Mathematics, 23.11.2019 02:31 lyn94

Zero product property to solve equations
1. a^2+8a=0
2. (3h-6)(2h-8)=0
3. 6m^2+17m+5=0
4. y^2-36=0
5. (3g+4)^2=0

Answers: 1

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1. a² + 8a = 0
a(a) + (8) = 0
a(a + 8) = 0
a = 0 or a + 8 = 0
- 8 - 8
a = -8

2.   (3h - 6)(2h - 8) = 0
(3(h) - 3(2))(2(h) - 2(4)) = 0
(3(h - 2))(2(h - 4)) = 0
6(h - 2)(h - 4) = 0
6 6
(h - 2)(h - 4) = 0
h - 2 = 0 or h - 4 = 0
+ 2 + 2 + 4 + 4
h = 2 h = 4

3. 6m² + 17m + 5 = 0
6m² + 15m + 2m + 5 = 0
3m(2m) + 3m(5) + 1(2m) + 1(5) = 0
3m(2m + 5) + 1(2m + 5) = 0
(3m + 1)(2m + 5) = 0
3m + 1 = 0 or 2m + 5 = 0
  - 1 - 1    - 5 - 5
3m = -1 2m = -5
3 3 3   3
m = ⁻¹/₃   m = -1²/₃

4. y² - 36 = 0
y² - 6y + 6y - 36 = 0
y(y) - y(6) + 6(y) - 6(6) = 0
y(y - 6) + 6(y - 6) = 0
  (y + 6)(y - 6) = 0
y + 6 = 0 or y - 6 = 0
- 6 - 6 + 6 + 6
y = -6 y = 6

5. (3g + 4)² = 0
(√(3g + 4))² = √(0)
  3g + 4 = 0
  - 4 - 4
3g = -4
3 3
g = -1¹/₃

Answer from: Quest

can i get the link to the assignment?

step-by-step explanation:

Answer from: Quest

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in $\delta bea$ and $\delta dea$ ;

$ba=da$ , as these are radii of our given circle.

$ea=ea$

therefore, $\delta bea\cong \delta dea$ by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

$3x-24=x+10$

$3x-x=24+10$

$2x=34$

$x=\frac{34}{2}$

$x=17$

now let us substitute x=17 in the expression for the length of be.

$\text{length of be}=3x-24$

$\text{length of be}=3\cdot 17-24$

$\text{length of be}=51-24$

$\text{length of be}=27$

therefore, length of be will be 27 units and option c is the correct choice.

In circle a, ∠bae ≅ ∠dae. what is the length of be? 14 units 17 units 27 units 34 units

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Zero product property to solve equations
1. a^2+8a=0
2. (3h-6)(2h-8)=0
3. 6m^2+17m...

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