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2 years ago

Order of operations : simplify completely (6-2X2)X3

Mathematics

1 answer:

ELEN [110]2 years ago

5 0

Answer:

6 is the answer.

Step-by-step explanation:

Using PEMDAS, let's solve.

(6 - 2 x 2) x 3
(6 - 4) x 3
2 x 3
6

Therefore, 6 is the answer.

Hoped this helped.

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If 3 pens cost 6 dollars how much is each pen? Please help I need this asap

NemiM [27]

Answer: $2

Step-by-step explanation:

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1 year ago

Audrey's bank account is currently at $75 a fee of 12.50 is deducted each month from Audreys she must maintain a balance greater

meriva

Answer:

5 months

Step-by-step explanation:

Current balance = $75

Monthly deductions = $12.50

75 - 12.5h

Where ,

h = number of months for deduction

she must maintain a balance greater than zero dollars

how many months will Audrey's account stay above zero dollars

0 > 75 - 12.5h

0 - 75 > -12.5h

-75 > - 12.5h

h < -75 / 12.5h

h < 6 months

h < 6 months means h must be less than 6 months which is 5 months

Audrey's account will stay above zero dollars for 5 months and will equal zero dollars in the 6th month

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3 years ago

Integration of sin^2 2x cos^3 2x dx

lilavasa [31]

$\int\left[\sin^2(2x)\cos^3(2x)\right]dx=\int\left[\sin^2(2x)\cos^2(2x)\cos(2x)\right]dx\\\\\text{Use the trigonometric identity}\sin^2(x)+\cos^2(x)=1\to\cos^2(x)=1-\sin^2(x)\\\\=\int\left\{\sin^2(2x)[1-\sin^2(2x)]\cos(2x)\right\}dx\\\\\text{substitute}\ \sin2x=t\ \to\ 2\cos2x\ dx=dt\ \to\ \cos2x\ dx=\dfrac{1}{2}\ dt\\\\=\dfrac{1}{2}\int\left[t^2(1-t^2)\right]dt=\dfrac{1}{2}\int(t^2-t^4)dt=\dfrac{1}{2}\left(\dfrac{1}{3}t^3-\dfrac{1}{5}t^5\right)+C$

$=\dfrac{1}{2}\left(\dfrac{1}{3}\sin^3(2x)-\dfrac{1}{5}\sin^5(2x)\right)+C$

4 0

3 years ago

Simplify. Evaluate the numerical bases. 3^2 b^2 c^-4 *3^-1 b^-5 c^7

photoshop1234 [79]

Hope I can be of assistance!

$3^2b^2c^{-4}\cdot \:3^{-1}b^{-5}c^7 \ \textgreater \ \mathrm{Apply\:exponent\:rule}:\ \:a^b\cdot \:a^c=a^{b+c}$
$3^{-1}\cdot \:3^2=\:3^{2-1}=\:3^1=\:3 \ \textgreater \ 3b^{-5}b^2c^{-4}c^7$

$\mathrm{Apply\:exponent\:rule}: \:a^b\cdot \:a^c=a^{b+c} \ \textgreater \ b^{-5}b^2=\:b^{2-5}=\:b^{-3}$
$3b^{-3}c^{-4}c^7$

$\mathrm{Apply\:exponent\:rule}: \:a^b\cdot \:a^c=a^{b+c} \ \textgreater \ c^{-4}c^7=\:c^{-4+7}=\:c^3$
$3b^{-3}c^3$

$\mathrm{Apply\:exponent\:rule}: \:a^{-b}=\frac{1}{a^b} \ \textgreater \ b^{-3}=\frac{1}{b^3} \ \textgreater \ 3\cdot \frac{1}{b^3}c^3$

$\mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \ \frac{1\cdot \:3c^3}{b^3}$

Finally
$\mathrm{Apply\:rule}\:1\cdot \:a=a \ \textgreater \ \frac{3c^3}{b^3}$

Hope this helps!

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3 years ago

Helaine graphed the equation 12 x -4y=3

koban [17]

Answer:-0.0625

Step-by-step explanation:

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3 years ago