A . 5 - 4/8 B. -1/8 - 1/4 C. -1/4 - (-1/8) or D. -1/6 + (-5/24)

Plz solve, free brainlest if you solve, this my little sis’s homework

faust18 [17]

Answer:

5x6=30

6x50=300

those are two lol.

3 0

2 years ago

The sum of an infinite geometric series is three times the first term. Find the common ratio of this series.

adelina 88 [10]

Answer:

$\frac{2}{3}$

Step-by-step explanation:

The sum of an infinite geometric series is expressed according to the formula;

$S_\infty = \dfrac{a}{1-r}$ where;

a is the first term of the series

r is the common ratio

If the sum of an infinite geometric series is three times the first term, this is expressed as $S_\infty = 3a$

Substitute $S_\infty = 3a$ into the formula above to get the common ratio r;

$3a = \dfrac{a}{1-r} \\\\$

$cross \ multiply\\\\3a(1-r) = a\\\\3(1-r) = 1\\$

open the parenthesis

$3 - 3r = 1\\\\$

subtract 3 from both sides

$3 - 3r -3= 1-3\\\\-3r = -2\\\\r = \frac{2}{3}$

Hence the common ratio of this series is $\frac{2}{3}$

3 0

2 years ago

Solve for x in the figure below

fredd [130]

Answer:

x = 35.4

Step-by-step explanation:

Sum of all interior angles in a triangle = 180°

Therefore:

(x + 2)° + (2x + 2)° + (2x - 1)° = 180°

x + 2 + 2x + 2 + 2x - 1 = 180

Collect like terms

x + 2x + 2x + 2 + 2 - 1 = 180

5x + 3 = 180

Subtract 3 from each side

5x + 3 - 3 = 180 - 3

5x = 177

Divide both sides by 5

5x/5 = 177/5

x = 35.4

3 0

2 years ago

If the original square had a side length of

irina [24]

Answer:

Part a) The new rectangle labeled in the attached figure N 2

Part b) The diagram of the new rectangle with their areas in the attached figure N 3, and the trinomial is $x^{2} +11x+28$

Part c) The area of the second rectangle is 54 in^2

Part d) see the explanation

Step-by-step explanation:

The complete question in the attached figure N 1

Part a) If the original square is shown below with side lengths marked with x, label the second diagram to represent the new rectangle constructed by increasing the sides as described above

we know that

The dimensions of the new rectangle will be

$Length=(x+4)\ in$

$width=(x+7)\ in$

The diagram of the new rectangle in the attached figure N 2

Part b) Label each portion of the second diagram with their areas in terms of x (when applicable) State the product of (x+4) and (x+7) as a trinomial

The diagram of the new rectangle with their areas in the attached figure N 3

we have that

To find out the area of each portion, multiply its length by its width

$A1=(x)(x)=x^{2}\ in^2$