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3 years ago

6

Which constant could each equation be multiplied by to eliminate the x variable in this system of equations? 4x + 5y = 62 5x − 3

y = 22

1 answer:

qwelly [4]3 years ago

7 0

Answer:

Multiply equation $4x+5y=62$ by 5 and $5x-3y=22$ by 4

Step-by-step explanation:

Given:

Equations are $4x+5y=62,5x-3y=22$

To find: constant by which each equation should be multiplied to eliminate the variable $x$ in the given system of equations

Solution:

$4x+5y=62...(i)\\5x-3y=22...(ii)$

Multiply equation (i) by 5 and (ii) by 4 to get the following equations:

$20x+25y=310\\20x-12y=88$

On subtracting these equations, we get

$(20x+25y)-(20x-12y)=310-88\\\\13y=222$

So, the variable $x$ gets eliminated

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The three side lengths of a triangle are represented by x, (x+10), and (x+5) units. The perimeter of the triangle is 75 units. W

ollegr [7]

Answer:

if the perimeter is 75 units and the side lengths are x + 10 and x + 5 it would probably be plus 60 to make it 75

8 0

3 years ago

What is an equation of the line that passes through the point ( 6 , − 2 ) (6,−2) and is perpendicular to the line 6 x + y = 2 6x

Diano4ka-milaya [45]

Answer:

An equation of the line that passes through the point (6, − 2) and is perpendicular to the line will be:

$y=\frac{1}{6}x-3$

Step-by-step explanation:

We know that the slope-intercept form of the line equation is

y=mx+b

where m is the slope and b is the y-intercept.

Given the line

6x+y=2

Simplifying the equation to write into the slope-intercept form

y = -6x+2

So, the slope = -6

As we know that the slope of the perpendicular line is basically the negative reciprocal of the slope of the line.

Thus, the slope of the perpendicular line will be: -1/-6 = 1/6

Therefore, an equation of the line that passes through the point (6, − 2) and is perpendicular to the line will be

$y-y_1=m\left(x-x_1\right)$

substituting the values m = 1/6 and the point (6, -2)

$y-\left(-2\right)=\frac{1}{6}\left(x-6\right)$

$y+2=\frac{1}{6}\left(x-6\right)$

subtract 2 from both sides

$y+2-2=\frac{1}{6}\left(x-6\right)-2$

$y=\frac{1}{6}x-3$

Therefore, an equation of the line that passes through the point (6, − 2) and is perpendicular to the line will be:

$y=\frac{1}{6}x-3$

5 0

2 years ago

Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

$S = \{1,1,2,2,2,3\}$

And the probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{6}$

$P(1) = \frac{1}{3}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{6}$

$P(2) = \frac{1}{2}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{6}$

P(Odd Number) is then calculated as:

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{6}$

$P(Odd\ Number) = \frac{3}{6}$

$P(Odd\ Number) = \frac{1}{2}$

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

$S = \{1,1,2,2,2,3,3\}$

The probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{7}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{7}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{7}$

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{7}$

$P(Odd\ Number) = \frac{3}{7}$

Comparing P(Odd Number) before and after

$P(Odd\ Number) = \frac{1}{2}$ --- Before

$P(Odd\ Number) = \frac{3}{7}$ --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

7 0

2 years ago

5. The temperature was

Vesna [10]

Answer:

6.2 is the difference

Step-by-step explanation:

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3 years ago

Consider the following graph of an absolute value function

Ahat [919]

A

The domain is -∞ < x < ∞

B

The range is -∞ < x ≤ 3

C

The graph is increasing from -∞ < y < 3

D

The graph is decreasing from 3 > y > -∞

E

The local maximum is at ( - 2, 3 )

F

There are no local minimums

7 0

3 years ago