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Likurg_2 [28]
3 years ago
7

Find the first term and the common difference of an arithmetic progression in which t5=3x-y and t11=5y

Mathematics
1 answer:
pishuonlain [190]3 years ago
3 0

first term: 5x-5y

common difference: -x/2 + y

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(2 tens 1 one)x10 in unit form and standard form
omeli [17]

When the number expression given as (2tens 1 one) x 10 is written in standard form, the standard form is 210 and the unit form is 2 hundred, and 1 ten

<h3>How to write the number in standard form?</h3>

The number expression is given as:

(2tens 1 one) x 10

2 tens is represented as:

2 * 10

1 one is represented as:

1 * 1

So, the number expression can be rewritten as:

(2tens 1 one) x 10 = (2 * 10 + 1 * 1) x 10

Evaluate the product

(2tens 1 one) x 10 = (20 + 1) x 10

Evaluate the sum

(2tens 1 one) x 10 = (21) x 10

Evaluate the product

(2tens 1 one) x 10 = 210

When the number expression given as (2tens 1 one) x 10 is written in standard form, the standard form is 210 and the unit form is 2 hundred, and 1 ten

Using the above steps as a guide, we have:

  • (5 hundreds 5 tens) * 10 ⇒ 5 thousands and 5 hundreds ⇒ 5500
  • (2 thousands 7 tens) / 10 ⇒ 2 hundreds and 7 units ⇒ 207
  • (4 ten thousands 8 hundred) / 10 ⇒ 4 thousands and 8 tens ⇒ 4080

Read more about standard form at

brainly.com/question/19169731

#SPJ1

4 0
2 years ago
​​someone pls help meeeee!!!!!!<br>​​
GuDViN [60]

Step-by-step explanation:

y = y

x - 2 = 2x - 3 - 3x²

3x² + x - 2x = - 3 + 2

3x² - x = - 1

.

a = 3, b = -1, c = 0

.

\sf{x = \frac{-b \pm \sqrt{b^{2} - 4ac} }{2a} }

\sf{x = \frac{- (-1) \pm \sqrt{-1^{2} - 4 (0)} }{2(3)} }

\sf{x = \frac{ 1 \pm \sqrt{1 - 0} }{6} }

\sf{x = \frac{ 1 \pm \sqrt{1} }{6} }

\sf{x = \frac{ 1 + 1 }{6} }

\sf{x = \frac{ 2 }{6} }

\sf{x = \frac{ 1 }{3} }

.

Substitution :

y = x - 2

y = ⅓ - 2

y = ⅓ - 6/3

y = - 5/3

y = - 1 ⅔

.

5 0
3 years ago
Your cell phone plan cost $24.99 per month plus $0.12 for each text message you send or receive. you have a most $30 to spend on
ioda

Answer:

The maximum number of text messages you can send per month would be 41.

Step-by-step explanation:

1. First, you want to create an inequality from the word problem.  You know that $24.99 is a constant, so we won't put any variables next to it.  You also know that each text message is $0.12, so that would need a variable since it can change each month.  You also know that you can spend $30 maximum each month, so that goes on the other side of the inequality.  Your inequality will look like this:  0.12x+24.99≤30 (x is the number of text messages).

2. Second, you want to solve for x.  To do this, subtract 24.99 from each side.  After this, the equation is 0.12x≤5.01

3. Now, divide each side by 0.12.  Now your inequality is x≤41.75.  This represents how many text messages you can send per month without going over 30.

4. Finally, you have to realize that is impossible to send exactly 41.75 text messages.  So, you have to round down to 41 to get an even number of text messages, and that's your answer!

Hope this helps!  Feel free to ask any questions :)

3 0
3 years ago
Read 2 more answers
Which fraction shows 32/48 in simplest form
Pie
32÷16=2
48÷16=3

so 2/3
3 0
4 years ago
Read 2 more answers
Which statements are true? Select all that apply.
KengaRu [80]

Answer:

Hectometers are smaller than kilometers.

<h2>                 OR</h2>

Grams are smaller than dekagrams.

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
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