A food scientist uses a mixture of the gases helium and argon to preserve packaged food. The amount of each gas the scientist uses for the mixture is modeled by the equation 5.6H + 0.56A = 22.4, where H and A
are the masses, in grams, of helium and argon, respectively. If the scientist uses 10 grams of argon, how
many grams of helium are used?

9514 1404 393

  3 grams of helium are used

Step-by-step explanation:

Solving the given equation for H, we have ...

  5.6H +0.56A = 22.4

  5.6H = 22.4 -0.56A

  H = (22.4 -0.56A)/5.6

  H = 4 -0.1A

Using A=10, we find ...

  H = 4 -0.1·10 = 3

3 grams of helium are used.

answer:   x = 3 + √11     and     x = 3 - √11

step-by-step explanation:

x² - 6x + 9 = 11

(x - 3)² = 11

x - 3 = ±√11

    x = 3 ± √11

x = 3 + √11     and     x = 3 - √11

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.