Answer
The second option

Explanation
Couterclockwise is rotating to the left, so if you rotate the green thing to the left it would be the same as the second option

6 0

3 years ago

Read 2 more answers

A horizontal line goes through the point (–7,10). Which statements are true about this line? Check all that apply.

Nata [24]

The slope of the line is the change in y values over the change in x values, since for horizontal line, the y values do not change then, the slope is zero.

If the line is horizontal, this means that whatever the value of the x-coordinate (abscissa) is, the y-coordinate (ordinate) will always be 10.

The intercept of the line is 10. So, third statement is not true.

Since the equation of the line is zero and its y-intercept is zero, using the equation of the line which is,
y = mx + b
where m is slope and b is y-intercept. The equation of the line is,
y = 0x + 10 ; y = 10

<em>The true statements for this item are 1st, 2nd, and 4th.</em>

7 0

3 years ago

Read 2 more answers

Will crown the correct answers brainliest 3x

Solnce55 [7]

Answer :

(1) $\frac{1}{x}+\frac{1}{y}=\frac{y+x}{xy}$

(2) $\frac{1}{5x}+\frac{1}{3y}=\frac{3y+5x}{15xy}$

(3) $\frac{1}{7x}-\frac{1}{2y}=\frac{2y-7x}{14xy}$

(4) $\frac{2}{5x}+\frac{3}{7y}=\frac{6y+15x}{21xy}$

(5) $\frac{7}{11x}-\frac{1}{33y}=\frac{231y-11x}{363xy}$

(6) $\frac{1}{2x}+\frac{3}{x}-\frac{1}{7y}=\frac{7y+42y-2x}{14xy}$

Step-by-step explanation :

(1) The given expression is: $\frac{1}{x}+\frac{1}{y}$

$\frac{1}{x}+\frac{1}{y}=\frac{y+x}{xy}$

(2) The given expression is: $\frac{1}{5x}+\frac{1}{3y}$

$\frac{1}{5x}+\frac{1}{3y}=\frac{3y+5x}{(5x)\times (3y)}=\frac{3y+5x}{15xy}$

(3) The given expression is: $\frac{1}{7x}-\frac{1}{2y}$

$\frac{1}{7x}-\frac{1}{2y}=\frac{2y-7x}{(7x)\times (2y)}=\frac{2y-7x}{14xy}$

(4) The given expression is: $\frac{2}{5x}+\frac{3}{7y}$

$\frac{2}{5x}+\frac{3}{7y}=\frac{(2\times 3y)+(3\times 5x)}{(5x)\times (7y)}=\frac{6y+15x}{21xy}$

(5) The given expression is: $\frac{7}{11x}-\frac{1}{33y}$

$\frac{7}{11x}-\frac{1}{33y}=\frac{(7\times 33y)-(1\times 11x)}{(11x)\times (33y)}=\frac{231y-11x}{363xy}$

(6) The given expression is: $\frac{1}{2x}+\frac{3}{x}-\frac{1}{7y}$

$\frac{1}{2x}+\frac{3}{x}-\frac{1}{7y}=\frac{(x\times 7y)+(3\times 2x\times 7y)-(2x\times x)}{(2x)\times (x)\times (7y)}=\frac{7xy+42xy-2x^2}{14x^2y}=\frac{7y+42y-2x}{14xy}$

7 0

3 years ago

Y-10=2 (x-8) in standard form, what is the a the b and the c (using the formula ax+by=c

jekas [21]

Y-10=2x -16
2x-y = -16+10

2x - y = - 6 ( ax + by = c )

6 0

3 years ago