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2 years ago

Line l is parallel to line e in the figure below.

Mathematics

2 answers:

VLD [36.1K]2 years ago

7 0

Answer:

A B D

Step-by-step explanation:

correct.

yulyashka [42]2 years ago

3 0

After the 2 angles the third and fifth line are great

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Diane needs 165 pieces of candy to hand out to guests at a party she is hosting. Each bag of candy contains 20 pieces. She has 2

Debora [2.8K]

20x + 25 = 165

x represents the number of bags. You can solve the equation to make sure the answer is 7.

20x + 25 = 165

Subtract 25 from both sides.

20x = 140

Divide 20 off of both sides.

x = 7

5 0

3 years ago

Make x the subject of the formula<br> х<br> +12c = 5d<br> 63

dezoksy [38]

X=5db^3-2cb^3
Workup in photo below.
Good luck

7 0

3 years ago

Using a phone card to make a long distance call costs a flat fee of $0.57 plus $0.31 per minute starting with the first

aliina [53]

Answer:

it's d

Step-by-step explanation:

multiply 0.31 by the amount of minutes used (27) and add the flat fee to it.

0.31×27+(0.57)= 8.94

7 0

2 years ago

Hindle McKingleberry is a young child and has $3 in his bank account on the first day of the

matrenka [14]

Answer:you got this

Step-by-step explanation:

7 0

3 years ago

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:

kondaur [170]

$\text{Proof by induction:}$
$\text{Test that the statement holds or n = 1}$

$LHS = (3 - 2)^{2} = 1$
$RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS$
$\text{Thus, the statement holds for the base case.}$

$\text{Assume the statement holds for some arbitrary term, n= k}$
$1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}$

$\text{Prove it is true for n = k + 1}$
$RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$

$LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}$
$= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}$
$= \frac{k(6k^{2} + 15k + 11) + 2}{}$
$= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$
$= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}$
$= RHS$

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.

3 0

3 years ago