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3 years ago

If two parallel lines are cut by a transversal, then alternate exterior angles are

Mathematics

2 answers:

Lina20 [59]3 years ago

6 0

The correct answer is A) Congruent. This means they are equal in size.

olasank [31]3 years ago

5 0

Answer:

<u>A: Congruent</u>

Step-by-step explanation:

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Change each improper fraction into an equal mixed number.

olchik [2.2K]

A.) 3 3/4

B.) 4 5/9

C.) 8. 375 - I couldn't solve this one

D.) 12 3/11

E.) 5 7/20

F.) 29

6 0

3 years ago

What is the range of the cubic function f(x)=2x^3+21

boyakko [2]

ANSWER

$( - \infty , + \infty )$

EXPLANATION.

The given function is

$f(x) = 2 {x}^{3} + 21$

Let

$y= 2 {x}^{3} + 21$

Solve for x.

$2 {x}^{3} = y - 21$

${x}^{3} = \frac{1}{2} y - \frac{21}{2}$

Take cubic root of both sides,

$x = \sqrt[3]{ \frac{1}{2} y - \frac{21}{2} }$

x is defined for all real values of y.

The range refers to the values of y, for which x is defined.

Hence the range is all real numbers.

In interval notation, we write this as,

$( - \infty , + \infty )$

The correct choice is A.

6 0

3 years ago

Read 2 more answers

A substance with a half life is decaying exponentially. If there are initially 12 grams of the substance and after 70 minutes th

77julia77 [94]

Answer: 233 min

Step-by-step explanation:

This problem can be solved by the following equation:

$A=A_{o} e^{-kt}$ (1)

Where:

$A=7 g$ is the quantity left after time $t$

$A_{o}=12 g$ is the initial quantity

$t=70 min$ is the time elapsed

$k$ is the constant of decay for the material

So, firstly we need to find the value of $k$ from (1) in order to move to the next part of the problem:

$\frac{A}{A_{o}}=e^{-kt}$ (2)

Applying natural logarithm on both sides of the equation:

$ln(\frac{A}{A_{o}})=ln(e^{-kt})$ (3)

$ln(\frac{A}{A_{o}})=-kt$ (4)

$k=-\frac{ln(\frac{A}{A_{o}})}{t}$ (5)

$k=-\frac{ln(\frac{7 g}{12 g})}{70 min}$ (6)

$k=0.00769995 min^{-1}$ (7) Now that we have the value of $k$ we can solve the other part of this problem: Find the time $t$ for $A=2 g$ .

In this case we need to isolate $t$ from (1):

$t=-\frac{ln(\frac{A}{A_{o}})}{k}$ (8)

$t=-\frac{ln(\frac{2 g}{12 g})}{0.00769995 min^{-1}}$ (9)

Finally:

$t=232.697 min \approx 233 min$

5 0

3 years ago

Xavier exercises by riding his bike. During his first week of training, he rides a total of 15 miles. Each week he increases the

Alenkinab [10]

Step-by-step explanation:

26 weeks are required hope in right

8 0

3 years ago

Is 5/5 greater than 99/100

AlladinOne [14]

5/5 is the same as 1 while 99/100 is only 0.99 (is less than 1).....So yes, 5/5 is greater than 99/100.

6 0

3 years ago

Read 2 more answers