-3 \cdot f(-8) + 7 \cdot g(2) =−3⋅f(−8)+7⋅g(2)=minus, 3, dot, f, left parenthesis, minus, 8, right parenthesis, plus, 7, dot, g,
dsp73
Answer:
-22
Step-by-step explanation:
The graph from which the information is to be read is attached below.
We want to find the value of −3⋅f(−8)+7⋅g(2).
From the graph:
Therefore:
−3⋅f(−8)+7⋅g(2)=−3(-2)+7(-4)
=6-28
=-22
−3⋅f(−8)+7⋅g(2)=-22
Two fractions are said to be equivalent if both the numerator and the numerator of one fraction can be obtained by multiplying or dividing the numerat.or and denominator of the other fraction by a single factor.
Dividing the numerator and denominator of 240/300 by 5 gives 48/60
Dividing the numerator and denominator of 240/300 by 60 gives 4/5
Dividing the numerator and denominator of 240/300 by 10 gives 24/30
Whereas dividing 300 by 2 gives 150 dividing 240 by 2 will not give 122. so 122/150 is not equivalent to 240/300.
Answer:happy
Step-by-step explanation:
Vg
◆ Define the variables:
Let the calorie content of Candy A = a
and the calorie content of Candy B = b
◆ Form the equations:
One bar of candy A and two bars of candy B have 774 calories. Thus:
a + 2b = 774
Two bars of candy A and one bar of candy B contains 786 calories
2a + b = 786
◆ Solve the equations:
From first equation,
a + 2b = 774
=> a = 774 - 2b
Put a in second equation
2×(774-2b) + b = 786
=> 2×774 - 2×2b + b = 786
=> 1548 - 4b + b = 786
=> -3b = 786 - 1548
=> -3b = -762
=> b = -762/(-3) = 254 calorie
◆ Find caloric content:
Caloric content of candy B = 254 calorie
Caloric content of candy A = a = 774 - 2b = 774 - 2×254 = 774 - 508 = 266 calorie
