3 years ago

Find the equation of a line given the point and slope (1,3) m=3

1 answer:

8 0

Simple....

we know the formula you need to use is y=mx+b....

plug 'n chug....

3=3(1)+b

3=3+b
-3 -3
b=0

Thus.....y=3x

Thus, your answer.

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Combine like terms.<br><br> y + 3 + 9(y + 5) =

Mnenie [13.5K]

Answer:

The answer is 10y + 48.

Step-by-step explanation:

First you have to get rid of brackets by expanding :

$y + 3 + 9(y + 5) \\ = y + 3 + 9y + 45$

Next you have to collect like-terms :

$y + 3 + 9y + 45 \\ = 10y + 48$

7 0

3 years ago

Read 2 more answers

PLEASE help me answer this question. Can anyone do this on paper, write step by step, and send the pile photo to me? Or just an

noname [10]

To find the diagonal you do
side x the square root of 2
so your answer would be

8 1/2

Hope this helps ;)

6 0

2 years ago

A spinner has eight equal sections numbered from 1–8. Predict how many times the pointer will land on a 2 or a 5 if you spin the

Alja [10]

Answer:

15 times

Step-by-step explanation:

By spinning once, the probability of landing on 2 is 1/8

The probability of landing on 5 is 1/8

The probability of landing on 2 or 5 is 1/8 + 1/8 = 2/8 = 1/4

So spinning 60 times, the number of times we expect a landing on 2 or 5 will be ;

1/4 * 60 = 15 times

8 0

2 years ago

A dress has a thirty percent discount applied and is on sale for 63 pound. What was the original price of the dress before the r

jekas [21]

Answer:

£90

Step-by-step explanation:

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3 years ago

If 150 grams of a radioactive isotope are present at 2:00 PM and 10 grams remain at 6:00 PM (the same day), what is the half-lif

pogonyaev

Initial amount, $A_o=150\ g$ .

Final amount, $A =10\ g$ .

Time taken, t = 6:00 - 2:00 = 4 hour.

We know,

$A=A_o(\dfrac{1}{2})^{\dfrac{t}{h}}\\\\10 = 150 \times \dfrac{1}{2}^{\dfrac{4}{h}}\\\\2^{\dfrac{4}{h}}=15\\\\\dfrac{4}{h}= log_215\\\\h = \dfrac{4}{log_215}\\\\h = 1.024 \ hours$

Therefore, the half-life of the isotope is 1.024 hours.

Hence, this is the required solution.

8 0

3 years ago