MIRB14

10 Questions
23 Answers

+1

1687

4

+115

Geometry proof

Let PA and PC be tangents from P to a circle. Let B and D be points on the circle such that B, D, and P are collinear. Prove that AB*CD=BC*DA.
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MIRB14 4 Mar 2018

+1

1498

1

+115

In the figure, $PA$ is tangent to semicircle $SAR$, $PB$ is tangent to semicircle $RBT$, and $SRT$ is a straight line. If arc $AS$ is $58^\c

In the figure, PA is tangent to semicircle SAR, PB is tangent to semicircle RBT, and SRT is a straight line. If arc AS is 58 degrees and arc BT is 37 degrees, then find angle APB, in degrees.
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MIRB14 19 Feb 2018

+3

3173

2

+115

In triangle $ABC$, $AB = 5$, $AC = 6$, and $BC = 7$. Circles are drawn with centers $A$, $B$, and $C$, so that any two circles are externall

In triangle ABC, AB=5, AC=6, and BC=7. Circles are drawn with centers A, B, and C, so that any two circles are externally tangent. Find the sum of the areas of the circles.
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MIRB14 7 Feb 2018

+2

2282

1

+115

In regular octagon $ABCDEFGH$, $M$ and $N$ are midpoints of $\overline{BC}$ and $\overline{FG}$ respectively. Compute $[ABMO]/[EDCMO]$. ($[A

In regular octagon ABCDEFGH, M and N are midpoints of line BC and line FG respectively. Compute [ABMO]/[EDCMO]. ([ABCD] denotes the area of polygon ABCD.)

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MIRB14 29 Jan 2018

+1

3139

4

+115

In square $ABCD$, $E$ is the midpoint of $\overline{BC}$, and $F$ is the midpoint of $\overline{CD}$. Let $G$ be the intersection of $\overl

In square ABCD, E is the midpoint of line BC, and F is the midpoint of line CD. Let G be the intersection of line AE and line BF. Prove that DG = AB.

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MIRB14 20 Jan 2018

+1

3251

2

+115

The smaller square in the diagram has side length 1. Equilateral triangles are constructed on its four sides, as shown. What is the area of

The smaller square in the diagram has side length 1. Equilateral triangles are constructed on its four sides, as shown. What is the area of the larger square?
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MIRB14 17 Jan 2018

+2

1614

4

+115

A circle is drawn that intersects all three sides of triangle PQR as shown below. Prove that if AB = CD = EF, then the center of the circle

A circle is drawn that intersects all three sides of triangle PQR as shown below. Prove that if AB = CD = EF, then the center of the circle is the incenter of triangle PQR.
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MIRB14 17 Des 2017

+2

2414

1

+115

Angle bisectors TX and UY of triangle TUV meet at point I. Find all possible values of angle V if angle TIU = 109 degrees. As your answer,

Angle bisectors TX and UY of triangle TUV meet at point I. Find all possible values of angle V if angle TIU = 109 degrees. As your answer, enter the number of degrees in angle V. If you find more than one possibility, list the possible values in increasing और पढ़ें ..

MIRB14 13 Des 2017

+3

3898

3

+115

A square $DEFG$ varies inside equilateral triangle $ABC,$ so that $E$ always lies on side $\overline{AB},$ $F$ always lies on side $\overlin

A square varies inside equilateral triangle so that always lies on side always lies on side and always lies on side The point starts on side and ends और पढ़ें ..

MIRB14 8 Des 2017

+2

2198

1

+115

In the staircase-shaped region below, all angles that look like right angles are right angles, and each of the eight congruent sides marked

In the staircase-shaped region below, all angles that look like right angles are right angles, and each of the eight congruent sides marked with a tick mark have length 1 foot. If the region has area 53 square feet, what is the number of feet in the perimeter और पढ़ें ..

MIRB14 4 Des 2017

+115

+2

Thanks, but is there a way to not use trig?

MIRB144 Mar 2018

+115

+1

Nevermind I solved it...

MIRB1419 Feb 2018

+115

+1

triangle BCD is similar to triangle KJL because of SSS similarity.

triangle FGH is not similar to triangle SRQ.

triangle CDE is not similar to triangle MLC.

triangle LMN is not similar to triangle FEG.

MIRB1419 Feb 2018

+115

+2

ANYWAYS... here's the answer if you don't want to do it

Simplify the original line: We get y=-(2/3)x+8

The line parallel to another line has the same slope.

Therefore, our line would be y=-(2/3)x+b, b being the y-intercept.

We also know that the line passes through point (12,-4). We can substitute the points into the line equation to solve b.

We get -4=-(2/3)*12+b => -4=-8+b => b=4

The line is y=-(2/3)x+4

Check it by using desmos.com

MIRB1429 Jan 2018

+115

+2

The formula for a circle is (x – h)^2 + (y – k)^2 = r^2.

Now we just make the formula into our problem: (x-7)^2+(y-8)^2=10^2

Then we substitute (13,16) into the formula. If the solution is right, then (13,16) is on circle O. If not, then it's not.

(13-7)^2+(16-8)^2=100

6^2+8^2=100

36+64=100

100=100

This equation is true, so the answer is yes.

MIRB1429 Jan 2018

+115

+1

Mathisfornerds, please, instead of making so many topics, put all of your questions in a numbered list for one topic.

Your question is similar to your other questions, just change the equation into slope-intercept form and do the samething and do the samething I did for your other question.

MIRB1429 Jan 2018

+115

+2

The line parallel to another line has the same slope.

Therefore, our line would be \(y=\frac{1}{4}+b\), b being the y-intercept.

We also know that the line passes through point (-8,-7). We can substitute the points into the line equation to solve b.

We get \(-7=\frac{1}{4}\times-8+b\Rightarrow-7=-2+b\Rightarrow b=-5\)

The line is \(y=\frac{1}{4}x-5\)

Check it by using desmos.com

MIRB1429 Jan 2018

+115

+1

Question answered here: https://web2.0calc.com/questions/math-postage-stamp-problem

MIRB1425 Jan 2018

+115

+3

Yes, press shift+backslash to make this: |

Surround a number with the absolute value signs, such as this: |-2|

The calculator should spit out 2

MIRB1425 Jan 2018

+115

+1

Thanks!

MIRB1422 Jan 2018

+115

+2

Thanks!

MIRB1420 Jan 2018

+115

+2

If you're using a mac, you can do alt+< or > to get ≤ or ≥

MIRB1427 Des 2017