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4 years ago

Find the gradient of the line segment between the points (0,2) and (2,-8).

Mathematics

2 answers:

valkas [14]4 years ago

5 0

Answer:

Step-by-step explanation:

Paladinen [302]4 years ago

5 0

Answer:

slope = - 5

Step-by-step explanation:

Calculate the slope m using the slope formula

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ ) = (0, 2) and (x₂, y₂ ) = (2, - 8)

m = $\frac{-8-2}{2-0}$ = $\frac{-10}{2}$ = - 5

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Find the area please

STatiana [176]

The correct should 356 yards^2
to solve this you simply add the are of the two squares after you calculate the area for each of them.

8 0

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:

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

3 0

3 years ago

Read 2 more answers

The table shows the linear relationship between the

mr Goodwill [35]

Answer:

Step-by-step explanation:

Y=8x + 0.25

3 0

3 years ago

Plz I need this thank you this is for 100 points luv u

Verizon [17]

Answer:

a. 120

b. -120

c. -120

d. -120

e. 120

Step-by-step explanation:

The amount of - will influence whether the value is positive or negative. Odd number of - is always negative. Even number of - is always positive. Hope this helped :)

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4 years ago

Find the points having x-coordinate of 9 whose distance from the point (3,-2) is 10.

kotegsom [21]

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{3}~,~\stackrel{y_1}{-2})\qquad (\stackrel{x_2}{9}~,~\stackrel{y_2}{y})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{d}{10}=\sqrt{[9-3]^2+[y-(-2)^2]}\implies 10=\sqrt{(9-3)^2+(y+2)^2} \\\\\\ 10=\sqrt{6^2+\stackrel{FOIL}{(y^2+4y+4)}}\implies 10^2=36+(y^2+4y+4)$

$\bf 100=y^2+4y+40\implies 0=y^2+4y-60 \\\\\\ 0=(y+10)(y-6)\implies y= \begin{cases} -10\\ 6 \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (9,-10)\qquad (9,6)~\hfill$

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4 years ago