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Scorpion4ik [409]
3 years ago
5

What is the name for facts in subtraction problems

Mathematics
1 answer:
12345 [234]3 years ago
6 0

The name for facts in subtraction problems: Fact Family is a group of math facts that uses the same numbers, for addition, and subtraction.

Answer: =====> I believe is, Subtraction Fact Strategies, or Fact Families.



1. ). Strategy: ===> Think Addition:

1.). Subtraction Description ====> Using the known addition fact to solve the subtraction problem. Ex. 13-5, think what goes with 5 to make 13?


2.). Strategy: ===> Fact Families

2.). Subtraction Description ====> Think of the fact family to recall the missing number.


3.). Strategy ===> Build Up Through

3.). Subtraction Description ===> Ten Used when either the subtrahend or minuend is 8 or 9. Ex: 14-9: start with 9 and work up through 10: 9 and 1 is 10 and 4 more makes 5.


4.). Strategy ==> Back Down Through Ten

4.). Subtraction Description: ==> Working backward with 10 as a “bridge”. Ex. 15-6, Take 5 away from 15 to get to ten. Then take one more away, leaving 9.




Hope that helps!!! : )

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For the polynomial function ƒ(x) = −x6 + 3x4 + 4x2, find the zeros. Then determine the multiplicity at each zero and state wheth
murzikaleks [220]

There is a multiple zero at 0 (which means that it touches there), and there are single zeros at -2 and 2 (which means that they cross). There is also 2 imaginary zeros at i and -i.


You can find this by factoring. Start by pulling out the greatest common factor, which in this case is -x^2.


-x^6 + 3x^4 + 4x^2

-x^2(x^4 - 3x^2 - 4)


Now we can factor the inside of the parenthesis. You do this by finding factors of the last number that add up to the middle number.


-x^2(x^4 - 3x^2 - 4)

-x^2(x^2 - 4)(x^2 + 1)


Now we can use the factors of two perfect squares rule to factor the middle parenthesis.


-x^2(x^2 - 4)(x^2 + 1)

-x^2(x - 2)(x + 2)(x^2 + 1)


We would also want to split the term in the front.


-x^2(x - 2)(x + 2)(x^2 + 1)

(x)(-x)(x - 2)(x + 2)(x^2 + 1)


Now we would set each portion equal to 0 and solve.


First root

x = 0 ---> no work needed


Second root

-x = 0 ---> divide by -1

x = 0


Third root

x - 2 = 0

x = 2


Forth root

x + 2 = 0

x = -2


Fifth and Sixth roots

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x = +/- \sqrt{-1}

x = +/- i

7 0
3 years ago
Find the intersection points using substitution or elimination for each system of equations:
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Answer:

Step-by-step explanation:

Okay, so I think I know what the equations are, but I might have misinterpreted them because of the syntax- I think when you ask a question you can use the symbols tool to input it in a more clear way, otherwise you can use parentheses and such.

Problem 1:

(x²)/4 +y²= 1

y= x+1

*substitute for y*

Now we have a one-variable equation we can solve-

x²/4 + (x+1)² = 1

x²/4 + (x+1)(x+1)= 1

x²/4 + x²+2x+1= 1

*subtract 1 from both sides to set equal to 0*

x²/4 +x^2+2x=0

x²/4 can also be 1/4 * x²

1/4 * x² +1*x² +2x = 0

*combine like terms*

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now, you can use the quadratic equation to solve for x

a= 5/4

b= 2

c=0

the syntax on this will be rough, but I'll do my best...

x= (-b ± √(b²-4ac))/(2a)

x= (-2 ±√(2²-4*(5/4)*(0))/(2*(5/4))

x= (-2 ±√(4-0))/(2.5)

x= (-2±2)/2.5

x will have 2 answers because of ±

x= 0 or x= 1.6

now plug that back into one of the equations and solve.

y= 0+1 = 1

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Hopefully this explanation was enough to help you solve problem 2.

Problem 2:

x² + y² -16y +39= 0

y²- x² -9= 0

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Answer:

round answer is : 10 ft

Step-by-step explanation:

The explanation is in the picture!

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