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2 years ago

Convert. _____seconds = 15 minutes 19 seconds

Mathematics

2 answers:

chubhunter [2.5K]2 years ago

8 0

Answer:

919

Step-by-step explanation:

bija089 [108]2 years ago

7 0

Answer: 919 seconds in 15 minutes and 19 seconds

Step-by-step explanation:

60 x 15 = 900

900 + 19 = 919

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Write the equation in exponential form. Assume that all constants are positive and not equal to 1.

algol13

Answer:

1. $16 = 4^2$

2. $2 = {16}^{\frac{1}{4}}$

3. $log_2 y=z$

Step-by-step explanation:

$1.\ log_2 16=4$

Write in exponential form

Using the law of logarithm which says if

$log_b A=x$

then

$A = b^x$

By comparison;

A = 16; b = 2 and x = 4

The expression $log_2 16=4$ becomes

$16 = 4^2$

$2.\ log_{16} 2=\frac{1}{4}$

Write in exponential form

Applying the same law as used in (1) above;

A = 2; b = 16 and $x = \frac{1}{4}$

The expression $log_{16} 2=\frac{1}{4}$ becomes

$2 = {16}^{\frac{1}{4}}$

$3.\ 2^z=y$

Write in logarithm form

Using the law of logarithm which says if

$b^x =A$

then

$log_b A=x$

By comparison;

b = 2; x = z and A = y

The expression $2^z=y$ becomes

$log_2 y=z$

3 0

2 years ago

Easy math, but not sure what I did wrong.

Ierofanga [76]

Answer:

Step-by-step explanation:

In his 5th year, he took 3 times as many exams as the first year. So the number of exams taken in the 5th year must be a multiple of 3.

If a₁ = 1, then a₅ = 3. However, this isn't possible because we need 4 integers between them, and a sum of 31.

If a₁ = 2, then a₅ = 6. Same problem as before.

If a₁ = 3, then a₅ = 9. This is a possible solution.

If a₁ = 4, then a₅ = 12. If we assume a₂ = 5, a₃ = 6, and a₄ = 7, then the sum is 34, so this is not a possible solution.

Therefore, Alex took 3 exams in his first year and 9 exams in his fifth year. So he took 19 exams total in his second, third, and fourth years.

3 < a₂ < a₃ < a₄ < 9

If a₂ = 4, then a₃ = 7 and a₄ = 8.

If a₂ = 5, then a₃ = 6 and a₄ = 8.

If a₂ = 6, then there's no solution.

So Alex must have taken 8 exams in his fourth year.

4 0

2 years ago

At what value of x does the graph of the following function F(x) have a vertical asymptote? F(x) = 4x/3x-6

riadik2000 [5.3K]

Answer:

vertical asymptote at x = 2

Step-by-step explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non zero for this value then it is a vertical asymptote.

3x - 6 = 0 ⇒ x = 2 ← is the vertical asymptote

6 0

3 years ago

a point is randomly chosen in the triangle shown below. which statement is true about the likelihood of the location of the poin

storchak [24]

Where is the diagram.
Just Repost it again.

3 0

2 years ago

7x + 9 = 58? solve for x

mars1129 [50]

The answer for the equation is x=7

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3 years ago

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