we can write:

49^(log 36) = 6^2^(2 log 3) = 6^4 log 3

49^(log 42) = 7^2^(2 log 3/2) = 7^2^(log 9) = 7^2

Substituting these values, we get:

3m + 4n = 6^4 log 3 * m = 49n

Simplifying, we get:

(6^4 log 3 - 4) n = 3m

Therefore, the possible values of n/m are given by:

n/m = (3m) / [(6^4 log 3 - 4) m] = 3 / (6^4 log 3 - 4)

Since there is only one possible value of n/m, we can directly compute it:

n/m = 3 / (6^4 log 3 - 4) = 3/12 = 1/4.