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2 years ago

I will mark u brainlest

Mathematics

2 answers:

hodyreva [135]2 years ago

3 0

Answer:

48/25

Step-by-step explanation:

$-1\frac{3}{5} / \frac{5}{-6}$

- $\frac{-8}{5} / \frac{5}{-6}$

$\frac{-8}{5} * \frac{-6}{5}$

$\frac{-8 * -6}{5 * 5}$

48/25

Julli [10]2 years ago

3 0

Answer:

48/25

Step-by-step explanation:

Change -1 3/5 into an improper fraction.
-8/5 ÷ 5/-6 = - 8/5 ÷ - 5/6

Keep Change Flip (KCF)

- 8/5 × - 6/5

Negative × Negative = Positive

8/5 × 6/5

Multiply the numerators to get 48.

Multiply the denominators to get 25.

Answer = 48/25

I hope this helped and please mark me as brainliest!

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Suppose an unknown radioactive substance produces 8000 counts per minute on a Geiger counter at a certain time, and only 500 cou

mariarad [96]

Answer:

The half-life of the radioactive substance is of 3.25 days.

Step-by-step explanation:

The amount of radioactive substance is proportional to the number of counts per minute:

This means that the amount is given by the following differential equation:

$\frac{dQ}{dt} = -kQ$

In which k is the decay rate.

The solution is:

$Q(t) = Q(0)e^{-kt}$

In which Q(0) is the initial amount:

8000 counts per minute on a Geiger counter at a certain time

This means that $Q(0) = 8000$

500 counts per minute 13 days later.

This means that $Q(13) = 500$ . We use this to find k.

$Q(t) = Q(0)e^{-kt}$

$500 = 8000e^{-13k}$

$e^{-13k} = \frac{500}{8000}$

$\ln{e^{-13k}} = \ln{\frac{500}{8000}}$

$-13k = \ln{\frac{500}{8000}}$

$k = -\frac{\ln{\frac{500}{8000}}}{13}$

$k = 0.2133$

$Q(t) = Q(0)e^{-0.2133t}$

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

$Q(t) = Q(0)e^{-0.2133t}$

$0.5Q(0) = Q(0)e^{-0.2133t}$

$e^{-0.2133t} = 0.5$

$\ln{e^{-0.2133t}} = \ln{0.5}$

$-0.2133t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.2133}$

$t = 3.25$

The half-life of the radioactive substance is of 3.25 days.

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3 years ago

Lizas cat had kittens when Liza and her brother weighed all of the kittens together they weighed 4 pounds 2 ounces since all the

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How many kittens are there?

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3 years ago

When printing an article of 2400 words, an entrepreneur decides to use two sizes of letters, using the largest one a printed pag

olga55 [171]

Step-by-step explanation:

x= Number of small pages

y= Number of full pages

1 x + 1 y = 21 .............1

Total words

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Eliminate y

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Multiply (2) by 1

-1500 x -1500 y = -31500

1200 x + 1500 y = 27000

Add the two equations

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6 0

3 years ago

How to solve Y+2=4 Y-2=8

lakkis [162]

Answer:

First question -> y = 2

Second question -> y = 10

Step-by-step explanation:

Step 1: Solve y + 2 = 4

Subtract 2 from both sides

y + 2 - 2 = 4 - 2

y = 2

Step 2: Solve y - 2 = 8

Add 2 to both sides

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4 years ago

Can i please get help on this question?

Kamila [148]

The answer would be 3.

In order to find this you must place the values in where you see the letters. So start with the letters and plug in.

mx - y

Since m = 1, put that in.

(1)x - y

Since x = 5, put that in

(1)(5) - y

Since y = 2, put that in

(1)(5) - (2)

Now follow the order of operations and solve

(1)(5) - 2

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5 0

3 years ago

Read 2 more answers