Solve the system of equations.

Mathematics

2 answers:

kotegsom [21]2 years ago

6 0

Answer:

c. no solution

Step-by-step explanation:

6d + 3f = 12

2d = 8 - f

Multiply the second equation by 3

3*2d = 3(8-f)

6d = 24-3f

Substitute into the first equation for 6d

(24-3f) +3f = 12

Combine like terms

24 =12

This is never true, so there are no solutions

Dmitry [639]2 years ago

5 0

Answer:

c. No Solution.

Step-by-step explanation:

6d + 3f = 12

2d = 8 - f

Rearranging the second equation:

2d + f = 8 Multiply this equation by 3:

6d + 3f = 24

Note that the left side of this equation = the left side of the first equation but the right sides are different. So this system does not make sense and there are No Solutions.

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Using the porpotions: WHIP= (Walks+Hits) divide by Innings Pitched. The pitcher wants to lower his WHIP from 1.53 to 1.3. How ma

BaLLatris [955]

This relates to ur previous question....there were 85 innings last question and 130 walks + hits which gave a whip score of 1.53......so if he wants to lower his whip (without changing his walks or hits) to 1.3....
130 / (85 + x) = 1.3....multiply both sides by (85 + x)
130 = 1.3(85 + x)
130 = 110.50 + 1.3x
130 - 110.50 = 1.3x
19.5 = 1.3x
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15 = x

he would have to pitch 15 more innings without any walks or hits to reach a goal of 1.3 whip score. This is not a reasonable goal.....have u ever seen a baseball game where a pitcher pitched 15 innings without any walks or hits ? And since a game consists of 9 innings....he would be 3 innings short of 2 games (thats if they dont go into overtime), and he would have to pitch a no hitter....and that is highly unlikely.

7 0

3 years ago

(1) a) 5270 cl= ? dal

Ivanshal [37]

Answer:

Check below

Step-by-step explanation:

1) These metric volume units can be easily converted by dividing or multiplying by 10 and its multiple. Like this: each step up on the ladder multiply by 10. Each step down divide by 10

$a)5270 \:cl=527\:dl\\b)kl=5.3 \times 10=530 hl\\c)58000ml=5.8 dal =(\frac{58000}{10000})\\d)0.39 dal=3.9 l\\e)7.8 dal=780 dl\\f)57.2 ml=5.72 cl$ .

2) When it comes to area, the "ladder scheme" remains valid but now we'll multiply or divide by

$10^3$

Bear in mind these useful relations:

$1litre =\frac{1}{10^3}m^{3}$

$1 cm^{3} =1 ml$

$a) 3.7 l=3700 cm^{3}\\b)29.6 dm^{3}\\29.6 dm^{3} \rightarrow 29.6\times10^{3} cm^3=29.6\times10^{3} ml = 29.6\times10^{2} cl \:or\:2960 cl\\c)8.1l=8100 cm^{3}\\d)5l=\frac{5}{10^{-3}} m^{3}\\e)5.2 cl=52ml=52 cm^{3}=52\times10^{3}=52000 \:mm^{3}\\f)375 ml=375 cm^{3}\\g) 1300 cm^{3}=1300 ml=13 dl\\h) 8.6 ml= 8.6 cm^{3}\times\: 10^{3}=8600 mm^{3}\\i) 51.8\times 10^{11} l \rightarrow 51.8\times 10^{-11}km^{3}$