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2 years ago

Is (a + b) and (a - b)? perfect square trinomials? O True O False

Mathematics

2 answers:

lisabon 2012 [21]2 years ago

8 0

Answer:

False

Step-by-step explanation:

Pos and negs cannot be perfect square trinomials

NeTakaya2 years ago

3 0

The answer should be false

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Triangle JKL has an area of in2 and a perimeter of in.

S_A_V [24]

Answer:

Area = 24 in²

Perimeter = 24in

Step-by-step explanation:

To find the area given the length of the 3 sides:

a= 8in

b= 6.4in + 3.6in= 10in

c= 6in

Add them together:

8in + 10in + 6in

= 24in²

To find the perimeter you do the same thing:

a= 8in

b= 10in

c= 6in

Add them together

=24in

DO <u>NOT</u> SQUARE THE ANSWER WHEN FINDING <u>PERIMETER</u>

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2 years ago

A customer wants to purchase a pair of shoes that cost $45 and the state sales tax is 7%.

Rasek [7]

Answer:

$22.84

Step-by-step explanation:

The tax is calculated by 0.075 * 21.25 = 1.59375 or $1.59. So the total bill is the cost of items purchased plus tax, which is $21.25 + $1.59 = $22.84.

3 0

3 years ago

What does 1 1/16-3/8 equal?

Artemon [7]

Answer:

$\frac{11}{16}$

6 0

2 years ago

R varies inversely as v2 if r=120 when v=1 ,find r when v=10

makvit [3.9K]

<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ r = k/v^2

Substitute the values in:

120 = k/1^2

Multiply both sides by 1^2 to find k

k = 120

Substitute new values in:

r = 120/10^2

Solve:

r = 1.2

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

6 0

3 years ago

How do I solve: 2 sin (2x) - 2 sin x + 2√3 cos x - √3 = 0

ziro4ka [17]

Answer:

$\displaystyle x = \frac{\pi}{3} +k\, \pi$ or $\displaystyle x =- \frac{\pi}{3} +2\,k\, \pi$ , where $k$ is an integer.

There are three such angles between $0$ and $2\pi$ : $\displaystyle \frac{\pi}{3}$ , $\displaystyle \frac{2\, \pi}{3}$ , and $\displaystyle \frac{4\,\pi}{3}$ .

Step-by-step explanation:

By the double angle identity of sines:

$\sin(2\, x) = 2\, \sin x \cdot \cos x$ .

Rewrite the original equation with this identity:

$2\, (2\, \sin x \cdot \cos x) - 2\, \sin x + 2\sqrt{3}\, \cos x - \sqrt{3} = 0$ .

Note, that $2\, (2\, \sin x \cdot \cos x)$ and $(-2\, \sin x)$ share the common factor $(2\, \sin x)$ . On the other hand, $2\sqrt{3}\, \cos x$ and $(-\sqrt{3})$ share the common factor $\sqrt[3}$ . Combine these terms pairwise using the two common factors:

$(2\, \sin x) \cdot (2\, \cos x - 1) + \left(\sqrt{3}\right)\, (2\, \cos x - 1) = 0$ .

Note the new common factor $(2\, \cos x - 1)$ . Therefore:

$\left(2\, \sin x + \sqrt{3}\right) \cdot (2\, \cos x - 1) = 0$ .

This equation holds as long as either $\left(2\, \sin x + \sqrt{3}\right)$ or $(2\, \cos x - 1)$ is zero. Let $k$ be an integer. Accordingly:

$\displaystyle \sin x = -\frac{\sqrt{3}}{2}$ , which corresponds to $\displaystyle x = -\frac{\pi}{3} + 2\, k\, \pi$ and $\displaystyle x = -\frac{2\, \pi}{3} + 2\, k\, \pi$ .
$\displaystyle \cos x = \frac{1}{2}$ , which corresponds to $\displaystyle x = \frac{\pi}{3} + 2\, k \, \pi$ and $\displaystyle x = -\frac{\pi}{3} + 2\, k \, \pi$ .

Any $x$ that fits into at least one of these patterns will satisfy the equation. These pattern can be further combined:

$\displaystyle x = \frac{\pi}{3} + k \, \pi$ (from $\displaystyle x = -\frac{2\,\pi}{3} + 2\, k\, \pi$ and $\displaystyle x = \frac{\pi}{3} + 2\, k \, \pi$ , combined,) as well as
$\displaystyle x =- \frac{\pi}{3} +2\,k\, \pi$ .

7 0

3 years ago