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3 years ago

8

What is the slope of a line if it is going straight across the 3rd dash line going up? It is a sraight line passing through I an

d II. Which is it Positive, negitive, undefined, or zero slope?

1 answer:

Firdavs [7]3 years ago

3 0

Do u have a picture or something I could look at because I don't understand the second part

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Write a verbal phrase for the following algebraic expression 12y

aleksandrvk [35]

The product of a number y and 12 or 12 times a number y. These are both equivalent expressions to 12y

6 0

2 years ago

Consider the following system of two linear equations:

harkovskaia [24]

Hello,

Answer:

The last graph (2;3)

Step-by-step explanation:

2x + 3y = 12

2x – 3y = 0

2x + 3y = 12

2x = 3y

3y + 3y = 12 ⇔ 6y = 12 ⇔ y = 12/6 = 2

2x + 3 × 2 = 12 ⇔ 2x + 6 = 12 ⇔ 2x = 6 ⇔ x = 6/2 = 3

4 0

1 year ago

Solve<br><img src="https://tex.z-dn.net/?f=%5Csf%20%5Cdfrac%7B1%7D%7Bp%7D%20%2B%20%5Cdfrac%7B1%7D%7Bq%7D%20%2B%20%5Cdfrac%7B1%7D

Nostrana [21]

Answer:

$\displaystyle \begin{cases} \displaystyle {x} _{1} = - p \\ \displaystyle x _{2} = - q \end{cases}$

Step-by-step explanation:

we would like to solve the following equation for x:

$\displaystyle \frac{1}{p} + \frac{1}{q} + \frac{1}{x} = \frac{1}{p + q + x}$

to do so isolate $\frac{1}{x}$ to right hand side and change its sign which yields:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{1}{p + q + x} - \frac{1}{x}$

simplify Substraction:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{x - (q + p + x)}{x(p + q + x)}$

get rid of only x:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{ - (q + p )}{x(p + q + x)}$

simplify addition of the left hand side:

$\displaystyle \frac{q + p}{pq} = \frac{ - (q + p )}{x(p + q + x)}$

divide both sides by q+p Which yields:

$\displaystyle \frac{1}{pq} = \frac{ -1}{x(p + q + x)}$

cross multiplication:

$\displaystyle x(p + q + x) = - pq$

distribute:

$\displaystyle xp + xq + {x}^{2} = - pq$

isolate -pq to the left hand side and change its sign:

$\displaystyle xp + xq + {x}^{2} + pq = 0$

rearrange it to standard form:

$\displaystyle {x}^{2} + xp + xq + pq = 0$

now notice we end up with a <u>quadratic</u><u> equation</u> therefore to solve so we can consider <u>factoring</u><u> </u><u>method</u><u> </u><u> </u>to use so

factor out x:

$\displaystyle x( {x}^{} + p ) + xq + pq = 0$

factor out q:

$\displaystyle x( {x}^{} + p ) +q (x + p)= 0$

group:

$\displaystyle ( {x}^{} + p ) (x + q)= 0$

by <em>Zero</em><em> product</em><em> </em><em>property</em> we obtain:

$\displaystyle \begin{cases} \displaystyle {x}^{} + p = 0 \\ \displaystyle x + q= 0 \end{cases}$

cancel out p from the first equation and q from the second equation which yields:

$\displaystyle \begin{cases} \displaystyle {x}^{} = - p \\ \displaystyle x = - q \end{cases}$

and we are done!

3 0

2 years ago

PLEASE HELP the information is below

liq [111]

Answer:

yellow: x=6

red: y=4

blue: y=-2

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

What is the value of i=

vaieri [72.5K]

$\sum\limits_{t=1}^3\left(4\cdot\left(\dfrac{1}{2}\right)^{t-1}\right)=4\cdot\left(\dfrac{1}{2}\right)^{1-1}+4\cdot\left(\dfrac{1}{2}\right)^{2-1}+4\cdot\left(\dfrac{1}{2}\right)^{3-1}\\\\=4\cdot\left(\dfrac{1}{2}\right)^0+4\cdot\left(\dfrac{1}{2}\right)^1+4\cdot\left(\dfrac{1}{2}\right)^2=4\cdot1+4\cdot\dfrac{1}{2}+4\cdot\dfrac{1}{4}\\\\=4+2+1=7\\\\Answer:\ \boxed{D.\ 7}$

5 0

2 years ago