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3 years ago

Simplify the expression. : 5 + 4 x (8 - 6) square

Mathematics

2 answers:

UNO [17]3 years ago

8 0

Answer:

8x+5

Step-by-step explanation:

8-6=2

2× 4x= 8x

8x+5

tensa zangetsu [6.8K]3 years ago

6 0

Answer:

<h3> $\boxed{ \huge{ \bold{ \sf{ \boxed{21}}}}}$ </h3>

Step-by-step explanation:

Use PEMDAS rule :

P = Parentheses

E = Exponents

M = Multiplication

D = Division

A = Addition

S = Subtraction

Let's solve :

$\sf{5 + 4 \times {(8 - 6)}^{2} }$

⇒ $\sf{5 + 4 \times {2}^{2} }$

⇒ $\sf{5 + 4 \times 4}$

⇒ $\sf{5 + 16}$

⇒ $\sf{21}$

Hope I helped!

Best regards!

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no links PLEASE HELP DUE IN 10 MINUTES you can use a calculator if you want. No links please help this is very important i will

Debora [2.8K]

Answer:

0.334

Step-by-step explanation:

Basically to show reasoning: 16*215= 3340

4 decimal places, so you would move it four decimal places from 3340.

Which equals to: 0.334

Hope this helped! :)

6 0

3 years ago

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.

Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Step-by-step explanation:

GIven that:

$f(x) = 5e^{-x^2} cos (4x)$

The Maclaurin series of cos x can be expressed as :

$\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+... \ \ \ (1)}$

$\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0} \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!} -\dfrac{x^6}{3!}+... \ \ \ (2)}$

From equation(1), substituting x with (4x), Then:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}- \dfrac{(4x)^6}{6!}+...}$

The first three terms of cos (4x) is:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}-...}$

$\mathtt{cos (4x) = 1 - \dfrac{16x^2}{2}+ \dfrac{256x^4}{24}-...}$

$\mathtt{cos (4x) = 1 - 8x^2+ \dfrac{32x^4}{3}-... \ \ \ (3)}$

Multiplying equation (2) with (3); we have :

$\mathtt{ e^{-x^2} cos (4x) = ( 1- x^2 + \dfrac{x^4}{2!} ) \times ( 1 - 8x^2 + \dfrac{32 \ x^4}{3} ) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1+ (-8-1)x^2 + (\dfrac{32}{3} + \dfrac{1}{2}+8)x^4 + ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + (\dfrac{64+3+48}{6})x^4+ ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Finally , multiplying 5 with $\mathtt{ e^{-x^2} cos (4x) }$ ; we have:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

7 0

3 years ago

"Which word is used to name the bold numbers and variables?

Kay [80]

If the bold numbers and variables are 2x + 5, then those would be B. addend, or the numbers that are added. Minuend is the number from which you subtract.
And since there is no dividing here, C or D are incorrect.

7 0

3 years ago

Please, my brain went blank and I didn't know how to answer this

Akimi4 [234]

The answer is 54.
The area of one square side is: 3 x 3 = 9
There are 6 sides on a cube so: 9 x 6 = 54

3 0

2 years ago

Read 2 more answers

Combine like terms to simplify the expression

Alex73 [517]

3.26+9.75=13.01. This is the value of the like term. So it would be 13.01d-2.65.

4 0

3 years ago

Read 2 more answers