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3 years ago

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Mathematics

1 answer:

adell [148]3 years ago

5 0

Answer:

1.) y=3x-7

2.) 2=5x-y

3.) y=-4-2x

4.) 1=3/4x+y

5.) y=4x+8

6.) 9=3/4x+8

7.) 9=y+x

8.) y=-2/6x+2

9.) 5=-2/3x+y

Step-by-step explanation:

Standard form is Ax=By=C and Slope form is y-mx+b

for all of the equations fiqure out which is standard and which is slope, 3x-y=7 is standard and just switch it to slope ie:

3x-y=7

+7 -7

3x-7=y

then switch around so y is in the front and vice versa, also divde any y or C by negative 1 if the answer is negative or if there is a number in front of it like in 9, and 8

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Which equation represents a line parallel to the line shown on the graph? On a coordinate plane, a line goes through (negative 6

ELEN [110]

Answer:

$\boxed{ \mathrm{y = \ negative \ 3 x + 3}}$

Step-by-step explanation:

The coordinates are (-6,0) and (-8,6)

Finding slope of the line

Slope = $\frac{rise}{run} = \frac{y2-y1}{x2-x1}$

Slope = $\frac{6-0}{-8+6}$

Slope = $\frac{6}{-2}$

Slope = -3

Parallel lines have equal slopes.

So, the line which is parallel to this line is y = negative 3 x + 3

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4 years ago

Read 2 more answers

What is a cubic polynomial function with zeros 3,3 and -3? show your work.

Brums [2.3K]

Answer:

Cubic polynomial function with zeros 3,3 and -3 is $\mathbf{x^3-3x^2-9x+27=0}$

Step-by-step explanation:

We need to find a cubic polynomial function with zeros 3,3 and -3.

If zeros of polynomial are: 3,3,and -3

we can write:

x=3, x=3, x=-3

We can write:

x-3=0, x-3=0, x+3=0

Now, we can write them as:

(x-3)(x-3)(x+3)=0

Multiplying the terms, we can find the polynomial:

$(x-3)(x-3)(x+3)=0\\(x(x-3)-3(x-3))(x+3)=0\\(x^2-3x-3x+9)(x+3)=0\\(x^2-6x+9)(x+3)=0\\x(x^2-6x+9)+3(x^2-6x+9)=0\\x^3-6x^2+9x+3x^2-18x+27=0\\x^3-6x^2+3x^2+9x-18x+27=0\\x^3-3x^2-9x+27=0$

So, cubic polynomial function with zeros 3,3 and -3 is $\mathbf{x^3-3x^2-9x+27=0}$

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3 years ago

Why can a variable be used to represent a number

irina [24]

Variables can be used to represent unknown numbers in a equation or expression that you will later find out what they are

6 0

3 years ago

Express the function as the sum of a power series by first using partial fractions. f(x) = 8 x2 − 4x − 12 f(x) = ∞ n = 0 find th

alexandr1967 [171]

I'm guessing the function is

$f(x)=\dfrac8{x^2-4x-12}=\dfrac8{(x-6)(x+2)}$

which, split into partial fractions, is equivalent to

$\dfrac1{x-6}-\dfrac1{x+2}$

Recall that for $|x|$ we have

$\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n$

With some rearranging, we find

$\dfrac1{x-6}=-\dfrac16\dfrac1{1-\frac x6}=\displaystyle-\frac16\sum_{n=0}^\infty\left(\frac x6\right)^n$

valid for $\left|\dfrac x6\right|$ , or $|x|$ , and

$\dfrac1{x+2}=\dfrac12\dfrac1{1-\left(-\frac x2\right)}=\displaystyle\frac12\sum_{n=0}^\infty\left(-\frac x2\right)^n$

valid for $\left|-\dfrac x2\right|$ , or $|x|$ .

So we have

$f(x)=\displaystyle-\frac16\sum_{n=0}^\infty\left(\frac x6\right)^n-\frac12\sum_{n=0}^\infty\left(-\frac x2\right)^n$

$f(x)=\displaystyle-\sum_{n=0}^\infty\left(\frac{x^n}{6^{n+1}}+\frac{(-x)^n}{2^{n+1}}\right)$

$f(x)=\displaystyle-\sum_{n=0}^\infty\frac{1+3(-3)^n}{6^{n+1}}x^n$

Taken together, the power series for $f(x)$ can only converge for $|x|$ , or $-2$ .

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3 years ago

2 math questions. Please show your work if you can, I struggle with this, Thanks !! 20pts

castortr0y [4]

Step-by-step explanation:

To find the solution, Tanisha can graph each side of the equation as its own function:

y₁ = log₃(5x)

y₂ = log₅(2x + 8)

The solution is where the two curves cross. From the graph, that happens at x ≈ 0.957.

desmos.com/calculator/lkukjhn3g2

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3 years ago