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2 years ago

Identify the coefficients in the following expression.3x + 4y - 7z + 9 *

Mathematics

1 answer:

igomit [66]2 years ago

8 0

Answer:

3, 4, -7

Step-by-step explanation:

Coefficients are the numbers in front of the variable.

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Please help me choose the correct answer to find the answer

Naddika [18.5K]

Answer:

y = -3x + 12

Step-by-step explanation:

6x + 2y = 24 *subtract 6x from both sides*

2y = -6x + 24 * divide both sides by 2*

y = -3x + 12

6 0

2 years ago

Mr. Lamb built a stone walkway in his back yard. Each stone was 5/6 foot long. How many stones did it take him if the walkway wa

hodyreva [135]

5/6 *10 = 8 1/3 stones

4 0

2 years ago

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Inga [223]

Answer:

$y=-\sqrt{3}x+2$

Step-by-step explanation:

We want to find the equation of a straight line that cuts off an intercept of 2 from the y-axis, and whose perpendicular distance from the origin is 1.

We will let Point M be (x, y). As we know, Point R will be (0, 2) and Point O (the origin) will be (0, 0).

First, we can use the distance formula to determine values for M. The distance formula is given by:

$\displaystyle d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Since we know that the distance between O and M is 1, d=1.

And we will let M(x, y) be (x₂, y₂) and O(0, 0) be (x₁, y₁). So:

$\displaystyle 1=\sqrt{(x-0)^2+(y-0)^2}$

Simplify:

$1=\sqrt{x^2+y^2}$

We can solve for y. Square both sides:

$1=x^2+y^2$

Rearranging gives:

$y^2=1-x^2$

Take the square root of both sides. Since M is in the first quadrant, we only need to worry about the positive case. Therefore:

$y=\sqrt{1-x^2}$

So, Point M is now given by (we substitute the above equation for y):

$M(x,\sqrt{1-x^2})$

We know that Segment OM is perpendicular to Line RM.

Therefore, their <em>slopes will be negative reciprocals</em> of each other.

So, let’s find the slope of each segment/line. We will use the slope formula given by:

$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$

Segment OM:

For OM, we have two points: O(0, 0) and M(x, √(1-x²)). So, the slope will be:

$\displaystyle m_{OM}=\frac{\sqrt{1-x^2}-0}{x-0}=\frac{\sqrt{1-x^2}}{x}$

Line RM:

For RM, we have the two points R(0, 2) and M(x, √(1-x²)). So, the slope will be:

$\displaystyle m_{RM}=\frac{\sqrt{1-x^2}-2}{x-0}=\frac{\sqrt{1-x^2}-2}{x}$

Since their slopes are negative reciprocals of each other, this means that:

$m_{OM}=-(m_{RM})^{-1}$

Substitute:

$\displaystyle \frac{\sqrt{1-x^2}}{x}=-\Big(\frac{\sqrt{1-x^2}-2}{x}\Big)^{-1}$

Now, we can solve for x. Simplify:

$\displaystyle \frac{\sqrt{1-x^2}}{x}=\frac{x}{2-\sqrt{1-x^2}}$

Cross-multiply:

$x(x)=\sqrt{1-x^2}(2-\sqrt{1-x^2})$

Distribute:

$x^2=2\sqrt{1-x^2}-(\sqrt{1-x^2})^2$

Simplify:

$x^2=2\sqrt{1-x^2}-(1-x^2)$

Distribute:

$x^2=2\sqrt{1-x^2}-1+x^2$

So:

$0=2\sqrt{1-x^2}-1$

Adding 1 and then dividing by 2 yields:

$\displaystyle \frac{1}{2}=\sqrt{1-x^2}$

Then:

$\displaystyle \frac{1}{4}=1-x^2$

Therefore, the value of x is:

$\displaystyle \begin{aligned}\frac{1}{4}-1&=-x^2\\-\frac{3}{4}&=-x^2\\ \frac{3}{4}&=x^2\\ \frac{\sqrt{3}}{2}&=x\end{aligned}$

Then, Point M will be:

$\begin{aligned} \displaystyle M(x,\sqrt{1-x^2})&=M(\frac{\sqrt{3}}{2}, \sqrt{1-\Big(\frac{\sqrt{3}}{2}\Big)^2)}\\M&=(\frac{\sqrt3}{2},\frac{1}{2})\end{aligned}$

Therefore, the slope of Line RM will be:

$\displaystyle \begin{aligned}m_{RM}&=\frac{\frac{1}{2}-2}{\frac{\sqrt{3}}{2}-0} \\ &=\frac{\frac{-3}{2}}{\frac{\sqrt{3}}{2}}\\&=-\frac{3}{\sqrt3}\\&=-\sqrt3\end{aligned}$