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3 years ago

1. 2X + 1 = Y 2Y = 6 Show your work

Mathematics

1 answer:

slamgirl [31]3 years ago

4 0

Answer:

x = 1

y = 3

Step-by-step explanation:

2x + 1 = y ------- eqn (1)

2y = 6--------eqn (2)

from eqn(2)

2y = 6

y = 3

when you divide 6 by 2y

substitute y = 3 into eqn (1)

2x + 1 = y

2x + 1 = 3

2x = 3-1

2x = 2

therefore

x = 1

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Y-5=3-9(y+2)what is the answer to this

FinnZ [79.3K]

Answer: y= -1

Step-by-step explanation:

Add 5 to both sides,

y-5+5=-9y-15+5

Simplify,

y=-9y-10

Add 9y to both sides,

y+9y=-9y-10+9y

Simplify,

10y=-10

Divide both sides by ten.

10y over 10 = -10 over 10

There you go!

7 0

3 years ago

Read 2 more answers

Find the other endpoint of the line segment with the given endpoint and midpoint. Endpoint: (9,8) Midpoint (0,9)

shepuryov [24]

Answer:

(-9, 10)

Step-by-step explanation:

The location of the midpoint of a line with endpoint at ( $x_1,y_1$ ) and ( $x_2,y_2$ ) is given as (x, y). The location of x and y are:

$x = \frac{x_1+x_2}{2},y=\frac{y_1+y_2}{2}$

Given the endpoint (9,8) and Midpoint (0,9), the location of the other endpoint can be gotten from:

$0=\frac{9+x_2}{2}\\ \\9+x_2=0\\\\x_2=-9\\\\Also,9=\frac{8+y_2}{2}\\ \\8+y_2=18\\\\y_2=18-8\\\\y_2=10$

Hence the endpoint is at (x2, y2) which is at (-9, 10)

5 0

4 years ago

A computer store buys a computer system at a cost of $453.00. The selling price was set at $755, but after a year the computer

Bas_tet [7]

Answer:

40percent of 755$

40/100 × 755

=302$

therefore selling proce is 302$

3 0

3 years ago

Please SOMEONE help me on this asap!!!

VARVARA [1.3K]

As x increases, the value of p(x) approaches a number that is y-int value of q(x)
y- int of the function p is -2 (one more) the y-int of the function q

What choices there are, then I can answer more exactly

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4 years ago

Read 2 more answers

2. If k is a constant, what is the value of k such that the polynomial k2x3 - 6kx+9 is divisible by x - 1?

marysya [2.9K]

Answer: The value of k =3.

Step-by-step explanation:

According to factor theorem, if g(x) =x-a is a factor of p(x) then p(a)=0.

Given: $x-1$ is a factor of $k^2x^3-6kx+9$ .

then

$k^2(1)^3-6k(1)+9=0\\\\\Rightarrow\ k^2-6k+9=0\\\\\Rightarrow\ k^2-3k-3k+9=0\\\\\Rightarrow\ k(k-3)-3(k-3)=0\\\\\Rightarrow\ (k-3)(k-3)=0\\\\\Rightarrow\ k=3$

Hence, the value of k =3.

4 0

3 years ago