All categories

3 years ago

What is equivalent to the expression 3x^2+27

Mathematics

1 answer:

MA_775_DIABLO [31]3 years ago

6 0

Please give brainliest

Answer: 3(x^2+9)

You might be interested in

311/500 as a decimal and 77/200 as a decimal

kogti [31]

311/500 as a decimal = 0.622

77/200 as a decimal = 0.385

You can solve this by doing long division.

4 0

3 years ago

Read 2 more answers

Select the correct answer. factor completely 12x^2+28x

hjlf

Answer:

B. 4x(3x+7)

Step-by-step explanation:

4 0

3 years ago

A house costs $600 more than 12 times the lot on which it was built. If the lot cost x dollars, what did the house cost?

Artemon [7]

Your answer is $600+12x

8 0

3 years ago

Read 2 more answers

A. 18 cm B. 40 cm C. 20 cm D. 36 cm

Doss [256]

Answer:

Step-by-step explanation:

108/(108-72) = 108/36 = 3

60/3 = 20

5 0

3 years ago

A jar contains 12 red marbles numbered 1 to 12 and 6 blue marbles numbered 1 to 6. A marble is drawn at random from the jar. Fin

lora16 [44]

Answer:

$(a)P(Red)=\dfrac{2}{3}\\(b)P(Odd) =\dfrac{1}{2}\\(c)P(\text{Red or Odd Numbered})=\dfrac{5}{6}\\(d)P(\text{Blue or Even Numbered})=\dfrac{2}{3}$

Step-by-step explanation:

Number of Red Marbles{1,2,3,4,5,6,7,8,9,10,11,12},n(R)=12

Number of Blue Marbles{1,2,3,4,5,6},n(B)=6

Total Number of Marbles, n(S)=6+12=18

(a)Probability that the Marble is Red

$P(R)=\dfrac{n(R)}{n(S)} =\dfrac{12}{18} =\dfrac{2}{3}$

(b)Probability that the marble is odd-numbered.

Number of Odd-Numbered Balls, n(O)=9

$P(Odd)=\dfrac{n(O)}{n(S)} =\dfrac{9}{18} =\dfrac{1}{2}$

(c)Probability that the marble is red or odd-numbered.

n(Red)=12

n(Odd Numbered marbles)=9

n(Red and Odd Numbered Marbles)=6

$P(\text{Red or Odd Numbered})=P(Red)+P(Odd\:Numbered)-P(\text{Red and Odd Numbered)}\\=\dfrac{12}{18} +\dfrac{9}{18}-\dfrac{6}{18} =\dfrac{15}{18}\\P(\text{Red or Odd Numbered})=\dfrac{5}{6}$ (d)Probability that the marble is blue or even-numbered.

n(Blue)=6

n(Even Numbered marbles)=9

n(Blue and Even Numbered Marbles)=3

$P(\text{Blue or Even Numbered})=P(Blue)+P(Even\:Numbered)-P(\text{Blue and Even Numbered)}\\=\dfrac{6}{18} +\dfrac{9}{18}-\dfrac{3}{18} =\dfrac{12}{18}\\P(\text{Blue or Even Numbered})=\dfrac{2}{3}$

5 0

3 years ago