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3 years ago

Please answer quickly

Mathematics

2 answers:

sertanlavr [38]3 years ago

8 0

Answer:

B. Subtraction

Step-by-step explanation:

Pie3 years ago

7 0

Answer: subtraction
Reasoning: You need to subtract 3 from both sides first.

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What is the area of the composite figure?

german

Answer:

16 + 6pi cm

Step-by-step explanation:

We have a square of length 4 cm

A = s^2 = 4^2 = 16

We have 3 semi circles

with radius 2

A semi circle has an area of

1/2 pi r^2 = 1/2 pi (2)^2 = 1/2 (4pi) = 2pi

There are 3 of them

3 * 2 pi = 6pi

Add the areas together for the square and the semicircles

16 + 6pi

6 0

3 years ago

Simplify the following<br>(4x + 3) (5x + 2)

lidiya [134]

Answer:

20x^2+23x+6

Step-by-step explanation:

(4x+3)(5x+2)

(4x)(5x)+2(4x)+(5x)3+3(2)

20x^2+8x+15x+6

20x^2+23x+6

Hope that helps :)

8 0

3 years ago

Read 2 more answers

Round 2 5/6 to the nearest half

Mekhanik [1.2K]

(2 5/6):2 = (2 5/6) * 1/2 = (12/6 + 5/6)*1/2 = 17/6*1/2 = 17/12 = 1 5/12

7 0

3 years ago

Read 2 more answers

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

HELP ME!! im so stumped. will give brainliest!

Olegator [25]

Answer:

$x=1+\sqrt{3}$ and $x=1-\sqrt{3}$