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3 years ago

Is (4,2) a solution of the system?

Mathematics

2 answers:

Triss [41]3 years ago

7 0

Answer:

Step-by-step explanation:

Plug x as 4 and y as 2.

2 = 4 - 2

2 = 2 True

2 = 3(4) + 4

2 = 12 + 4

2 = 16 False

lara [203]3 years ago

3 0

Answer:

No.

Step-by-step explanation:

Substitute 4 (as x) and 2 (as y) into the 2 equations to see if they fit.

y = x - 2

2 = 4 - 2

2 = 2

The first equation is true for (4,2).

Now try the 2nd one.

y = 3x + 4

2 = 3(4) + 4

2 ≠ 16

So the 2nd equation is not true for (4,2).

Either one not true makes the solution incorrect.

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What is the volume of a container with sides 1ft, 2ft, and 2.5 ft

almond37 [142]

Answer:

5ft^3

Step-by-step explanation:

The formula for volume is length × width × height.

so you multiply 1×2×2.5, which equals 5

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3 years ago

What is the GCF of the polynomials terms? 14a^3b^4-7ab^7+21a^2b

cupoosta [38]

Answer:

$\large\boxed{7ab}$

Step-by-step explanation:

$14a^3b^4-7ab^7+21a^2b\\\\14a^3b^4=2\cdot\boxed7\cdot\boxed{a}\cdot\boxed{b}\cdot b\cdot b\cdot b\\\\7ab^7=\boxed7\cdot\boxed{a}\cdot\boxed{b}\cdot b\cdot b\cdot b\cdot b\cdot b\cdot b\\\\21a^2b=3\cdot\boxed7\cdot\boxed{a}\cdot a\cdot\boxed{b}\\\\GCF(14a^3b^4,\ -7ab^7,\ 21a^2b)=\boxed7\cdot\boxed{a}\cdot\boxed{b}=7ab$

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3 years ago

Which of the following is not a way to determine if a relation is a function?

Dima020 [189]

D. Determine if the relations’ graph forms a line.

Step-by-step explanation:

A function is a relation if the input values lead to only one output value.The input values, normally x values make up the domain, where as the output values form the domain.A function is a relation where the input values are associated with a single output.The vertical test is used to determine if a curve is a function.The line graph should only hit a single point on the curve.

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3 years ago

Evaluate the interval (Calculus 2)

Darya [45]

Answer:

$2 \tan (6x)+2 \sec (6x)+\text{C}$

Step-by-step explanation:

Fundamental Theorem of Calculus

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{12}{1-\sin (6x)}\:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\$\displaystyle \int a\:\text{f}(x)\:\text{d}x=a \int \text{f}(x) \:\text{d}x$\end{minipage}}$

If the terms are multiplied by constants, take them outside the integral:

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)}\:\:\text{d}x$

Multiply by the conjugate of 1 - sin(6x) :

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)} \cdot \dfrac{1+\sin(6x)}{1+\sin(6x)}\:\:\text{d}x$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{1-\sin^2(6x)} \:\:\text{d}x$

$\textsf{Use the identity} \quad \sin^2 x+ \cos^2 x=1:$

$\implies \sin^2 (6x) + \cos^2 (6x)=1$

$\implies \cos^2 (6x)=1- \sin^2 (6x)$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

Expand:

$\implies 12\displaystyle \int \dfrac{1}{\cos^2(6x)}+\dfrac{\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

$\textsf{Use the identities }\:\: \sec \theta=\dfrac{1}{\cos \theta} \textsf{ and } \tan\theta=\dfrac{\sin \theta}{\cos \theta}:$

$\implies 12\displaystyle \int \sec^2(6x)+\dfrac{\tan(6x)}{\cos(6x)} \:\:\text{d}x$

$\implies 12\displaystyle \int \sec^2(6x)+\tan(6x)\sec(6x) \:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}$

$\boxed{\begin{minipage}{6 cm}\underline{Integrating $ \sec kx \tan kx$}\\\\$\displaystyle \int \sec kx \tan kx\:\text{d}x= \dfrac{1}{k}\sec kx\:\:(+\text{C})$\end{minipage}}$

$\implies 12 \left[\dfrac{1}{6} \tan (6x)+\dfrac{1}{6} \sec (6x) \right]+\text{C}$

Simplify:

$\implies \dfrac{12}{6} \tan (6x)+\dfrac{12}{6} \sec (6x)+\text{C}$

$\implies 2 \tan (6x)+2 \sec (6x)+\text{C}$

Learn more about indefinite integration here:

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3 0

2 years ago

Determine which of the lines, if any, are parallel or perpendicular. Explain.

pishuonlain [190]

Answer:

See below

Step-by-step explanation:

Let's rewrite all three equations is standard slope-intercept format of y = mx + b, where m is the slope and b the y-intercept (the value of y when x = 0).

Line a: -x+2y=3

2y = x + 3

y = (1/2)x + 3

Line b: -6x=3y-1

-3y = 6x - 1

y = -2x + (1/3)

Line c: 4x-8y=5

-8y = -4x + 5

y = (1/2)x - (5/8)

Parallel lines have the same slope (m). Perpendicular lines have slopes that are the negative inverse (-1/m)of each other.

Slopes, m, for the lines are;

a) (1/2)
b) -2
c) (1/2)

The negative inverse of (1/2) is -2.

Lines a and c are parallel (same slope)

Line b is perpendicular since it's slope is the negative inverse of both a and b (-1/(1/2)) = -2

8 0

3 years ago