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3 years ago

Please help it's due tmrw

Mathematics

1 answer:

Makovka662 [10]3 years ago

5 0

Answer:

If you just use you finger to keep track of one point and follow each set of instructions you will find that B is correct.

I realize that in a previous comment I had said im bad at geometry but I immediately assumed this had to do with ASA and such.

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Solve for x.<br> 9<br> 2x-111<br> X<br> = [?]

VashaNatasha [74]

Answer:

x=94

Step-by-step explanation:

I hope I'm correct. Here's my explanation: x+9 and 2x-111 would equal 180, hence the line (supplementary). So x+9+2x-111=180

x=94

hope this helps.

5 0

3 years ago

Read 2 more answers

Pls help me i need it

Delvig [45]

Answer:

13=x

Step-by-step explanation:

Side-spliting theorem

21/(x-4)=14/6

cross multiply

21*6=14*(x-4)

distribute 14

126=14x-56

add 56 to both sides

126+56=14x-56+56

leaving you with

182=14x

Divide both sides by 14 to get x by itself

182/14=14x/14

13=x

4 0

2 years ago

Emily has $19 more dollars than Juan. Juan has $65. How much money does Emily have? Type your answer in the box below

kozerog [31]

Answer:

65+19=84

Step-by-step explanation:

Emily has $84

6 0

2 years ago

Read 2 more answers

PLZ HELP 20 POINTS AND CHOOSING BRAINLIEST

kow [346]

The answer is C (301.6)

8 0

2 years ago

Using the given information, give the vertex form equation of each parabola.

Amanda [17]

Answer:

The equation of parabola is given by : $(x-4) = \frac{-1}{3}(y+3)^{2}$

Step-by-step explanation:

Given that vertex and focus of parabola are

Vertex: (4,-3)

Focus:( $\frac{47}{12}$ ,-3)

The general equation of parabola is given by.

$(x-h)^{2} = 4p(y-k)$ , When x-componet of focus and Vertex is same

$(x-h) = 4p(y-k)^{2}$ , When y-componet of focus and Vertex is same

where Vertex: (h,k)

and p is distance between vertex and focus

The distance between two points is given by :

L= $\sqrt{(X2-X1)^{2}+(Y2-Y1)^{2}}$

For value of p:

p= $\sqrt{(X2-X1)^{2}+(Y2-Y1)^{2}}$

p= $\sqrt{(4-\frac{47}{12})^{2}+((-3)-(-3))^{2}}$

p= $\sqrt{(\frac{1}{12})^{2}}$

p= $\frac{1}{12}$ and p= $\frac{-1}{12}$

Since, Focus is left side of the vertex,

p= $\frac{-1}{12}$ is required value

Replacing value in general equation of parabola,

Vertex: (h,k)=(4,-3)

p= $\frac{-1}{12}$

$(x-h) = 4p(y-k)^{2}$

$(x-4) = 4(\frac{-1}{12})(y+3)^{2}$

$(x-4) = \frac{-1}{3}(y+3)^{2}$

8 0

2 years ago