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3 years ago

Two distinct coplanar lines that do not intersect are known as ____ lines.

Mathematics

2 answers:

irakobra [83]3 years ago

6 0

The answer is parallel

madreJ [45]3 years ago

5 0

Coplanar
do not intersect

answer is paralell lines

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X * 2 - 21 x + 12 x - 12 x * 2 - 17

BaLLatris [955]

Answer:

The functions is;

x - 12

Step-by-step explanation:

Input is (x)

(2) - 12 = -10

(6) - 12 = -6

(10) - 12 = -2

(5) - 12 = -7

3 0

3 years ago

What is the "least common denominator" of the following two fractions: 1/6 and 1/2

astra-53 [7]

Answer:

it is 6

Step-by-step explanation:

2 and 6 are the denominators and the least common denominator is the number in which 2 & 6 can go in without leaving a remender and 6 is that number

8 0

3 years ago

Read 2 more answers

Can u solve this problem in show your work

aksik [14]

Hey there!

We can solve this problem by doing PEMDAS, which lists:

$Parentheses \\ Exponents \\ Multiplication \\ Division \\ Addition \\ Subtraction$

We have in this equation

$Parentheses \\ Exponents \\ Dvision$

Let's sp;ve this equation now!

Parentheses ↓

$(6 -3) = 3$

Exponents ↓

$3^2 = 9$

Division ↓

$\frac{36}{9} = 9$

Okay now , I show you the steps so it can be easier to solve!

$\frac{36}{(6-3)^2} \\ \\ 6-3 = 3 \\ \\ \frac{36}{3^2} \\ \\ 3^2 = 9 \\ \\ \frac{36}{9} \\ \\ \\ \\ Answer: 4$

Good luck on your assignment and enjoy your day!

~ $LoveYourselfFirst:)$

5 0

3 years ago

Read 2 more answers

Describe and sketch the suraface in R3 represented by the equation x+y=2.

Talja [164]

Answer:

The three-dimensional surface represented by the equation x+y=2 is an inclined plane, as seen in the image.

5 0

3 years ago

Water whose temperature is at 100∘C is left to cool in a room where the temperature is 60∘C. After 3 minutes, the water temperat

Tju [1.3M]

Answer:

21.68 minutes ≈ 21.7 minutes

Step-by-step explanation:

Given:

$T=60+40e^{kt}$

Initial temperature

T = 100°C

Final temperature = 60°C

Temperature after (t = 3 minutes) = 90°C

Now,

using the given equation

$T=60+40e^{kt}$

at T = 90°C and t = 3 minutes

$90=60+40e^{k(3)}$

$30=40e^{3k}$

$e^{3k}=\frac{3}{4}$

taking the natural log both sides, we get

3k = $\ln(\frac{3}{4})$

3k = -0.2876

k = -0.09589

Therefore,

substituting k in 1 for time at temperature, T = 65°C

$65=60+40e^{( -0.09589)t}$

$5=40e^{( -0.09589)t}$

$e^{( -0.09589)t}=\frac{5}{40}$

$e^{( -0.09589)t}=0.125$

taking the natural log both the sides, we get

( -0.09589)t = ln(0.125)

( -0.09589)t = -2.0794

t = 21.68 minutes ≈ 21.7 minutes

6 0

3 years ago