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3 years ago

In quadrilateral XWZY, name the angles which are consecutive with angle W

Mathematics

1 answer:

Aleksandr [31]3 years ago

3 0

The consecutive angles are

Angle x

And

Angle z

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Good morning can u help???

svp [43]

Answer:

Select the proportional relationship.

That is the only thing that is in the checkbox. It won't let you continue unless you get something.

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3 years ago

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Help me plz quick only 5 mins

solniwko [45]

Answer:

line B'C' = 1

m<C'D'A' = 90 degrees (it is still a rectangle)

Step-by-step explanation:

Dilation of coordinates:

Original: d:(3,2), a:(3,4), b:(7,4), c:(7,2)

Divide all numbers by 2.

Altered: d:(1.5,1), a:(1.5,2), b:(3.5,2), c:(3.5,1)

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3 years ago

Log(x-2)+log2=2logy<br>log(x-3y+3)=0

In-s [12.5K]

Answer:

Step-by-step explanation:

$\left\{\begin{array}{ccc}\log(x-2)+\log2=2\log y&(1)\\\log(x-3y+3)=0&(2)\end{array}\right$

$===========================\\\text{DOMAIN:}\\\\x-2>0\to x>2\\y>0\\x-3y+3>0\to x-3y>-3\\=====================\\\\\text{We know:}\ \log_ab=c\iff a^c=b,\\\\\text{therefore}\ \log_ab=0\iff a^0=b\to b=1\\\\\text{From this we have}\\\\\log(x-3y+3)=0\iff x-3y+3=1\qquad\text{add}\ 3y\ \text{to both sides}\\\\x+3=3y+1\qquad\text{subtract 3 from both isdes}\\\\\boxed{x=3y-2}\qquad(*)\\\\\text{Substitute it to (1)}$

$\log(3y-2-2)+\log2=2\log(y)\qquad\text{use}\ \log_ab^n=n\log_ab\\\\\log(3y-4)+\log2=\log(y^2)\qquad\text{subtract}\ \log(y^2)\ \text{from both sides}\\\\\log(3y-4)+\log2-\log(y^2)=0\\\\\text{Use}\ \log_ab+\log_ac=\log_a(bc)\ \text{and}\ \log_ab-\log_ac=\log_a\dfrac{b}{c}\\\\\log\dfrac{(3y-4)(2)}{y^2}=0\qquad\text{use}\ \log_ab=0\Rightarrow b=1\\\\\dfrac{(3y-4)(2)}{y^2}=1\iff y^2=(3y-4)(2)\\\\\text{Use the distributive property}\ a(b+c)=ab+ac$

$y^2=(3y)(2)+(-4)(2)\\\\y^2=6y-8\qquad\text{subtract}\ 6y\ \text{from both sides}\\\\y^2-6y=-8\qquad\text{add 8 to both sides}\\\\y^2-6y+8=0\\\\y^2-2y-4y+8=0\\\\y(y-2)-4(y-2)=0\\\\(y-2)(y-4)=0\iff y-2=0\ \vee\ y-4=0\\\\\boxed{y=2\ \vee\ y=4}\in D$

$\text{Put the values of}\ y\ \text{to }\ (*):\\\\\text{for}\ y=2:\\\\x=3(2)-2\\\\x=6-2\\\\\boxed{x=4}\in D\\\\\text{for}\ y=4:\\\\x=3(4)-2\\\\x=12-2\\\\\boxed{x=10}\in D$

4 0

3 years ago

How do you solve and check x/4 = -5

Mashutka [201]

Rearrange x/4-(-5)=0

Simplify x/4

-5 = -5/1 = -5*4/4

x- (-5*4)/4 = x+20/4

x+ 20/4 = 0

x+20/4 * 4 = 0*4

The equation now takes the shape: x+20=0

subtract 20 from both sides of the equation: x=-20

Answer: x=-20

6 0

3 years ago

You bought a magazine for $5 and four erasers. You spend a total of $25. How much did each eraser cost?

Lesechka [4]

Subtract magazine from total:

25 -5 = 20

Divide by number of erasers:

20/4 = 5

Each eraser cost $5

5 0

3 years ago

Read 2 more answers