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log 96889010407 / log7   = 13 

Realize that 7^13=96889010407.

therefore your answer is 13

After the And we get I mean And we get 13

We know that 96889010407 is 7^13 so \(log_7(96889010407)=13 \)and we get 

cos(300)=1/2 or \(\cos(300)=\dfrac12\)

we have 2(x-1)^2+(x-1)((x-1)+(x+2)) we can distribute to get 2x^2+−4x+2+x^2+−x+x^2+2x+−x+1+−x+−2 and then combine like terms to get (2x^2+x^2+x^2)+(−4x+−x+2x+−x+−x)+(2+1+−2) which can be simplfied as 4x^2+−5x+1 then becomes (D)4x^2−5x+1 so we have that the area is \(\boxed{(D)4x^2−5x+1}\)

Center rectangle   (x+2    + x-1)  *   (x-1)  = (2x+1)(x-1) = 2x^2 -x-1

     plus TWO squares    2 *  ( x-1)(x-1)   = 2 (x^2-2x+1)

Add the reds for area  

Angle opposite 57    is   180 - 57 = 123^o

ENTIRE lower arc from 57  to B is then   2 x 123 = 246^o

Arc encompassed by angle next to 'b' is thus   246 - 100 = 146^o

    meaning the supplementary angle with b   is     73^o  

                      b + 73 = 180       b = 107^o

we know that it has the same ration as 63:57 so we have 57/63*21=1197/63=19 so we have that the length of CG is \(\boxed{19}\)

we seperate them into a rectangle and a triangle where the rectangle's area is 9*20=180 and the triangle's area is (14-9)*(20-11)/2=5*9/2=45/2=22.5 so the area of the shape is 180+22.5=202.5 so we get that the area of the shape is \(\boxed{202.5}\)

We know that this is a right triangle since it has pythagorean triples 

see here:

https://www.geogebra.org/calculator/hty44jpe

I plotted here

we can use algebra to form the equation 25x+5(x+16)=590 we distribute to get 25x+(5)(x)+(5)(16)=590 then it becomes 25x+5x+80=590 and we combine like terms to get (25x+5x)+(80)=590 then 30x+80=590 we then subtract 80 from both sides to get 30x+80−80=590−80 then 30x=510 finally we divide both sides by 30 to get 30x/30=510/30 then we have x=17 so we get that she has \(17\text{ Quarters}\)

and 33 nickels :)

To check we do 5*33+17*25 which simplifies into 165+425=590 so that means that our answer 

I don't know how to solve it mathematically, but I can write a short computer code to find out how many numbers are there between 1 and 100,000 whose digits sum up to 22. Here is the code:

%%time;# Function to return the  sum of digits of x between L and R # 
def sumOfDigits(x): 
    sum = 0
    while x != 0: 
        sum += x % 10
        x = x//10
    return sum
  # Function to return the count 
# of required numbers 
def countNumbers(L, R): 
    count = 0
    for i in range(L, R + 1): 
  # ENTER SUM OF DIGITS HERE # 
     if sumOfDigits(i)==22: 
            count += 1
    return count 
  L =1; R = 99999
print("Total = ", f"{countNumbers(L, R):,d}") 

OUTPUT =6,000 [Exactly!]. Therefore, the probability is: 6,000 / 100,000 ==6%.

AREA:   10*8/2 for the area of the triangle and π4^2 so we add them up in a equation to get that the area 80/2+π4^2=40+16π so the area of the shaded region is 40+16π or \(\boxed{40+16π}\)

Circumfrence=8π/2 for the rounded part and 2*(sqrt(10^2+(8/2)^2)) again we make them into an equation 4π+2*(sqrt(10^2+4^2))=4π+2*(sqrt(116)) then we simplify it to get 4π+4*(sqrt(29)) or \(4π+4\sqrt{29}\)

This question has been asked multiple times...you could look for it as well Guest.

We know Julius has 87 stamps. Liam has 10. Then let the number of stamps Mario have be x. We know Julius has three times as many stamps as Liam+Mario so we form an equation.

87=3(10+x)

87=30+3x

3x=57

x=19.

Therefore Mario has 19 stamps.

We know that the perimeter is equal to 2x-1+2x-1+3x+3+3x+3=144 which can be simplfied to 4x-2+6x+6=114 and then 10x+4=114 we then subtract 4 to get 110 and divide by 10 to get x=11 so PQ=11*3+3=33+3=36 and we have that PQ=36 or \(\boxed{PQ=36}\)

let's say that y=-0.5 since that works with -1 < y < 0 so we compute that  A=0.25, B=1.5, C=0.5, D=-1 so we eventually see that B is the greatest so we get B or \(\color{green}{\boxed{B}}\)

SP + PQ = 114/2      solve for x 

an angle in a regular pentagon is 108^o and the triangle is isosceles so the measure of DEB is 108-((180-108)/2)=108-(72/2)=108-36=72^o sp we have that \(\angle DEB\) is 72^o or \(\boxed{72^\circ}\).

Great job, geno!!!

Thanks Guest I forgot the final step

\(z-\sqrt{-2i},z=-\sqrt{2i},z=\sqrt{-2i},\text{ and }z=\sqrt{2i}.\) since we quad root to get \(\pm\sqrt{\pm2i}\)

We have C(25,3)=2300 possible ways and there are C(3,1)*12=3*12=36 ways so the probability that at least two of the three had been sitting next to each other is \(\dfrac{36}{2300}=\dfrac9{575}\)

Perimeter is the sum of the two sides given  * 2

(2x-1  +  3x+3) * 2 = 114     

                    solve for 'x'   then   PQ = 3x+3