The X and Y intercepts and the Slope are called the line properties. We shall now graph the line -3x-4y-9 = 0 and calculate its properties. Graph of a Straight Line : Calculate the Y-Intercept : Notice that when x = 0 the value of y is 9/-4 so this line "cuts" the y axis at y=-2.25000. y-intercept = 9/-4 = -2.25000.

Step-by-step explanation:

Find the X and Y Intercepts 3x-4y=9. 3x − 4y = 9 3 x - 4 y = 9. Find the x-intercepts. Tap for more steps... To find the x-intercept (s), substitute in 0 0 for y y and solve for x x. 3 x − 4 ⋅ 0 = 9 3 x - 4 ⋅ 0 = 9. Solve the equation. Tap for more steps... Simplify 3 x − 4 ⋅ 0 3 x - 4 ⋅ 0.

6 0

3 years ago

I need help with 12 13 and 14

nikdorinn [45]

Answer: Lines $\frac{}{BC}$ and $\frac{}{EF}$ are different lengths.

Step-by-step explanation:

The distance formula is $\sqrt{(x_1-x_2) ^{2}+(y_1-y_2) ^{2} }$ , and you can use this formula to solve for the lengths of both lines $\frac{}{BC}$ and $\frac{}{EF}$ .

For line $\frac{}{BC}$ , let $x_{1}$ = the x at point B, or 1, and let $x_{2}$ = the x at point C, or 2.

Now, let $y_{1}$ = the y at point B, or 4, and let $y_{2}$ = the y at point C, or -1.

Now, solve the formula to find the length $\frac{}{BC}$ = $\sqrt{(x_1-x_2) ^{2}+(y_1-y_2) ^{2} }\\$ .

$\frac{}{BC}$ = $\sqrt{(1-2)^{2} +(4-(-1))^2$

$\frac{}{BC}$ = $\sqrt{(-1)^{2} +(4+1)^2$

$\frac{}{BC}$ = $\sqrt{1 +5^2$

$\frac{}{BC}$ = $\sqrt{(1+25)$

$\frac{}{BC}$ = $\sqrt{26} \\$

Now, for line $\frac{}{EF}$ , let $x_{1}$ = the x at point E, or -4, and let $x_{2}$ = the x at point F, or -1.

Let $y_{1}$ = the y at point E, or -3, and let $y_{2}$ = the y at point F, or 1.

Now, solve the formula to find the length $\frac{}{EF}$ = $\sqrt{(x_1-x_2) ^{2}+(y_1-y_2) ^{2} }\\$ .

$\frac{}{EF}$ = $\sqrt{(-4-(-1))^{2} +(-3-1)^2$

$\frac{}{EF}$ = $\sqrt{(-4+1)^{2} +(-4)^2$

$\frac{}{EF}$ = $\sqrt{(-3)^2+16$

$\frac{}{EF}$ = $\sqrt{(9+16)$

$\frac{}{EF}$ = $\sqrt{25}$

$\frac{}{EF}$ = 5

Now, look back at $\frac{}{BC}$ . The two lines have different lengths, so you have now justified the fact that they are not the same.

Questions 13 and 14 would be solved in much the same way- but please let me know if you want me to show the work for those as well!

7 0

3 years ago