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Ann [662]
3 years ago
13

Which graph represents the function? f(x)=−x2+x+6

Mathematics
1 answer:
sergiy2304 [10]3 years ago
4 0
First page , second image is the correct answer :)
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A 20 pound turkey serves 28 how many will a 30 pound turkey serve
mariarad [96]

Answer:

42

Step-by-step explanation:

half of 20 is 10

half of 28 is 14

so 10 pounds=14 people

if 10 pounds=14 people and theres 30 people

you add 14 3 times for each ten

hope this helps give me my piece of turkey now

3 0
2 years ago
Ryan drove 480 miles and used 20 gallons of gas. How many miles did Ryan's car get per gallon? Write an equation that would solv
lisabon 2012 [21]

Answer:

24 miles per gallon

Step-by-step explanation:

Step one:

given data

total distance covered= 480miles

The total amount of gallons used is = 20

The miles per gallon

= 480/20

=24 miles per gallon

Step two:

hence the equation that we used in solving the problem is stated as

miles per gallon= total distance/total gallon sused

3 0
3 years ago
From the set {5, 15, 23), which of the values can be substituted for m to make the
Arlecino [84]

Answer:

its A and i am sure her is the explanation

Step-by-step explanation:

  1. m<19-4
  2. m<15
7 0
3 years ago
Help!<br> This is the problem <br><br><br> 1+10k+1-7k
ivolga24 [154]

Answer:

3002

Step-by-step explanation:

Very simple, just add them up.

5 0
2 years ago
Read 2 more answers
85.87 J or heat energy is added to a 34.8 g mass of substance the temperature rises from 21.76°C
Andru [333]

Answer:

The required specific heat is 196.94 joule per kg per °C  

Step-by-step explanation:

Given as :

The heat generated = Q = 85.87 J

Mass of substance  (m)= 34.8 gram = 0.0348 kg

Change in temperature = T2 - T1 = 34.29°C - 21.76°C = 12.53°C

Let the specific heat = S

Now we know that

Heat = Mass × specific heat × change in temperature

Or, Q = msΔt

Or, 85.87 = (0.0348 kg ) × S × 12.53°C

Or , 85.87 = 0.4360 × S

Or, S = \frac{85.87}{0.4360}

∴ S = 196.94 joule per kg per °C

Hence the required specific heat is 196.94 joule per kg per °C   Answer

7 0
2 years ago
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