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2 years ago

If you have a GCF of x is that a root?

Mathematics

1 answer:

Mekhanik [1.2K]2 years ago

3 0

Answer:

No,

Step-by-step explanation:

No, GCF is not a root, it is the coefficient of x.

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Reason for 4 pleaseeee

Marina CMI [18]

The reason for statement 4 is CPCTC.

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3 years ago

Help me with number 15 and 16

loris [4]

Answer:

Problem 15)

$40.91

Problem 16)

12.446 cm^2

Step-by-step explanation:

Problem 15)

Gala.

12.8 * 1.35 = 17.28

Honeycrisp.

9 * 1.68 = 15.12

Fuji

7.4 * 1.15 = 8.51

17.28 + 15.12 + 8.51 = 40.91

Problem 16)

(area1) + (area2) + (area3) = (total area)

(1.25*4.38) + (1.36*2.15) + (1.14*3.55) = x

5.475 + 2.924 + 4.047 = x

12.446 = x

3 0

2 years ago

. Given ????(5, −4) and T(−8,12):

damaskus [11]

Answer:

a) $y=\dfrac{13x}{16}-\dfrac{129}{16}$

b) $y = \dfrac{13x}{16}+ \dfrac{37}{2}$

Step-by-step explanation:

Given two points: $S(5,-4)$ and $T(-8,12)$

Since in both questions,a and b, we're asked to find lines that are perpendicular to ST. So, we'll do that first!

Perpendicular to ST:

the equation of any line is given by: $y = mx + c$ where, m is the slope(also known as gradient), and c is the y-intercept.

to find the perpendicular of ST <u>we first need to find the gradient of ST, using the gradient formula.</u>

$m = \dfrac{y_2 - y_1}{x_2 - x_1}$

the coordinates of S and T can be used here. (it doesn't matter if you choose them in any order: S can be either x_1 and y_1 or x_2 and y_2)

$m = \dfrac{12 - (-4)}{(-8) - 5}$

$m = \dfrac{-16}{13}$

to find the perpendicular of this gradient: we'll use:

$m_1m_2=-1$

both $m_1$ and $m_2$ denote slopes that are perpendicular to each other. So if $m_1 = \dfrac{12 - (-4)}{(-8) - 5}$ , then we can solve for $m_2$ for the slop of ther perpendicular!

$\left(\dfrac{-16}{13}\right)m_2=-1$

$m_2=\dfrac{13}{16}$ :: this is the slope of the perpendicular

a) Line through S and Perpendicular to ST

to find any equation of the line all we need is the slope $m$ and the points $(x,y)$ . And plug into the equation: $(y - y_1) = m(x-x_1)$

side note: you can also use the $y = mx + c$ to find the equation of the line. both of these equations are the same. but I prefer (and also recommend) to use the former equation since the value of 'c' comes out on its own.

$(y - y_1) = m(x-x_1)$

we have the slope of the perpendicular to ST i.e $m=\dfrac{13}{16}$