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2 years ago

In a sale, a microwave cost Php.2,999.99. It’s price has been reduced by Php 500.00 What was its price before the sale.

Mathematics

1 answer:

slava [35]2 years ago

8 0

The price was 3,499.99

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PLSSSSSSSSSSSSSS HELP <3

Komok [63]

1. Given

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4. transitive property of congruence

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3 years ago

What is the value of a in the equation 5a -10b = 45 when b = 3

leva [86]

Salutations!

$5a-10b = 45$

value of $b = 3$

Multiply 10 and 3

$10*3=30$

Now, you can solve the equation --

$5a-30=45$

separate the variables aside, and the numbers aside.

$5a=45 +30$

$5a=75$

$a=75/5$

$a=15$

Hope I helped (:

Have a great day!

6 0

3 years ago

4(2x−3)=4(2x)−4(3) what is the proptery

Nikitich [7]

9514 1404 393

Answer:

distributive property

Step-by-step explanation:

The <em>distributive property</em> of multiplication over addition lets you expand the product in the manner shown.

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3 years ago

Complete the graphic organizer by dragging each expression to the exponent law it illustrates. Assume that x and y are nonzero r

Agata [3.3K]

Answer:

Step-by-step explanation:

i got it correct:)

7 0

3 years ago

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

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3 years ago