Factorise xy(x-y)+yz(y-z)+zx(z-x)

Answers

Answer 1

Answer: expand first:-
yx^2 - xy^2 + zy^2 -yz^2 + xz^2 - zx^2
there is no common factor so we cant factorize this

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Which statement about the given equation is NOT true?3x + 4y = 12 a. It is in slope-intercept form with a positive slope of 4/3. b. It is an example of a linear equation. c. It has two variables and has coefficients of 3 and 4. d. It represents a line which, when graphed, doesn't pass through the origin.
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It takes 8 minutes for Byron to fill the kiddie pool in the backyard using only a handheld hose. When his younger sister is impatient, Byron also uses the lawn sprinkler to add water to the pool so it is filled more quickly. If the hose and sprinkler are used together, it takes 5 minutes to fill the pool. Which equation can be used to determine r, the rate in parts per minute, at which the lawn sprinkler would fill the pool if used alone?

Answers

i think 8 or 7 because its -3 so +3 to the 5 minutes or 2 since it is a sprinkler hope i help $\alpha \alpha \alpha \alpha \beta x_(123) \lim_(n \to \infty) a_n$

Final answer:

To find the rate at which the lawn sprinkler would fill the pool if used alone, subtract the rate of the hose from the combined rate. The equation is rs = 1/5 - 1/8.

Explanation:

To determine the rate at which the lawn sprinkler would fill the pool if used alone, we can set up an equation using the concept of rates. Let r be the rate at which the sprinkler fills the pool. If it takes 8 minutes for Byron to fill the pool with just the hose, then the rate of the hose alone is 1 pool/8 minutes, or rh = 1/8. If it takes 5 minutes to fill the pool when both the hose and sprinkler are used together, then the combined rate is 1 pool/5 minutes, or rc = 1/5.

The rate of the sprinkler alone, rs, can be determined by subtracting the rate of the hose from the combined rate. Thus, we have rs = rc - rh. Substituting the given values, we have rs = 1/5 - 1/8.

Therefore, the equation that can be used to determine the rate at which the lawn sprinkler would fill the pool if used alone is rs = 1/5 - 1/8.

Learn more about Rate of filling pool with hose and sprinkler here:

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Wat does y2-y1 / x2-x1 have to do with sequences??!!

Answers

It doesn't. That's the formula to calculate the gradient of a line (of the form y=mx+c).

the area of a rectangle is 54 cm. The length is 2 cm more than x and the width is 5 cm less than twice x. Slove for x. round your answer to the nearest whole number.

Answers

area=LW
54=LW
L is 2 more than x L=2+x
w is 5 less than 2x W=-5+2x

54=LW
L=2+x
W=-5+2x
input
54=(2+x)(-5+2x)
distribute/FOIL
54=-10+4x-5x+2x^2
add like terms
54=-10-x+2x^2
minus 54
2x^2-x-64=0
quadratic formula
x= $\frac{ -b+/-\sqrt{b^(2)-4ac} }{2a}$
x= $\frac{ -(-1)+/-\sqrt{(-1)^(2)-4(2)(-64)} }{2(2)}$
x= $( 1+/-√(1-(-256)) )/(4)$
x= $( 1+/-√(1-(-256)) )/(4)$
x=4.257 or -3.75
we disregard the negative one because that would make legnth and width negative

x=4 (rounded)

The sum of two prime numbers is 85. what is the product of these two prime numbers

Answers

$2 +83 =85 \n \n 2 * 83 = \boxed {\bf {166}}$

Write in vertex form:

y = x2 - 8x + 19

Answers

$y=x^2-8x+19\ny=x^2-8x+16+3\ny=(x-4)^2+3$

In parallelogram ABCD, E is the midpoint of AB and F is the midpoint of DC . Let G be the intersection of the diagonal DB and the line segment EF . Prove that G is the midpoint of EF.

Answers

The midpoint of the line $\overline{EF}$ is the point that divides $\overline{EF}$ in two halves of the same length.

ΔDFG ≅ ΔBGE and $\overline{FG}$ ≅ $\overline{EG}$ by CPCTC, therefore, Gis the midpoint of $\overline{EF}$

Reasons:

The given parameters are;

The midpoint of AB in parallelogram ABCD = E

The midpoint of DC = F

Point of intersection of EF and DB = Point G

Required:

To prove that point G is the midpoint of EF.

Solution:

Statement ${}$ Reason

1. m∠BDC ≅ m∠ABD ${}$ 1. Alternate angles theorem

2. m∠DGF ≅ m∠BGE ${}$ 2.Vertical angles theorem

3. $\overline{DC}$ = $\overline {AB}$ ${}$ 3. Opposite sides of a parallelogram ABCD

4. $\overline{CF}$ ≅ $\overline{DF}$ ${}$ 4. Definition of midpoint of DC

5. $\overline{CF}$ = $\mathbf{\overline{DF}}$ ${}$ 5. Definition of congruency

6. $\overline{CF}$ + $\overline{DF}$ = DC ${}$ 6. Segment addition property

7. $\overline{CF}$ + $\overline{CF}$ = DC ${}$ 7. Substitution property

8. 2· $\overline{CF}$ = DC ${}$ 8. Addition

9. $\overline{CF}$ = 0.5· $\overline{DC}$ = $\overline{DF}$ ${}$ 9. Division property

Similarly;

10. $\overline{AE}$ = 0.5· $\overline{AB}$ = $\overline{EB}$ ${}$ 10. Division property

11. 0.5· $\overline{DC}$ = 0.5· $\overline{AB}$ ${}$ 11. Multiplication property of equality

12. $\overline{AE}$ = $\overline{EB}$ ${}$ 12. Substitution property

13. ΔDFG ≅ ΔBGE ${}$ 13. Angle-Angle-Side rule of congruency

14. $\overline{FG}$ ≅ $\overline{EG}$ ${}$ 14. CPCTC ${}$

15. $\overline{FG}$ = $\overline{EG}$ ${}$ 15. Definition of congruency

16. Point G is the midpoint of $\overline{EF}$ ${}$ 17. Definition of midpoint

Learn more about the midpoint of a line here:

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Answer:

GF = GE that prove G is the mid-point of EF

Step-by-step explanation:

In the Parallelogram ABCD

∵E is the mid-point of AB

∵F is the mid-point of CD

∵AB = CD opposite sides in the parallelogram

∴EB = DF⇒(1)

∵AB // CD opposite sides in the parallelogram

∴m∠EBD = m∠FDB alternate angles ⇒(2)

∵BD intersects EF at G

∴m∠BGE = m∠DGF vertically opposite angles ⇒(3)

By using (1) , (2) and (3) you can prove:

ΔBGE is congruent to ΔDGF ⇒ AAS

∴GF = GE

∴G is the mid-point of EF

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