The prime factorization of 293 in expanded form

Answers

Answer 1

Answer: You've come to the right place. Here is the answer: None, because 293 is a prime number.

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Answer 2

Answer:

Answer:

hello

Step-by-step explanation:

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flora is packing her baby sister's blocks in a box that has a volume of 24 cubic inches.if the box is 2 inches tall,how long and wide could the box be?

How do I set this up

Answers

To justify the yearly membership, you want to pay at least the same amount as a no-membership purchase, otherwise you would be losing money by purchasing a yearly membership. So set the no-membership cost equal to the yearly membership cost and solve.

no-membership costs $2 per day for swimming and $5 per day for aerobic, in other words, $7 per day. So if we let d = number of days, our cost can be calculated by "7d"

a yearly membership costs $200 plus $3 per day, or in other words, "200 + 3d"

Set them equal to each other and solve:

7d = 200 + 3d

4d = 200

d = 50

So you would need to attend the classes for at least 50 days to justify a yearly membership. I hope that helps!

How do u divide 34.4 by 4 and show your work.

Answers

34.4 ÷ 4 = 8.6, which 8.6 as a fraction is 43/5

(3 1/2 - 9 3/4) divided by (-2.5)

Answers

Answer:

Step-by-step explanation:

Let's take the numerator first

3 1/2 - 9 3/4

Convert to improper fraction

7/2 - 39/4

Find the LCM

$(14 - 39)/(4)$

= -25/4

Now let's take the denominator

-2.5

Convert to fraction

-5/2

Combine the two parts of the fraction

-25/4 / (-5/2)

This becomes;

-25/2 * -2/5

= +5

the pet store sells bags of pet food. there are 4 bags of cat food. one sixth of the bags of food are bags of cat food. how many bags of pet food does the pet store have?

Answers

28 bags if you include the cat food (which you would)

The answer is 24.

Work:
4 x 6=24

4=cat food bags
6=denominator of fraction
24=total bags of food

1/6 + 3/8 in simplest form

Answers

13/24 is the correct answer

13/24 because the least common denominator is 24. write 1/6 as 4/24 and 3/8 as 9/24.

9/24+4/24= 13/24 and 13/24 cannot be simplified/reduced

Find all possible value of the given variable 1.h²+5h=0
2.z²-z=0
3.m²+13m+40=0
4.z²-3z=0
5.q²+7q=0
6.k²+2k=0
7.x²-3x-70=0
8.q²+7q-60=0
9.z²+9z-36=0
10.d²-13d+22=0

Answers

$1.\n \n h^2+5h=0 \n \nh(x+5)=0\n \nx=0 \ \ \ or \ \ \ x+5 =0\ \ |-5\n \nx+5-5=0-5\n \nx=0 \ \ \ or \ \ \ x=-5$

$2.\n \n z^2-z=0\n \nz(x-1)=0\n \nz=0 \ \ \ or \ \ \ z-1 =0 \ \ | +1\n \nz-1+1 =0 +1 \n \nx=0 \ \ \ or \ \ \ z=1$

$3.\n \nm^2+13m+40=0 \n \na=1 ,\ b=13, \ c=40 \n \n\Delta =b^2-4ac =13^2-4\cdot 1\cdot 40=169 - 1600=-1431 \n \nand \ we \ know \ when \ \Delta \ is \ negative, \ theres \ no \solution$

$4.\n \nz^2-3z=0 \n \n (z-3)=0\n \nz=0 \ \ \ or \ \ \ z-3 =0\ \ |+3\n \n z-3+3=0+3\n \nz=0 \ \ \ or \ \ \ z=3$

$5.\n \nq^2+7q=0 \n \nq(q+7)=0\n \nq=0 \ \ \ or \ \ \ q+7 =0\ \ |-7\n \nq+7-7=0-7\n \nq=0 \ \ \ or \ \ \ q=-7$

$6.\n \nk^2+2k=0\n \nk(k+2)=0\n \nk=0 \ \ \ or \ \ \ k+2 =0\ \ |-2\n \nk+2-2=0-2\n \nk=0 \ \ \ or \ \ \ k=-2$

$7. \n \n x^2-3x-70=0 \n \na=1,\ b=-3, \ c=-70 \n \n\Delta =b^2-4ac = (-3)^2-4\cdot 1\cdot (-70)= 9+280=289\n \n x_(1)=(-b-√(\Delta) )/(2a)=(3-√(289))/(2 )=( 3-17)/(2)=(-14)/(2)=-7$

$x_(2)=(-b+√(\Delta) )/(2a)=(3+√(289))/(2 )=( 3+17)/(2)=(20)/(2)=10\n \n(x+7)(x-10)=0$

$8.\n \nq^2+7q-60=0 \n \na=1,\ b=7, \ q=-60 \n \n\Delta =b^2-4ac = 7^2-4\cdot 1\cdot (-60)=49+240=289 \n \n x_(1)=(-b-√(\Delta) )/(2a)=(-7-√(289))/(2 )=( -7-17)/(2)=(-24)/(2)=-12$

$x_(2)=(-b+√(\Delta) )/(2a)=(-7+√(289))/(2 )=( -7+17)/(2)=( 10)/(2)= 5\n \n(x+12)(x-5)=0$

$9.\n \nz^2+9z-36=0 \n \na=1,\ b=9, \ q=-36 \n \n\Delta =b^2-4ac = 9^2-4\cdot 1\cdot (-36)= 81+144=225\n \n x_(1)=(-b-√(\Delta) )/(2a)=(-9-√(225))/(2 )=( -9-15)/(2)=(-24)/(2)=-12$

$x_(2)=(-b+√(\Delta) )/(2a)=(-9+√(225))/(2 )=( -9+15)/(2)=(6)/(2)=3\n \n(x+11)(x-3)=0$

$10.\n \nd^2-13d+22=0 \n \na=1,\ b=-13, \ q=22 \n \n\Delta =b^2-4ac = (-13)^2-4\cdot 1\cdot 22= 169-88=81\n \n d_(1)=(-b-√(\Delta) )/(2a)=(13-√(81))/(2 )=( 13-9)/(2)=(4)/(2)=2$

$d_(2)=(-b+√(\Delta) )/(2a)=(13+√(81))/(2 )=( 13+9)/(2)=(22)/(2)=11\n \n(d-2)(d-11)=0$

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