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yawa3891 [41]
3 years ago
13

What are the leading coefficient, constant term and degree, if any, of the algebraic expression 1+x^3

Mathematics
1 answer:
MA_775_DIABLO [31]3 years ago
4 0
The degree refers to the x to some exponent where the highest exponent is the degree.
Degree: 3
The leading coefficient is the coefficient of the term we just found(of the highest exponent)
Leading Coefficient: 1
The constant term is the term without an x in it.
Constant Term: 1
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HELP PLEASE WILL GIVE BRAINLIST
rodikova [14]

Answer:

Let R be the greater radius and r be the smaller radius

A) Area of the sidewalk = R^2 - r^2   - This can be the expression

B)  = 3.14

=(R^2-r^2)

=(11^2-9^2)

=(121-81)

=*40

That was the simplified expression

Answer = 3.14 * 40 = 125.6m^2

Step-by-step explanation:

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3 years ago
What is the y-intercept for the graph of 7x – 3y = –5?
AysviL [449]
2 ways to find the y int.

(1) put the equation in y = mx + b form and the y int will be in the b position
7x - 3y = -5
-3y = -7x - 5
y = 7/3x + 5/3....so 5/3 is ur y int

(2) another way is to sub in 0 for x and solve for y
7(0) - 3y = -5
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y = -5/-3
y = 5/3...ur y int
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3 years ago
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I will rate you brainliest
slava [35]

Answer:

(3x+11)/ (5x-9)

Step-by-step explanation:

The numerator is what is on the top of the bar in the middle

(3x+11)/ (5x-9)

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Solve the equation. StartFraction dx Over dt EndFraction equals StartFraction 1 Over x e Superscript t plus 7 x EndFraction An i
MAVERICK [17]

Answer:

\frac{1}{49}(7x-1)e^{7x}+e^{-t}=C

Step-by-step explanation:

We are given that

\frac{dx}{dt}=\frac{1}{xe^{t+7x}}

We have to find the implicit function

Using separation variable method

\frac{dx}{dt}=\frac{1}{xe^t\cdot e^{7x}}

By using property x^a\cdot x^y=x^{a+y}

xe^{7x}dx=e^{-t}dt

By using property \frac{1}{x^a}=x^{-a}

Taking integration on both sides

\int xe^{7x}dx=\int e^{-t}dt

Parts integration method

\int u\cdot v dx=u\int vdx-\int (\frac{du}{dx}\int vdx)dx

By parts integration method

x\int e^{7x}dx-\int (\frac{dx}{dx}\int e^{7x}dx)dx=-e^{-t}+C

Using formula \int e^{ax} dx=\frac{e^{ax}}{a}+C

\frac{xe^{7x}}{7}-\frac{1}{7}\int e^{7x}dx=-e^{-t}+C

\frac{xe^{7x}}{7}-\frac{1}{49}e^{7x}+e^{-t}=C

\frac{1}{49}(7x-1)e^{7x}+e^{-t}=C

We are given that

F(x,t)=C

F(x,t)=\frac{1}{49}(7x-1)e^{7x}+e^{-t}=C

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3 years ago
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