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Papessa [141]
3 years ago
8

9x^2 - 12x + 4 = 0 Solve for x

Mathematics
2 answers:
hram777 [196]3 years ago
4 0

Answer:

x = 2/3

Step-by-step explanation:

9x2−12x+4=0

Step 1: Factor left side of equation.

(3x−2)(3x−2)=0

Step 2: Set factors equal to 0.

3x−2=0 or 3x−2=0

x=  2 /3

Dvinal [7]3 years ago
3 0

Answer:

Step-by-step explanation:

hello :

9x² - 12x + 4 = 0

a=9   b=-12  c= 4

delta = b²-4ac

delta =  (-12)²- 4(9)(4)=144 - 144 = 0

single solution x = -b/2a = -(-12)/18  = 12/18 = 2/3

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7 0
3 years ago
The ratio of the measures of the three angles in a tangle 3107. Find the measures of the angles and write them in size order,
N76 [4]

Answer:

27, 90 and 63

Step-by-step explanation:

Given

Ratio of triangle sides

Ratio = 3 : 10 : 7

Required:

The length of each side

Triangles in a triangle add up to 180.

The side with ratio 3 is:

S_1 = \frac{3}{3 + 10 + 7} *180

S_1 = \frac{3 *180}{20}

S_1 = \frac{540}{20}

S_1 = 27

The side with ratio 10 is:

S_2 = \frac{10}{3 + 10 + 7} *180

S_2 = \frac{10 *180}{20}

S_2 = \frac{1800}{20}

S_2 = 90

Lastly:

The side with 7 as its ratio

S_3 = \frac{7}{3 + 10 + 7} *180

S_3 = \frac{7 *180}{20}

S_3 = \frac{1260}{20}

S_3 = 63

Hence, the angles are: 27, 90 and 63

5 0
3 years ago
Atul has 2/3 lb of candy. Jose has 3/5 lb and Maria has 1/2 lb less then José. How many more pounds of candy does Atul have than
liraira [26]

Given:

Atul has \dfrac{2}{3} lb of candy.

Jose has \dfrac{3}{5} lb of candy.

Maria has \dfrac{1}{2} lb less than Jose.

To find:

How many more pounds of candy does Atul have than Maria?

Solution:

Since, Maria has \dfrac{1}{2} lb less than Jose, therefore

Maria has = \dfrac{3}{5}-\dfrac{1}{2} lb

Maria has = \dfrac{6-5}{10} lb

Maria has = \dfrac{1}{10} lb

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Difference = \dfrac{2}{3}-\dfrac{1}{10} lb

Difference = \dfrac{20-3}{30} lb

Difference = \dfrac{17}{30} lb

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Answer:

(22 + 26) ÷ 8 = g (or groups)

3 0
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Dmitry_Shevchenko [17]

Answer:

-4y^2 - 2y + 10

Step-by-step explanation:

8 0
2 years ago
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