MATHEMATICS COLLEGE

2# Find the missing variable (s). Round to the nearest tenth.

Answers

Answer 1

Answer:

15 cos(33°) m

Step-by-step explanation

Presumably m stands for meters.

If RS were 1, x would be cos(33°)

Multiply sides of triangle by 15 m

When calculating cos(33°) first make sure cos(90°) comes out zero, not -0.448...

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Mike buys a pack of markers for $2.85. He uses a 20% off coupon, what is the price of the markers with the coupon?A $2.28
B $3.42
C .57
D $2.53

Answers

The answer would be C

Answer:

c.57

Step-by-step explanation:

What is the measure of 0 in radians?

Answers

Answer:

Its 7 pi over 6

Step-by-step explanation:

Answer:

180 degree?

Step-by-step explanation:

Solve each inequality for x. (Enter your answers using interval notation.)(a) ln(x) < 0

(b) ex > 2

Solve each equation for x.

(a) ln(3x − 17) = 5

Express the given quantity as a single logarithm.

ln(a + b) + ln(a − b) − 5 ln(c)

Answers

Using exponential and logarithmic functions, it is found that:

a) The solution of the inequality $ln((x)) < 0$ is $(-infty, 1)$ .

b) The solution of the inequality $e^x > 2$ is $(ln(2), \infty)$

a) The solution to the equation $ln((3x - 17)) = 5$ is x = 55.14.

As a single logarithm, the expression is:

$\ln{\left((a^2 - b^2)/(c^5)\right)}$

Inequality a:

$ln((x)) < 0$

Applying the exponential to both sides:

$e^(ln((x))) < e^(0)$

$x < 1$

Hence, in interval notation:

$(-infty, 1)$

Inequality b:

$e^x > 2$

Applying ln to both sides:

$ln(e^x) > ln(2)$

$x > ln(2)$

Hence, in interval notation, the solution is:

$(ln(2), \infty)$

Equation a:

$ln((3x - 17)) = 5$

$e^(ln((3x - 17))) = e^5$

$3x - 17 = e^5$

$3x = 17 + e^5$

$x = (17 + e^5)/(3)$

$x = 55.14$

The quantity given is:

$ln((a + b)) + ln((a - b)) - 5ln(c)$

To express as a single logarithm, these following properties are applied:

$ln(a) + ln(b) = ln(ab)$

$ln(a) - ln(b) = \ln{\left((a)/(b)\right)}$

$aln(b) = ln(b^a)$

Hence:

$ln((a + b)) + ln((a - b)) - 5ln(c) = ln((a + b)(a - b)) - ln(c^5)$

As a single logarithm, the expression is:

$\ln{\left((a^2 - b^2)/(c^5)\right)}$

A similar problem is given at brainly.com/question/21506771

Answer:

(a) ln(x) = 0

Then 0 < x < 1

(b) e^x > 2

Then ln2 < x < ∞

(a) ln(3x - 17) = 5

x = 55.1377197

ln(a + b) + ln(a - b) - 5ln(c)

= ln[(a² - b²)/c^5]

Step-by-step explanation:

First Part.

(a) ln(x) < 0

=> x < e^(0)

x < 1 ....................................(1)

But the logarithm of 0 is 1, and the logarithm of negative numbers are undefined, we can exclude the values of x ≤ 0.

In fact the values of x that satisfy this inequalities are between 0 and 1.

Therefore, we write:

0 < x < 1

(b) e^x > 2

This means x > ln2

and must be finite.

We write as:

ln2 < x < ∞

Second Part.

(a) ln(3x - 17) = 5

3x - 17 = e^5

3x = 17 + e^5

x = (1/3)(17 + e^5)

= 55.1377197

Third Part.

We need to write

ln(a + b) + ln(a - b) - 5ln(c)

as a single logarithm.

ln(a + b) + ln(a - b) - 5ln(c)

= ln(a + b) + ln(a - b) - ln(c^5)

= ln[(a + b)(a - b)/(c^5)]

= ln[(a² - b²)/c^5]

Look at the set of ordered pairs.{(4, 6), (−7,−15), (13, 15), (−21, 8), (?, ?)}

Which of the following could replace the missing ordered pair to make the set not a function?

(−2, 4)
(21, 9)
(−7, 15)
(6, 8)
(13,−15)

Answers

The ordered pair that could replace the missing ordered pair to make the set not a function is given as follows:

(−7, 15).

When does a set of ordered pairs represent a function?

A set of ordered pairs represents a function if every input (first component of the ordered pair) is associated with exactly one output (second component of the ordered pair). In other words, if no two ordered pairs in the set have the same first component but different second components.

Hence the ordered pair (-7,15) would make the relation not a function, as the set already has the ordered pair (-7,-15), in which the input 7 is already mapped to an output of -15, hence it cannot be mapped to an output of 15.

More can be learned about relations and functions at brainly.com/question/12463448

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What's the answer? It's not a fraction.