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3 years ago

Which is an equation in point slope form for the given point and slope? Point: (-3,7) Slope: 4

Mathematics

2 answers:

melisa1 [442]3 years ago

6 0

Answer:

y-7=4x(x+3)

Step-by-step explanation:

(The 4x is 4 times ...)

kipiarov [429]3 years ago

4 0

Answer:

$y -7 = 4(x+3)$

Step-by-step explanation:

Point-slope form is in the format of $y-y_1 = m (x-x_1)$ . When knowing a line's slope and a point it passes through, you can write an equation in that format. The $m$ , $x_1$ , and $y_1$ must be substituted for with real values.

The $m$ represents the slope, thus substitute 4 for $m$ . $x_1$ represents the x-value of one point the line passes through, while $y_1$ represents that same point's y-value. The line passes through (-3, 7), so substitute -3 for $x_1$ and 7 for $y_1$ . This gives the following answer and equation:

$y-(7) = 4(x-(-3))\\y -7 = 4(x+3)$

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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 $\frac{12}{\sqrt{3}} \ m.$

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