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if two quantities are proportional, then their relationship can be repr…

Question

if two quantities are proportional, then their relationship can be represented by an equation of the form $y = kcdot x$
(a) assume that the number of miles driven is proportional to the gas used by the car. if a car can drive 66 miles on 3 gallons of gas, find an equation that represents this relationship
(b) assume that the water used in a pancake recipe is proportional to the flour used at a ratio of 6 cups of water to 7 cups of flour. find an equation that represents this relationship
(c) assume that the ratio of currency a to currency b to currency c is 5 : 6 : 900 and that these currencies are proportional to each other. find equations that represent relationships between pairs of these quantities.

(b) choose the correct answer below
a. $w = 7cdot f$
b. $w = 6cdot f$
c. $w=\frac{6}{7}cdot f$
d. $w=\frac{7}{6}cdot f$

(c) select all that apply
a. $a = 6cdot b$
b. $c = 150cdot a$
c. $a = 180cdot c$
d. $b=\frac{6}{5}a$
e. $b = 150cdot c$
f. $b=\frac{1}{150}cdot c$
g. $c = 150cdot b$

Explanation:

Response
Part (a)

Step 1: Identify variables and constant of proportionality

Let \( y \) be the number of miles driven and \( x \) be the gas used (in gallons). The proportional relationship is \( y = kx \). We know when \( x = 3 \), \( y = 66 \). To find \( k \), we substitute these values into the equation: \( 66 = k \times 3 \).

Step 2: Solve for \( k \)

Divide both sides by 3: \( k=\frac{66}{3}=22 \). So the equation is \( y = 22x \) (where \( y \) is miles and \( x \) is gallons of gas).

Brief Explanations

We have water (\( W \)) proportional to flour (\( F \)) with a ratio of 6 cups water to 7 cups flour. The proportional equation is \( W = kF \), where \( k=\frac{\text{water}}{\text{flour}}=\frac{6}{7} \). So \( W=\frac{6}{7}F \), which matches option C.

Brief Explanations

Using the ratio \( A:B:C = 5:6:900 \), we express relationships as \( A = 5k \), \( B = 6k \), \( C = 900k \). Checking each option:

  • D: \( B=\frac{6}{5}A \) (substitute \( A = 5k \), \( B = 6k \), holds).
  • F: \( B=\frac{1}{150}C \) (substitute \( C = 900k \), \( \frac{1}{150}(900k)=6k = B \), holds).
  • G: \( C = 150B \) (substitute \( B = 6k \), \( 150(6k)=900k = C \), holds).

Answer:

\( y = 22x \) (or using appropriate variable names like \( \text{miles} = 22 \times \text{gas} \))

Part (b)