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s2008m [1.1K]
3 years ago
7

Find domain and range of 2x+3y=4

Mathematics
1 answer:
jeka943 years ago
7 0

Answer:

The domain is (-infinity, infinity)

The range is (-infinity, infinity)

Step-by-step explanation:

In order to turn this equation into a linear function, you must solve for y, to get y=. First you subtract 2x from both sides. You will then have 3y=4-2x. Next, you divide by 3 on both sides to isolate the "y".  Then you'll get y=-2/3x+4/3. The domain and range of this function once it is graphed is (-infinity, infinity).

Hope I helped :)

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2. Find the perimeter of each of the composite figures: please explain how you got the anser please
stiv31 [10]

Answer:

a) 52

b) 21 + 6.25π

Step-by-step explanation:

Part (a)

the sides which we don't have:

(longer side): 7+6 = 13m

(shorter side): 13—8= 5m

So:

Perimeter = 2P = 13+5+7+8+6+13=52

Part (b)

perimeter of the half circle:

\pi  \times {r}^{2}  =  \pi  \times {2.5}^{2}  = 6.25\pi

So:

Perimeter = 2P = 2×8 + 5 + 6.25π = 21 + 6.25π

5 0
3 years ago
Solve 1/3(3.2×1.4)2.8<br><br>Simplify your answer​
pochemuha

4 68⁄375 is the purpose answer

6 0
3 years ago
(6.1x10^30) (5.1x10^34)
Ilia_Sergeevich [38]
=31.11×10⁶⁴=3.111×10⁶⁵.
3 0
4 years ago
By what number should 9 4upon5 be multiplied to get 42 ​
kondaur [170]

Answer:

4 2/7

Step-by-step explanation:

  • x * 9 4/5 = 42
  • x * 49/5 = 42
  • x = 42 : 49/5
  • x = 42 * 5/49
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7 0
3 years ago
And a certain watery an urn contains balls number 1 to 34 from this are in five balls are chosen randomly without replacement fo
lina2011 [118]

The probability of winning the lottery with one ticket is mathematically given as

P(W)=2.99484408*10^{−8}

<h3>What is the probability of winning the lottery with one ticket?</h3>

Generally, the equation for the probability is  mathematically given as

P(1st ball)=1/34

one ball drawn remain

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P(2nd ball)=1/33

In conclusion, the probability of winning the lottery with one ticket is

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P(W)=2.99484408*10^{−8}

Read more about probability

brainly.com/question/11234923

#SPJ1

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