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2 years ago

9

A and B represent whole numbers, and A▯B means A² + B² - 2AB. What is the value of 9▯(5▯2)?

1 answer:

andreev551 [17]2 years ago

3 0

It’s 54 because I did the math to get the answer

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What is 2/5 as percentage

Alina [70]

First, change 2/5 into a decimal by dividing 2 by 5. Decimal= .4. Now move the decimal 2 places to the right to find percentage. Answer is 40%

3 0

3 years ago

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Use the triangle below to find cos V.

Semenov [28]

Answer:

$cos\ V = \frac{2\sqrt{29}}{29}$

Step-by-step explanation:

Given

The attached triangle

Required

Find cos V

In trigonometry:

$cos\theta = \frac{Adjacent}{Hypotenuse}$

In this case:

$cos V= \frac{TV}{UV}$

Where

$TV = 6$

and

$UV^2 = TV^2 + TU^2$ -- Pythagoras

$UV^2 = 6^2 + 15^2$

$UV^2 = 261$

Take square roots

$UV = \sqrt{261$

$UV = \sqrt{9*29$

$UV = \sqrt{9} *\sqrt{29$

$UV = 3\sqrt{29$

So:

$cos V= \frac{TV}{UV}$

$cos\ V = \frac{6}{3\sqrt{29}}$

$cos\ V = \frac{2}{\sqrt{29}}$

Rationalize:

$cos\ V = \frac{2}{\sqrt{29}}*\frac{\sqrt{29}}{\sqrt{29}}$

$cos\ V = \frac{2\sqrt{29}}{29}$

8 0

2 years ago

A^2 + b^2 + c^2 = 2(a − b − c) − 3. (1) Calculate the value of 2a − 3b + 4c.

Verdich [7]

Answer:

$2a - 3b + 4c = 1$

Step-by-step explanation:

Given

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Required

Determine $2a - 3b + 4c$

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Open bracket

$a^2 + b^2 + c^2 = 2a - 2b - 2c - 3$

Equate the equation to 0

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 3 = 0$

Express 3 as 1 + 1 + 1

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 1 + 1 + 1 = 0$

Collect like terms

$a^2 - 2a + 1 + b^2 + 2b + 1 + c^2 + 2c + 1 = 0$

Group each terms

$(a^2 - 2a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

Factorize (starting with the first bracket)

$(a^2 - a -a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$(a(a - 1) -1(a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1) (a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + b+b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b(b + 1)+1(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + c+c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c(c + 1)+1(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)^2) = 0$

Express 0 as 0 + 0 + 0

$(a - 1)^2 + (b + 1)^2 + (c + 1)^2 = 0 + 0+ 0$

By comparison

$(a - 1)^2 = 0$

$(b + 1)^2 = 0$

$(c + 1)^2 = 0$

Solving for $(a - 1)^2 = 0$

Take square root of both sides

$a - 1 = 0$

Add 1 to both sides

$a - 1 + 1 = 0 + 1$

$a = 1$

Solving for $(b + 1)^2 = 0$

Take square root of both sides

$b + 1 = 0$

Subtract 1 from both sides

$b + 1 - 1 = 0 - 1$

$b = -1$

Solving for $(c + 1)^2 = 0$

Take square root of both sides

$c + 1 = 0$

Subtract 1 from both sides

$c + 1 - 1 = 0 - 1$

$c = -1$

Substitute the values of a, b and c in $2a - 3b + 4c$

$2a - 3b + 4c = 2(1) - 3(-1) + 4(-1)$

$2a - 3b + 4c = 2 +3 -4$

$2a - 3b + 4c = 1$

7 0

3 years ago

Lenny sold 576 tacos in 48 hours what was lenny's average rate of tacos sales

cupoosta [38]

If your looking for hourly they sold a total of 12 tacos per hour.
If your looking for daily they sold a total of 288 tacos per day.

4 0

3 years ago

Read 2 more answers

If 4C + 3 = 15, what is the value of C?

Soloha48 [4]

4c+3=15
You need to get rid of the 3
So you minus it from each side so you
4c=12
12/4=3
So c =3

4 0

3 years ago

Read 2 more answers