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3 years ago

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In the figure below, parallel lines AB and CD are cut by transversals AC and BD They intersect at E. Two angle measures are give

n. What is the measure of angle ABD?
20 points to help me

2 answers:

ratelena [41]3 years ago

7 0

Answer:

Step-by-step explanation:

????????

Hoochie [10]3 years ago

3 0

Answer:

A : 146

Step-by-step explanation:

Since the lines AB and CD are parallel, we know that the angle opposite to BDC is also 34. To find ABD we can do 180-34 , which gives us 146

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An artist used a scale factor of 1 in=8 ft to create a model of a building. if the actual building has a height of 64 feet, how

nalin [4]

Answer:

<em>Okay, so what I think you mean to ask what the scale factor between the model and actual building</em> is, so i crunched the numbers. if 1 in equaled 8 feet and the actual thing was 64 ft, then we can just say that 64 divided by 12 is 5.3, so the model could be

5.3 ft. is the size of the model

BUT

you said how many feet are they apart so in inches its 705, in ft its 58.75

Step-by-step explanation:

a foot is 12 in.

64/12 is 5.3

64 ft is 768 inches, and 5.3 is 63.

768-63 = 705.

there are 705 inches in difference, in feet its 58.75

8 0

3 years ago

Translate the sentence into an inequality.<br><br> Four subtracted from c is less than - 20 .

11Alexandr11 [23.1K]

Answer:

$\displaystyle -4 + c < -20$

Step-by-step explanation:

You could either do what I did in the above answer, or you could do this:

$\displaystyle c - 4< -20$

It does not matter how you write it, as long as you understand the concept!

I am joyous to assist you anytime.

3 0

3 years ago

Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

$Sin\ C = \frac{h}{a}$

Make h the subject:

$h = aSin\ C$

Also, in BOC (Using Pythagoras)

$a^2 = h^2 + x^2$

Make $x^2$ the subject

$x^2 = a^2 - h^2$

Substitute $aSin\ C$ for h

$x^2 = a^2 - h^2$ becomes

$x^2 = a^2 - (aSin\ C)^2$

$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

$x= aCos\ C$

In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

6 0

3 years ago

If \: x + y + z = 0 \: then \: show \: that \: {x}^{3} + {y}^{3 } + {z}^{3} =3xy<br>

tankabanditka [31]

Answer:

Given

x+y+z=0

⟹x+y=−z

Cubing on both sides

(x+y) 3 =(−z) 3

⟹x 3 +y 3 +3x 2y+3xy 2 =−z 3

⟹x 3 +y 3 +3xy(x+y)=−z 3

⟹x 3+y 3+3xy(−z)=−z 3

⟹x 3 +y 3−3xyz=−z 3

⟹x 3 +y 3 +z 3 =3xyz

Step-by-step explanation:

Hope it is helpful.....

8 0

3 years ago

Solve the system of equations using substitution

Sauron [17]

y = −2x + 1

4x + 2y = −1

Replace Y in the second equation with the value of Y in the first one.

4x + 2(-2x +1) = -1

Distributive Property:

4x + -4x + 2 = -1

Combine like terms:

2 = -1 which is not true, so there are no solutions.

4 0

3 years ago

Read 2 more answers