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3 years ago

Write an equation of the line that passes through (2, 2) and is parallel to the line y = 7x

Mathematics

1 answer:

zimovet [89]3 years ago

4 0

Answer:

y = 7x - 12

Step-by-step explanation:

Parallel lines have same slope.

y = mx + b

m - > slope & b -> y-intercept

y = 7x

So, slope = 7 & (2 , 2)

y - y1 = m(x -x1)

y - 2 = 7(x - 2)

y - 2 = 7x - 14

y = 7x - 14 + 2

y = 7x - 12

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Solve equation. 7/8-2/3

erma4kov [3.2K]

Answer:

$\frac{7}{8} - \frac{2}{3 } \\ \frac{7 \times 3}{8 \times 3} - \frac{2 \times 8}{3 \times 8 } \\ \frac{21 - 16}{24 } \\ = \frac{5}{24}$

3 0

3 years ago

Read 2 more answers

Alejandro made an error in the steps below when determining the equation of the line that is perpendicular to the line 4x – 3y =

vazorg [7]

The slope of the line given its equation is calculated through, m = -A / B. The slope of the given line is 4/3. The line perpendicular to it has the slope of -3/4. The slope-point form of the equation is,
y - y1 = m(x - x1)
where m is the slope and x1 and y1 the abscissa and ordinate of the point, respectively.
Substituting the values above,
y --2 = (-3/4)(x - 3)

Simplifying the equation gives 3x + 4y = 1.

4 0

3 years ago

Read 2 more answers

The sum of first three terms of a finite geometric series is -7/10 and their product is -1/125. [Hint: Use a/r, a, and ar to rep

Alchen [17]

Ooh, fun

geometric sequences can be represented as
$a_n=a(r)^{n-1}$
so the first 3 terms are
$a_1=a$
$a_2=a(r)$
$a_2=a(r)^2$

the sum is -7/10
$\frac{-7}{10}=a+ar+ar^2$
and their product is -1/125
$\frac{-1}{125}=(a)(ar)(ar^2)=a^3r^3=(ar)^3$

from the 2nd equation we can take the cube root of both sides to get
$\frac{-1}{5}=ar$
note that a=ar/r and ar²=(ar)r
so now rewrite 1st equation as
$\frac{-7}{10}=\frac{ar}{r}+ar+(ar)r$
subsituting -1/5 for ar
$\frac{-7}{10}=\frac{\frac{-1}{5}}{r}+\frac{-1}{5}+(\frac{-1}{5})r$
which simplifies to
$\frac{-7}{10}=\frac{-1}{5r}+\frac{-1}{5}+\frac{-r}{5}$
multiply both sides by 10r
-7r=-2-2r-2r²
add (2r²+2r+2) to both sides
2r²-5r+2=0
solve using quadratic formula
for $ax^2+bx+c=0$
$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$
so
for 2r²-5r+2=0
a=2
b=-5
c=2

$r=\frac{-(-5) \pm \sqrt{(-5)^2-4(2)(2)}}{2(2)}$
$r=\frac{5 \pm \sqrt{25-16}}{4}$
$r=\frac{5 \pm \sqrt{9}}{4}$
$r=\frac{5 \pm 3}{4}$
so
$r=\frac{5+3}{4}=\frac{8}{4}=2$ or $r=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2}$

use them to solve for the value of a
$\frac{-1}{5}=ar$
$\frac{-1}{5r}=a$
try for r=2 and 1/2
$a=\frac{-1}{10}$ or $a=\frac{-2}{5}$

test each
for a=-1/10 and r=2
a+ar+ar²= $\frac{-1}{10}+\frac{-2}{10}+\frac{-4}{10}=\frac{-7}{10}$
it works

for a=-2/5 and r=1/2
a+ar+ar²= $\frac{-2}{5}+\frac{-1}{5}+\frac{-1}{10}=\frac{-7}{10}$
it works

both have the same terms but one is simplified

the 3 numbers are $\frac{-2}{5}$ , $\frac{-1}{5}$ , and $\frac{-1}{10}$

6 0

3 years ago

What is the sum of the given polynomials in standard form?

Aneli [31]

Answer:

D. -x2 + 2x -3

Step-by-step explanation:

the first step is we open all the brackets then it will be :

x2 - 3x - 2x2 + 5x - 3

= x2 - 2x2 - 3x + 5x - 3

= -x2 + 2x -3

4 0

3 years ago

What is the value of y in this system of equations: 3x + y = 9 y = –4x + 10 ?

Ahat [919]

3x + y = 9 (I)
y = –4x + 10 (II)
------------------------
 $\left \{ {{3x + y = 9 } \atop {y = -4x + 10 }} \right.$

Pass the incognito "4x" to the first term, changing the signal when changing sides.
-------------------------
simplify by (-1)
 $\left \{ {{3x + y = 9.(-1)} \atop {4x + y = 10 }} \right.$ 

-------------------------
 $\left \{ {{- 3x - \diagup\!\!\!\!y =-9} \atop {4x + \diagup\!\!\!\!y = 10}} \right. 
$

-------------------------
 $\left \{ {{- 3x = - 9} \atop {4x = 10}} \right.$ 

-----------
 $\boxed{x = 1}$ 

Substitute in equation (I) to find the value of "Y".

3x + y = 9 (I)
3*(1) + y = 9
3 + y = 9
y = 9 - 3
$\boxed{y = 6}$

Answer:
$\boxed{\boxed{ \left \ {{x=1} \atop {y=6}} \right. }}$

6 0

3 years ago