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Makovka662 [10]
3 years ago
14

Given that 3m + 2n = 17, findthe value of n when m= 3.​

Mathematics
1 answer:
Tanya [424]3 years ago
6 0

Answer:

N= 4

Step-by-step explanation:

3(3)+2n=17

9+2n=17

-9. -9

2n=8

2. 2

N=4

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11+m>15 what’s the answer for this problem?
hjlf

Answer:

m > 4

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Equality Properties

Step-by-step explanation:

<u>Step 1: Define</u>

11 + m > 15

<u>Step 2: Solve for </u><em><u>m</u></em>

  1. Subtract 11 on both sides:                    m > 4

Here we see that any value <em>m</em> greater than 4 would work as a solution to the inequality.

6 0
3 years ago
this is a shorthand way of writing very large or very small numbers in which the number is expressed as a number between 1 and 1
Flauer [41]

Answer:

function

Step-by-step explanation:

3 0
3 years ago
Which algebraic expresion is a polynomial with a degree of 2
MariettaO [177]
A polynomial with a degree of 2 can be:

x² + 1

y² + 2

etc..

hope this helps
3 0
3 years ago
I will give brainliest!!
photoshop1234 [79]

Answer:

SA: 664 in²

LA: it’s supposed to be 1040cm???

Step-by-step explanation:

SA: find the area of all sides of the prism first (I suppose it’s rectangular?)

80+80+112+112+140+140 = 664

LA: the lateral area formula is (perimeter of base)*height...

perimeter of base = (6+6+14+14) = 40

40 * 26 = 1040

I may be wrong on the second one since I don’t know what the prism’s shape is... Hope it helps though ;)

8 0
3 years ago
What is the limit of f (×) as x approaches - infinty
tatyana61 [14]
If the degree of numerator and denominator are equal, then limit will be leading coefficient of numerator divided by the leading coefficient of denominator.

So then the limit would be 3/1 = 3.

Alternatively,

\displaystyle \lim_{x\to\infty}\dfrac{3x^2+6}{x^2-4}=\displaystyle \lim_{x\to\infty}\dfrac{3x^2+6}{x^2-4}\cdot\dfrac{1/x^2}{1/x^2}=\lim_{x\to\infty}\dfrac{3+\frac6{x^2}}{1-\frac4{x^2}} = \dfrac{3+0}{1-0}=\boxed{3}

Hope this helps.
5 0
3 years ago
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