All categories

3 years ago

15-4t=3 how do i solve this in words

Mathematics

1 answer:

Ulleksa [173]3 years ago

3 0

-4t = -15 + 3

Divide both sides by -4.

T= -15/4 -1

You might be interested in

A team played 38 matched in a football league. They won 22 matches and Lost nine matches. The rest of the matches were draws. Th

pychu [463]

Answer:

The answer is 73.

Step-by-step explanation:

To find the total number of points for a win, you would do 3*22 which equals 66.

Then the number of draws, which is 7 draws, has one point per draw, so it would be 7.

Add 66 by 7=73

Hope this helps!

4 0

3 years ago

What's 2/5 divided by 4 ( simplest form)

Lelechka [254]

Answer:

1/10

Step-by-step explanation:

$\frac{2}{5} \div 4 = \frac{2}{5} * \frac{1}{4}$

we multiply by the reciprocal when we work with fractions and division.

Now you just multiply across.

Therefore, you have 2/20, which simplifies to 1/10.

6 0

3 years ago

Read 2 more answers

Alyssa, Dana, and Judy are out to lunch. They have a 15% off coupon for their bill which was a total of $130. They live in Calif

Nady [450]

Answer

Hold on I quickly need to calculate then ill be done

Step-by-step explanation:

4 0

3 years ago

What is the value of the expression 4 × (2a − 1) if a = 3?

dlinn [17]

Answer:

Step-by-step explanation:

4 * (2a − 1)

Let a=3

4 * (2*3 − 1)

Simplify inside the parentheses first

4 * (6 − 1)

4 * (5)

4 0

3 years ago

You roll a six-sided die twice. What is the probability of rolling an even number and then an odd number? A) 1 16 B) 1 8 C) 1 4

jeyben [28]

<u>Answer:</u>

The probability of rolling an even number and then an odd number is $\frac{1}{4}$ Option C is correct

<u>Solution:</u>

Given, You roll a six-sided die twice.

We have to find what is the probability of rolling an even number and then an odd number?

We know that, rolling single die two times is equivalent to two dice rolling at a time.

$\text { Now, probability of an event }=\frac{\text { number of favourable outcomes }}{\text { total possible outcomes }}$

So, now, total possible outcomes = 6 x 6 = 36

And, number of favourable outcomes = 3 even on 1st die x 3 odd on 2nd die = 3 x 3 = 9

Then, probability = $\frac{9}{36} = \frac{1}{4}$

Hence, the probability of given condition is $\frac{1}{4}$ and option c is correct.

5 0

3 years ago

Read 2 more answers