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3 years ago

Brainliestto who helps me plz I need it

Mathematics

1 answer:

Mashutka [201]3 years ago

3 0

AB || CD and AC is oblique ;

So : Angle ( BAC ) = Angle ( DCA )

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∆ ABC - ∆ ACD :

(1) AB = CD

(2) Angle ( BAC ) = Angle ( DCA )

(3) AC = AC

So According to (1) , (2) , (3) :

∆ ABC =~ ∆ ACD

Well, if two triangles are deposited, all their components are equal peer to peer.

So : AD = BC

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Simplify this expression: -(x+2)+4 i have no idea im not good with math ;(

topjm [15]

ANSWER :

- ( x+ 2 ) + 4

-x - 2 + 4

-x + 2

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HELP ME , I CANNOT POST PICTURE & I HAVE 2 THINGS THAT REQIRE A PICTURE WHAT DO I DO ??!

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Answer:

Step-by-step explanation:

so when you post your picture you wanna click the thing that looks like a paper clip and attach a file or picture. or your can copy your photo and paste it in your question if you dont know how to take a picture just screenshot it *depends what kind of computer your using*

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This is finding exact values of sin theta/2 and tan theta/2. I’m really confused and now don’t have a clue on how to do this, pl

Lostsunrise [7]

First,

tan(θ) = sin(θ) / cos(θ)

and given that 90° < θ < 180°, meaning θ lies in the second quadrant, we know that cos(θ) < 0. (We also then know the sign of sin(θ), but that won't be important.)

Dividing each part of the inequality by 2 tells us that 45° < θ/2 < 90°, so the half-angle falls in the first quadrant, which means both cos(θ/2) > 0 and sin(θ/2) > 0.

Now recall the half-angle identities,

cos²(θ/2) = (1 + cos(θ)) / 2

sin²(θ/2) = (1 - cos(θ)) / 2

and taking the positive square roots, we have

cos(θ/2) = √[(1 + cos(θ)) / 2]

sin(θ/2) = √[(1 - cos(θ)) / 2]

Then

tan(θ/2) = sin(θ/2) / cos(θ/2) = √[(1 - cos(θ)) / (1 + cos(θ))]

Notice how we don't need sin(θ) ?

Now, recall the Pythagorean identity:

cos²(θ) + sin²(θ) = 1

Dividing both sides by cos²(θ) gives

1 + tan²(θ) = 1/cos²(θ)

We know cos(θ) is negative, so solve for cos²(θ) and take the negative square root.

cos²(θ) = 1/(1 + tan²(θ))

cos(θ) = - 1/√[1 + tan²(θ)]

Plug in tan(θ) = - 12/5 and solve for cos(θ) :

cos(θ) = - 1/√[1 + (-12/5)²] = - 5/13

Finally, solve for sin(θ/2) and tan(θ/2) :

sin(θ/2) = √[(1 - (- 5/13)) / 2] = 3/√(13)

tan(θ/2) = √[(1 - (- 5/13)) / (1 + (- 5/13))] = 3/2

3 0

3 years ago

Please help me with those question please help with out number 1

alina1380 [7]

Answer:

1) For each value of x, a value of y is increased by 5.

x = 0, y = 5

x = 1, y = 10

x = 2, y = 15

x = 3, y = 20

------------------------------------------------------------------------------------------

2) For each two values of x, a value of y is increased by 10.

x = 0, y = -2

x = 2, y = 8

x = 4, y = 18

x = 6, y = 28

-------------------------------------------------------------------------------------------

x = 0, y = 1

x = 1, y = $\frac{7}{5}$

x = 5, y = 3

x = 10, y = 5

-------------------------------------------------------------------------------------------------

x = 0, y = 2

x = 1, y = 17

x = 2, y = 32

x = 5, y = 77

Step-by-step explanation:

This is as easy as replacing x for the actual value show on the table.

$y=5x+5$

When x = 0, y = ?

$y=5(0)+5\\y=0+5\\y=5$

When x = 1, y = ?

$y=5(1)+5\\y=5+5\\y=10$

When x = 2, y = ?

$y=5(2)+5\\y=10+5\\y=15$

When x = 3, y = ?

$y=5(3)+5\\y=15+5\\y=20$

-------------------------------------------------------------------------------------------------------

$y=5x-2$

When x = 0, y = ?

$y=5(0)-2\\y=0-2\\y=-2$

When x = 2, y = ?

$y=5(2)-2\\y=10-2\\y=8$

When x = 4, y = ?

$y=5(4)-2\\y=20-2\\y=18$

When x = 6, y = ?

$y=5(6)-2\\y=30-2\\y=28$

----------------------------------------------------------------------------------------------------

$y=\frac{2}{5}x +1$

When x = 0, y = ?

$y=\frac{2}{5}(0) +1$

$y=0 +1\\y=1$

When x = 1, y = ?

$y=\frac{2}{5}(1) +1\\y=\frac{2}{5} +1$

$y=\frac{2}{5}+1(\frac{5}{5})\\ y=\frac{2}{5}+\frac{5}{5} \\y=\frac{7}{5}$

When x = 5, y = ?

$y=\frac{2}{5}(5) +1\\y=2+1\\y=3$

When x = 10, y = ?

$y=\frac{2}{5}(10) +1\\y=2(2) +1\\y=4+1\\y=5$

-------------------------------------------------------------------------------------------------

$y=15x+2$

When x = 0, y = ?

$y=15(0)+2\\y=0+2\\y=2$

When x = 1, y = ?

$y=15(1)+2\\y=15+2\\y=17$

When x = 2, y = ?

$y=15(2)+2\\y=30+2\\y=32$

When x = 5, y = ?

$y=15(5)+2\\y=75+2\\y=77$

4 0

3 years ago

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What is the solution to the system of equations graphed below

aliina [53]

The point where the lines intersect is the point that satisfies both equations.

It is (1, 5), selection C.

3 0

3 years ago

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