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3 years ago

1. Write the equation of the line for the points (2,-1) and (-4,11).

Mathematics

1 answer:

lyudmila [28]3 years ago

7 0

Answer:

$y=-2x+3$

Step-by-step explanation:

Slope-intercept form of an equation is written as $y=mx+b$ , where $m$ is the slope and $b$ is the y-intercept.

The slope of a line that passes through the points $(x_1,\: y_1)$ and $(x_2, \: y_2)$ is $m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}$ . Using the coordinates $(2, -1)$ and $-4,11)$ as given in the problem, we have slope of this line to be:

$m=\frac{11-(-1)}{-4-2}=\frac{12}{-6}=-2$ .

Now using this slope we've found and any point the line passes through, we can find the y-intercept of this equation:

$-1=-2(2)+b, \\ b=3$

Therefore, the equation of this line in slope-intercept form is $\fbox{$y=-2x+3$}$ .

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The equation y=−15x+12 is the slope-intercept form of which equation?

Sliva [168]

I honestly don't know but I think it would come out to be 4/5

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3 years ago

Rewrite the function in standard form, intercept form, find the vertex, find the y-intercept, and find the x-intercepts.

jasenka [17]

Answer:

We have the function, $f(x)=(x+3)^{2}-4$

On simplifying, we get,

$f(x)=x^2+6x+9-4$

i.e. $f(x)=x^2+6x+5$

Thus, the standard form of the function is $f(x)=x^2+6x+5$ .

Now, the factors of the given functions are (x+1) and (x+5).

Since, the intercept form of the function is the factored form of the function.

So, we have, intercept form of the function is $f(x)=(x+1)(x+5)$ .

Now, we know that,

Value of x-coordinate of the vertex is $\frac{-b}{2a}$ i.e. $\frac{-6}{2\times 1}$ i.e. $\frac{-6}{2}$ i.e. x= -3

Then, $f(-3)=(-3)^2+6\times (-3)+5$ i.e. $f(-3)=9-18+5$ i.e. f(-3)=-4

So, the vertex of the function is (-3,-4).

Further, we know that, 'the y-intercept of a function is the point where the function crosses y-axis'.

So, when x=0, we have, $f(0)=0^2+6\times 0+5$ i.e. f(0) = 5

Thus, the y-intercept is (0,5)

Also, 'the x-intercept of a function is the point where the function crossese x-axis'.

Then, for f(x)=0, we have $0=x^2+6x+5$ i.e. $0=(x+1)(x+5)$ i.e. x= -1 and x= -5

Thus, the x-intercept are (-1,0) and (-5,0).

8 0

3 years ago

I’ll give brainliest!!

nignag [31]

Answer:

If A is wrong then B I think

Step-by-step explanation:

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3 years ago

Looking at the top of tower A and base of tower B from points C and D, we find that ∠ACD = 60°, ∠ADC = 75° and ∠ADB = 30°. Let t

katrin2010 [14]

Answer:

$\text{Exact: }AB=25\sqrt{6},\\\text{Rounded: }AB\approx 61.24$

Step-by-step explanation:

We can use the Law of Sines to find segment AD, which happens to be a leg of $\triangle ACD$ and the hypotenuse of $\triangle ADB$ .

The Law of Sines states that the ratio of any angle of a triangle and its opposite side is maintained through the triangle:

$\frac{a}{\sin \alpha}=\frac{b}{\sin \beta}=\frac{c}{\sin \gamma}$

Since we're given the length of CD, we want to find the measure of the angle opposite to CD, which is $\angle CAD$ . The sum of the interior angles in a triangle is equal to 180 degrees. Thus, we have:

$\angle CAD+\angle ACD+\angle CDA=180^{\circ},\\\angle CAD+60^{\circ}+75^{\circ}=180^{\circ},\\\angle CAD=180^{\circ}-75^{\circ}-60^{\circ},\\\angle CAD=45^{\circ}$

Now use this value in the Law of Sines to find AD:

$\frac{AD}{\sin 60^{\circ}}=\frac{100}{\sin 45^{\circ}},\\\\AD=\sin 60^{\circ}\cdot \frac{100}{\sin 45^{\circ}}$

Recall that $\sin 45^{\circ}=\frac{\sqrt{2}}{2}$ and $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$ :

$AD=\frac{\frac{\sqrt{3}}{2}\cdot 100}{\frac{\sqrt{2}}{2}},\\\\AD=\frac{50\sqrt{3}}{\frac{\sqrt{2}}{2}},\\\\AD=50\sqrt{3}\cdot \frac{2}{\sqrt{2}},\\\\AD=\frac{100\sqrt{3}}{\sqrt{2}}\cdot\frac{ \sqrt{2}}{\sqrt{2}}=\frac{100\sqrt{6}}{2}={50\sqrt{6}}$

Now that we have the length of AD, we can find the length of AB. The right triangle $\triangle ADB$ is a 30-60-90 triangle. In all 30-60-90 triangles, the side lengths are in the ratio $x:x\sqrt{3}:2x$ , where $x$ is the side opposite to the 30 degree angle and $2x$ is the length of the hypotenuse.

Since AD is the hypotenuse, it must represent $2x$ in this ratio and since AB is the side opposite to the 30 degree angle, it must represent $x$ in this ratio (Derive from basic trig for a right triangle and $\sin 30^{\circ}=\frac{1}{2}$ ).

Therefore, AB must be exactly half of AD:

$AB=\frac{1}{2}AD,\\AB=\frac{1}{2}\cdot 50\sqrt{6},\\AB=\frac{50\sqrt{6}}{2}=\boxed{25\sqrt{6}}\approx 61.24$

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