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3 years ago

Find the equation of the line parallel to y=-11-5, that passes through the point

Mathematics

2 answers:

Elanso [62]3 years ago

7 0

Answer:

y=41−11x

Step-by-step explanation:

The equation of the line in the slope-intercept form is y=5−11x.

The slope of the parallel line is the same: m=−11.

So, the equation of the parallel line is y=−11x+a.

To find a, we use the fact that the line should pass through the given point: −3=(−11)⋅(4)+a.

Thus, a=41.

Therefore, the equation of the line is y=41−11x.

RSB [31]3 years ago

3 0

Answer:

y=41−11x

Step-by-step explanation:

The equation of the line in the slope-intercept form is y=5−11x.

The slope of the parallel line is the same: m=−11.

So, the equation of the parallel line is y=−11x+a.

To find a, we use the fact that the line should pass through the given point: −3=(−11)⋅(4)+a.

Thus, a=41.

Therefore, the equation of the line is y=41−11x.

Answer: y=41−11x.

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3 years ago

A triangle has a base of x 1/2 m and a height of x 3/4 m. If the area of te triangle is 16m to the power of 2, what are the base

Olenka [21]

Answer:

The base of triangle is $\frac{8}{\sqrt{3}} \ m$ and the height of triangle is $\frac{12}{\sqrt{3}} \ m.$

Step-by-step explanation:

Given:

A triangle has a base of x 1/2 m and a height of x 3/4 m. If the area of the triangle is 16m to the power of 2.

Now, to find the base and height of the triangle.

The base of triangle = $x\times\frac{1}{2} =\frac{x}{2} \ m.$

The height of triangle = $x\times \frac{3}{4} =\frac{3x}{4}\ m.$

The area of triangle = $16\ m^2.$

Now, we put the formula of area to solve:

$Area=\frac{1}{2} \times base\times height$

$16=\frac{1}{2} \times \frac{x}{2} \times \frac{3x}{4}$

$16=\frac{3x^2}{16}$

Multiplying both sides by 16 we get:

$256=3x^2$

Dividing both sides by 3 we get:

$\frac{256}{3} =x^2$

Using square root on both sides we get:

$\frac{16}{\sqrt{3}}=x$

$x=\frac{16}{\sqrt{3}}$

Now, by substituting the value of $x$ to get the base and height:

$Base=\frac{x}{2}\\\\Base=\frac{\frac{16}{\sqrt{3}}}{2} \\\\Base=\frac{8}{\sqrt{3}} \ m.$

So, the base of triangle = $\frac{8}{\sqrt{3}} \ m.$

$Height=x\times\frac{3}{4} \\\\Height=\frac{16}{\sqrt{3}}\times \frac{3}{4} \\\\Height=\frac{12}{\sqrt{3}} \ m.$

Thus, the height of triangle = $\frac{12}{\sqrt{3}} \ m.$

Therefore, the base of triangle is $\frac{8}{\sqrt{3}} \ m$ and the height of triangle is $\frac{12}{\sqrt{3}} \ m.$

7 0

3 years ago