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Zina [86]
3 years ago
10

What is 4x to the power of 2 - x = 18 in standard form

Mathematics
1 answer:
Marianna [84]3 years ago
3 0

Recall that exponents are a way of representing repeated multiplication. For example, the notation 54 can be expanded and written as 5 • 5 • 5 • 5, or 625. And don’t forget, the exponent only applies to the number immediately to its left, unless there are parentheses.

What happens if you multiply two numbers in exponential form with the same base? Consider the expression (23)(24). Expanding each exponent, this can be rewritten as (2 • 2 • 2) (2 • 2 • 2 • 2) or  2 • 2 • 2 • 2 • 2 • 2 • 2. In exponential form, you would write the product as 27. Notice, 7 is the sum of the original two exponents, 3 and 4.

What about (x2)(x6)? This can be written as (x • x)(x • x • x • x • x • x) = x • x • x • x • x • x • x • x  or x8. And, once again, 8 is the sum of the original two exponents.

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B

Step-by-step explanation:

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and adding them would be

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Name ____________________________ Class________ Date ____________________________ Directions: Each diagram is formed by two para
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Step-by-step explanation:

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Triangles ABC and ADC are shown below. Paul claims that triangles ABC and ADC are congruent.
guapka [62]

Answer:

1 no

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Read 2 more answers
PLEASE HELP I NEED A ASNSWER ASAP
stiks02 [169]
7 yes is a function
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7 0
3 years ago
4x^2 y+8xy'+y=x, y(1)= 9, y'(1)=25
jarptica [38.1K]

Answer with explanation:

\rightarrow 4x^2y+8x y'+y=x\\\\\rightarrow 8xy'+y(1+4x^2)=x\\\\\rightarrow y'+y\times\frac{1+4x^2}{8x}=\frac{1}{8}

--------------------------------------------------------Dividing both sides by 8 x

This Integration is of the form ⇒y'+p y=q,which is Linear differential equation.

Integrating Factor

 =e^{\int \frac{1+4x^2}{8x} dx}\\\\e^{\log x^{\frac{1}{8}+\frac{x^2}{2}}\\\\=x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}

Multiplying both sides by Integrating Factor  

x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\times [y'+y\times\frac{1+4x^2}{8x}]=\frac{1}{8}\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\\\\ \text{Integrating both sides}\\\\y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\frac{1}{8}\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx \\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=-[x^{\frac{9}{8}}]\times\frac{ \Gamma(0.5625, -x^2)}{(-x^2)^{\frac{9}{16}}}\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=(-1)^{\frac{-1}{8}}[ \Gamma(0.5625, -x^2)]+C-----(1)

When , x=1, gives , y=9.

Evaluate the value of C and substitute in the equation 1.

6 0
3 years ago
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