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Sindrei [870]
3 years ago
6

Which best explains whether or not triangles RST and ACB are congruent?

Mathematics
1 answer:
irga5000 [103]3 years ago
7 0
Two triangles are congruent if and only if their corresponding parts are congruent (CPCTC).
You might be interested in
Is it true that the planes x + 2y − 2z = 7 and x + 2y − 2z = −5 are two units away from the plane x + 2y − 2z = 1?
zhuklara [117]

Lets Find It Out..

First we'll find the equation of ALL planes parallel to the original one.

As a model consider this lesson:

Equation of a plane parallel to other

The normal vector is:
<span><span>→n</span>=<1,2−2></span>

The equation of the plane parallel to the original one passing through <span>P<span>(<span>x0</span>,<span>y0</span>,<span>z0</span>)</span></span>is:

<span><span>→n</span>⋅< x−<span>x0</span>,y−<span>y0</span>,z−<span>z0</span>>=0</span>
<span><1,2,−2>⋅<x−<span>x0</span>,y−<span>y0</span>,z−<span>z0</span>>=0</span>
<span>x−<span>x0</span>+2y−2<span>y0</span>−2z+2<span>z0</span>=0</span>
<span>x+2y−2z−<span>x0</span>−2<span>y0</span>+2<span>z0</span>=0</span>

Or

<span>x+2y−2z+d=0</span> [1]
where <span>a=1</span>, <span>b=2</span>, <span>c=−2</span> and <span>d=−<span>x0</span>−2<span>y0</span>+2<span>z0</span></span>

Now we'll find planes that obey the previous formula and at a distance of 2 units from a point in the original plane. (We should expect 2 results, one for each half-space delimited by the original plane.)
As a model consider this lesson:

Distance between 2 parallel planes

In the original plane let's choose a point.
For instance, when <span>x=0</span> and <span>y=0</span>:
<span>x+2y−2z=1</span> => <span>0+2⋅0−2z=1</span> => <span>z=−<span>12</span></span>
<span>→<span>P1</span><span>(0,0,−<span>12</span>)</span></span>

In the formula of the distance between a point and a plane (not any plane but a plane parallel to the original one, equation [1] ), keeping <span>D=2</span>, and d as d itself, we get:

<span><span>D=<span><span>|a<span>x1</span>+b<span>y1</span>+c<span>z1</span>+d|</span><span>√<span><span>a2</span>+<span>b2</span>+<span>c2</span></span></span></span></span>
<span>2=<span><span><span>∣∣</span>1⋅0+2⋅0+<span>(−2)</span>⋅<span>(−<span>12</span>)</span>+d<span>∣∣</span></span><span>√<span>1+4+4</span></span></span></span>
<span><span>|d+1|</span>=2⋅3</span> => <span><span>|d+1|</span>=6</span>First solution:
<span>d+1=6</span> => <span>d=5</span>
<span>→x+2y−2z+5=0</span>Second solution:
<span>d+1=−6</span> => <span>d=−7</span>
<span>→x+2y−2z−7=<span>0</span></span></span>
8 0
3 years ago
****Brainliest to best answer!!***<br> Thank u to everyone who helps!
Dmitrij [34]

Answer:

A

Step-by-step explanation:

When you read the question it says that her average speed from the library to the gym is 15 mph less than her speed from her home to the library. So x - 15 is what describes that situation.

If it is A it cannot be anything else.

B is incorrect. x - 15 is not a time. It is a rate. It can be used to develop the amount of time, but by itself it is still A.

C is incorrect. Her average speed from her house to the library is x

D is incorrect. The distance is given as the same as from her home to the library as  from the library to the gym. Both are 4

7 0
3 years ago
52 points will mark brainliest for correct answers
Sedbober [7]
18 and 17 should have a line through
6 0
3 years ago
Read 2 more answers
Factor the expression: -x - 6 =
tia_tia [17]

Answer:

- ( x + 6) I think

7 0
3 years ago
Read 2 more answers
2. The Geo Air pilot is looking at SCCA from the plane. From the aircraft the angle of depression is 17 degrees. If the plane is
lana66690 [7]

Answer:

The distance  from the plane to SCCA is 34,203.0 feet approximately,  and the horizontal distance is 32,708.5 feet approximately.

Step-by-step explanation:

You can draw a right triangle like the one shown in the figure attached, where:

x: horizontal distance.

y: distance from the plane to SCCA.

You can calculate x as following:

tan\alpha=\frac{opposit}{adjacent}

Where:

\alpha=17\°\\opposite=10,000\\adjacent=x

Substitute and solve for x:

tan(17\°)=\frac{10,000}{x}\\\\x=\frac{10,000}{tan(17\°)}\\\\x=32,708.5ft

You can calculate y as following:

sin\alpha=\frac{opposit}{hypotenuse}

Where:

\alpha=17\°\\opposite=10,000\\hypotenuse=y

Substitute and solve for y:

sin(17\°)=\frac{10,000}{y}\\\\y=\frac{10,000}{sin(17\°)}\\\\y=34,203.0ft

7 0
3 years ago
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