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GarryVolchara [31]
3 years ago
5

Write the slope-intercept equation of the line parallel to 5y = 2x + 20 that goes through (-1, 3)

Mathematics
1 answer:
katen-ka-za [31]3 years ago
8 0

Answer:

y = \frac{2}{5}x + 3\frac{2}{5}

Step-by-step explanation:

This line was never finished, so you finish it like this:

\frac{5y}{5} = \frac{2x + 20}{5} → y = \frac{2}{5}x + 4

3 = −⅖ + b

3\frac{2}{5} = b \\ \\ y = \frac{2}{5}x + 3\frac{2}{5}

Parallel Lines have SIMILAR <em>RATE OF CHANGES</em> [<em>SLOPES</em>], so ⅖ remains the way it is.

I am joyous to assist you anytime.

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<h3>Answer:  11.2 inches</h3>

Work Shown:

(real height)/(scale height) = (real length)/(scale length)

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8 0
2 years ago
12=n/5 <br><br> what does n equal <br> also n/5 is supposed to be a fraction
blsea [12.9K]
60 do the inverse by multiplying the numbers you have
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3 years ago
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Rename 2/5 as a percent_______________________
loris [4]

It can be helpful to remember that "%" is just a shorthand way to write "/100" (divided by 100, or per 100, or per cent). ("Cent" is often used to refer to 100 or 1/100. A "cent" is 1/100 of a dollar, or 1/100 of an acre, for example. A <em>cent</em>ury is 100 years.)

Make the denominator of your fraction be 100. Multiply numerator and denominator by 20/20.

(2/5)×(20/20) = 40/100 = 40%

You can also get there via decimals.

... 2/5 = 0.4 = 0.40 (40 hundredths) = 40/100 (40 hundredths) = 40%

Or, you can multiply by 100%.

... (2/5)×(100%) = (200/5)% = 40%

7 0
3 years ago
Describe how both the Rational Root Theorem and Descartes’ Rules of Signs help you to find the zeros of a polynomial? Give me an
MrRa [10]

Answer:

Step-by-step explanation:

Rational Root Theorem: If the polynomial

P(x) = a n x n + a n – 1 x n – 1 + ... + a 2 x 2 + a 1 x + a 0

has any rational roots, then they must be of the form of (factors of a0/factors of an).

Example: F(x) = 4x² + 5x +2

If this polynomial has any rational roots, then they must be (factors of 2)/(factors of 4), so (±1, ±2)/(±1, ±2, ±4). So if this polynomial has any rational roots, they must be either: ±1, ±1/2, ±1/4, or ±2. Notice that this polynomial doesn't have to have any rational roots, but if it does, then the roots must fit the Rational Root Theorem.

Descartes' Rules of Signs:

a). In a polynomial, how many time the sign changes is how many positive roots the polynomial will have.

Example: 5x³ + 6x² - 2x - 1

In this expression, the sign only changed once, between 6x² and 2x, so it will only have one positive root.

Example 2: 6x³ - 4x² + x - 6

In this expression, the sign changed 3 times (remember there is a invisible "+" sign before the 6x³), so it will have 3 positive roots.

b). In a polynomial, if you plug in "-x" for all "x", then how many times the new polynomial changes sign is how many negative roots the old polynomial have.

Example: 5x³ + 6x² - 2x - 1.

If we plug in "-x" for all "x", then we get 5(-x)³ + 6(-x)² - 2(-x) -1, which simplifies to -5x³ + 6x² + 2x -1. In this new expression, the sign changed twice, so we have two negative roots for the expression. Notice how we got one positive root the first time and two negative roots the second time, and 1 + 2 = 3. The Fundamental Theorem of Algebra states that for a nth degree polynomial, it will have n complex roots. The polynomial we worked with was a 3rd degree polynomial, and we got 1 + 2 = 3 roots in the end.

4 0
3 years ago
This is a hard one for me. Can someone solve this one.
asambeis [7]

Answer:

100

Step-by-step explanation:

3 0
3 years ago
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