(-5,-1); 2y=2x-4 help me ASAP please

List 5 fractions that are between 1/3 and 4/5

koban [17]

9514 1404 393

Answer:

2/5, 7/15, 8/15, 3/5, 2/3

Step-by-step explanation:

If these fractions are expressed with a common denominator, that would be 3×5 = 15. Then the given fractions are 1/3 = 5/15, and 4/5 = 12/15. The numerators 5 and 12 differ by 7, so we can easily choose 5 fractions in that range:

6/15 = 2/5

7/15

8/15

9/15 = 3/5

10/15 = 2/3

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<em>Alternate solutions</em>

There is no requirement for the fractions to be written any particular way or with any particular spacing. The limits in decimal are 1/3 = 0.3333...(repeating) and 4/5 = 0.8. We could choose the decimal fractions ...

0.34, 0.40, 0.50, 0.60, 0.70

or

0.41, 0.52, 0.63, 0.74, 0.79

7 0

3 years ago

20 points

Slav-nsk [51]

Answer:

grade 5 palang po ako di ko po alam yan sowey po

8 0

2 years ago

Helllllp >_< Compare: 1.2057 O 1.1421

vladimir2022 [97]

Answer:

Widal Kit Salmonella Ha, Hb, Hc, Hd, OA, OB, OC, OD

Rapid Labs Widal antigen set contains ready to use concentrated smooth antigen suspensions of the bacilli; S. typhi ‘O’. S. typhi ‘H’. S, paratyphi ‘AO’. S. paratyphi ’BO‘. S. paratyphi ’AH’, S. paratyphi ‘BH’. S. paratyphi ‘CH’. S. paratyphi ‘CO’ and/or polyspecific positive control reactive with these antigens.

When the Widal antigen suspensions are mixed and incubated with the patient’s serum, anti salmonella antibodies present in the patient serum react with the antigen suspensions to give agglutination. Agglutination is a positive test result, indicating presence of anti-salmonella antibodies in the patient serum. No agglutination is a negative test result indicating absence of anti-salmonella antibodies.

Pack size – 8 x 5ml

This kit does NOT contain positive and negative controls

Widal Kit Salmonella Ha, Hb, Hc, Hd, OA, OB, OC, OD

Rapid Labs Widal antigen set contains ready to use concentrated smooth antigen suspensions of the bacilli; S. typhi ‘O’. S. typhi ‘H’. S, paratyphi ‘AO’. S. paratyphi ’BO‘. S. paratyphi ’AH’, S. paratyphi ‘BH’. S. paratyphi ‘CH’. S. paratyphi ‘CO’ and/or polyspecific positive control reactive with these antigens.

When the Widal antigen suspensions are mixed and incubated with the patient’s serum, anti salmonella antibodies present in the patient serum react with the antigen suspensions to give agglutination. Agglutination is a positive test result, indicating presence of anti-salmonella antibodies in the patient serum. No agglutination is a negative test result indicating absence of anti-salmonella antibodies.

Pack size – 8 x 5ml

This kit does NOT contain positive and negative control

Step-by-step explanation:

6 0

3 years ago

What is the f(n) runtime of the following pseudocode: sum = 0 for A = N/2 downto 1 for B = 1 to 4N increment sum by B

nikitadnepr [17]

Answer:

f(N) = 2+ N/2 + 6N² units of time.

Step-by-step explanation:

Assigning 0 to the variable sum takes one unit of time.

Each time you increment sum by B, you need to call the value of sum, sum it to B and assign it to sum, which takes three units of time in total. You are repeating this process for each value of B which ranges from 1 to 4n and for each value of A which ranges from 1 to n/2. Opening the FOR takes also another unit of time, so, as a result, we have

f(N) 1 + 1 (open the FOR in A)+ N/2*(1 (open the FOR in B) + 4N*3) = 2+ N/2 + 6N² units of time. It has order complexity O(N²).

5 0

4 years ago

Kaya is saving money at a rate of $30 per month. Edgardo is saving money at a rate of $35 per month. They both started saving at

ladessa [460]

Let Kaya's savings be 30x and Edgardo's savings be 35x If they both started saving at the same time: f(x)=30x f(x)=35x Now, sub in values for x in to the function starting with 0. Subtract y2-y1 and x2-x1 for both functions. For slope: m=y2-y1/x2-x1 so your result will be m=30/1=30 for f(x) = 30x and m=35/1=35 for f(x) = 35x so the slopes are m=30 and m=35 respectively!

7 0

3 years ago