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You are lucky this got answered.    Thanks to guest answerer.

Next time proof read your question properly in the first place!

I did it partly by logic and partly by trial and error

999 base10  =  1 111 100 111 in base 2   there are 10 digits but not all can be a 1

so the maximum sum of the base 10 digits is 9

900 base 10  = 1 110 000 100    that is no good so it is less than 900

800 base 10 =  1 100 100 000   that is no good

the most I can add to it is 10 base10  so that is   1010 base two

which means that  1 100 will be the highest four base 2 digits which means that the max number of ones is 8 which is too small

so the number is less than 800

700, 701,702,710,711,720 are all no good  so it has to be less than 700

600,601,602,603,610, 611, 612, 620, 621, 630   all no good so it is less than 600

500 base 10 = 111 110 100 this doesn't work but it looks promising

there are 6 ones here  ... if I add 2 base 10 that will add 1 extra 1 ... sounds good

502 = 111 110 110    PERFECT

503 = 111 110 111    PERFECT

504,510,511,512,513,520,521,522,530,531,540  all no good

So the largest one is    503

Note: I cold easily have made a careless error.

First we list the two equations.

\(x + 3y = -4 \)                    (1)

\(3x - 7y = 36 + 5x - 2y\) we must simplify the second equation to get:

\(-2x - 5y = 36\)               (2)

Now we have a system of equations that looks much nicer. On to the math:

First we multiply equation(1) by -2 to get:

\(-2x - 6y = 8\)                  (3) We will call this equation 3 for the sake of clarity.

Next we subtract equation(3) from equation(2) to cancel out variable x.

\(-2x - 5y - (-2x - 6y) = 36 - 8\) This simplifies to:

\(y = 28\)

Now we can plug the value of \(y\) back into equation(1) to get \(x\):

\(x + 84 = -4\) Then we subtract 84 from both sides of the equation to get x:

\(x = -88\)

Now we have the ordered pair of (x,y)

\(x = -88\)

\(y = 28\)

NO NO NO SCRATCH THAT  

216 + 27  =  243  

Darn, that's what comes of trying to do addition in my head.  

_.

If x+y=9 and xy=18, what is the value of x^3+y^3?    

You can solve for x and y the long way around,

but we can see intuitively that they are 6 and 3.  

so         x³ + y³  =  6³ + 3³  =  216 + 27  =  238  

_.

Given two real numbers p and q such that 1/p + 1/q = 1 and pq = 9, what is p + q?  

                                                   1/p + 1/q  =  1  

multiply both sides by pq                  q + p  =  pq  

but the problem says pq = 9  

therefore                                            q + p  =  9  

_.

Megan baked an apple pie for her family for dinner. She cut the pie into eight pieces. Her father ate one-fourth of the pie, her mother ate one-eighth of the pie, her older brother ate one pieces of pie and Megan ate the rest. How many pieces of pie did Megan eat?

Father 1/4 • 8  =  2 pieces  

Mother 1/8 • 8  =  1 piece  

Brother             =  1 piece  

That totals 4 pieces, which left 4 pieces,  

so Megan ate 4 pieces.  The little p*g.  

_.

At 8:00 a.m. Mr.Hernandez started packing quarts of milk in crates. By 9:00 a.m., he had packed 72 quarts. He took a break from 9:00 a.m. to 9:30 a.m. From 9:30 a.m. to 10:30 a.m., he packed 85 quarts. How many quarts of milk did he pack from 8:00 a.m. to 10:30 a.m.?  

72 + 85 = 157  

David washes his father's car every seven days. Jason washes his mother's car every three days. Both of them washed their cars on October 28. On what date will David and Jason both wash cars again? Make an organized list to solve the problem.  

Date          David          Jason  

10/28            x                  x    

10/29  

10/30  

10/31                               x  

11/01  

11/02  

11/03                               x  

11/04            x  

11/05  

11/06                               x  

11/07  

11/08  

11/09                               x  

11/10  

11/11            x  

11/12                               x  

11/13  

11/14  

11/15                               x  

11/16  

11/17  

11/18            x                 x  

21 days, to November 18. 

You could get the number of days by multiplying 3 times 7 but the problem said make a list. 

_.

I'm assuming that "attendance will be down 25%" means:

108  - [ 25% x 108] ==108 - 27 ==81 students attending the concert.

Since each student will eat an average of 2 cookies, then that means:

Walter and Gretel will have to bake: 81  x  2 ==162 cookies. Since their recipe, which make one pan holding 15 cookies, then the number of recipes they will have to make should be:

162 cookies / 15 cookies per one pan ==10.8 pans or recipes. Since they have to make full recipes, then that means they will have to bake a total of 11 recipes for the 81 students attending the concert.

If you wish to calculate the total ingredients needed for the 11 recipes, then you will simply multiply:

11  x  1 1/2 cups of flour ==16.50 cups

11 x 2 eggs ==22 eggs

11 x  3 tabespoons of butter ==33 tablespoons.

11  x  3/4 cups of sugar ==8.25 cups.

11  x  1 package of chocolate drops ==11 packages.

15 cookies in a pan.

Each term is the previous term multiplied by 1/4

128 * 1/4 = 32 = x

32 * 1/4 = 8 = y

x + y = 32 + 8 = 40

The 21st of January would be Sunday.

\(27a=42b+17\\ 27a-42b=17\\ \)

The lowest common denominator of 27 and -42  is  3   and 3 does not go into 17.

Therefore I do not think there are any integer solutions here. 

a is meant to be an integer.

1/39 ==0.025641  025641  025641  025641  025641

Since the fraction has a "period" of 6, or repeating the same 6 digits, therefore the 291st digit should be:

291 mod 6==3, or the 3rd digit from the decimal point, or 5.

How many cookies in a pan?  There is a number missing.

Hi Melody, I just noticed your post...

Vielleicht könnte ich zu einem See wandern, und einen Fisch fangen!

I’ll say that, in your wandering soliloquy, your comment is in perfectly composed German syntax.  I can read and speak German reasonably well at the social level. When the conversation becomes technical, such as in complex science or math, I’ll spend more time using a dictionary or translator than writing or speaking. 

Once, while I was staying at a vacation rental near Bonn, the owners’ 22-year-old son, Hans, who was principal caretaker and bellhop, asked if I needed anything.  I wanted the bed moved a few feet closer to the window, so I could see the near full moon setting in the predawn morning. So, I said, 

“Bitte hilf mir mein Bett in den Mondstrahlen zu erleben.”

(“Please help me experience my bed in the moonbeams.”)

What I should have said is,

"Bitte helfen Sie mir, mein Bett in den Mondstrahlen zu bewegen" 

("Please help me move my bed into the moonbeams")

Hans could tell what I meant by my gestures and countenance, and I could tell he was greatly amused by my request.  He also said he’d be very happy to do both. 

----------

When people take 3 or more spoons of sugar in their coffee I often make a snide remark regarding them having coffee with their sugar.

LOL!!  I make similar snarky remarks, and sometimes I ask if I can bring them insulin and a syringe. I usually reserve the insolent-insulin remarks for those who put a third-cup or more of sugar in a twelve ounce mug of coffee.   

----

Giving birth to water sounds somewhat less painful than giving birth to a baby.

Years ago, I read a science fiction story where the earth-visiting aliens referred to the humans as “big bags of mostly water.”   We are LOL!

GA

--. .-

Because ABCD is a rhombus, then AB = BC = CD = AD.

AB = BC = CD = AD = 37 units.

One of rhombus ABCD's diagonals has length 28. We can divide 28 by 2 to get 14, which is the leg of a right triangle when rhombus ABCD is split into 4 congruent triangles.

The hypotenuse is 37, and the leg is 14, so that means the other leg is:\(\sqrt{37^2 - 14^2}\)  = \(\sqrt{1173}\). Ew that looks ugly, but it is correct. That means the other diagonal has length \(2\sqrt{1173}\). 

Since the area of a rhombus is one diagonal length times the other diagonal length over 2, then the area of this rhombus would be \(28\sqrt{1173}\) units squared.

This figure is not drawn to scale. 

We can exclude BD and AB because using the Hinge Theorem, AD is the smallest line segment out of triangle ADB.

Angle C = 65, Angle CDA = 60, and Angle DAC = 55.

Then using the Hinge Theorem, CD is the included side of the two largest angles in triangle ACD:

Therefore CD is the shortest side.

We can list them since we know that a multiple of 3 has to have a digit sum that is divisible by 3:

24

42

222

444

That's it... So we have 4 positive multiples of 3 less than 1000 using only the digits 2 and/or 4.

pr

We can first label the radii of each of the circles to a variable.

Radius of circle A = x

Radius of circle B = y

Radius of circle C = z

Now we can write some equations:

\(x + y = 14\)

\(y + z = 18\)

\(x + z = 28\)

Then we add up the three equations and divide by two to get:

\(x + y + z = 30\)

We take the equation above us (equation4) and subtract it from equation2 to get x = 12.

We take the equation above us (equation4) and subtract it from equation3 to get y = 2.

We take the equation above us (equation4) and subtract it from equation1 to get z = 16.

So the radius of circle A is 12

So the radius of circle B is 2

So the radius of circle C is 16

GEMETRY