Question

and 350 calories at lunch. she consumes \\(\frac{2}{3}\\) of her daily calories at\
consumed at dinner, which statements describe the situation? check all that\
apply.\
\\(\bigcirc\\) leena consumed 1 500 calories at dinner\
\\(\bigcirc\\) the equation \\(\frac{2}{3}(x + 400 + 350) = x\\) can be used to model the situation.\
\\(\bigcirc\\) leena consumed 500 calories at dinner.\
\\(\bigcirc\\) the equation \\(\frac{2}{3}(x) = x(400 + 300)\\) can be used to model the situation.\
\\(\bigcirc\\) leena consumed 1 000 calories at dinner\
\\(\bigcirc\\) the equation \\(\frac{2}{3}x(400 + 300) = x\\) can be used to model the situation.

Explanation:

Step1: Define variable for daily calories

Let $x$ = total daily calories.

Step2: Identify meal calorie breakdown

Breakfast = 400, Lunch = 350, Dinner = $x - 400 - 350$.

Step3: Set up correct equation

$\frac{2}{3}x = 400 + 350$ (since $\frac{2}{3}$ of daily calories are breakfast + lunch)

Step4: Solve for total daily calories

$\frac{2}{3}x = 750$
$x = 750 \times \frac{3}{2} = 1125$

Step5: Calculate dinner calories

Dinner = $1125 - 750 = 375$ (Wait, correction: Re-express the relationship: The problem states she consumes $\frac{2}{3}$ of daily calories at dinner? No, re-read: "She consumes $\frac{2}{3}$ of her daily calories at [breakfast? The first part is cut off, but the equation options show: Correct model is that $\frac{2}{3}$ of daily calories is breakfast + lunch, so $\frac{2}{3}x = 400 + 350$. Wait no, if dinner is $\frac{2}{3}x$, then breakfast + lunch = $\frac{1}{3}x$. Let's re-express:
If dinner = $\frac{2}{3}x$, then $400 + 350 = \frac{1}{3}x$
$x = 750 \times 3 = 2250$
Dinner = $2250 - 750 = 1500$. Now match equations:
The equation $\frac{2}{3}(x) = x - (400 + 350)$ → $\frac{2}{3}x = x - 750$ → $750 = x - \frac{2}{3}x = \frac{1}{3}x$ → $x=2250$, dinner=1500. Now check options:
Option 1: Leena consumed 1500 calories at dinner (matches)
Option 2: $\frac{2}{3}(x + 400 + 350) = x$ → $\frac{2}{3}(x+750)=x$ → $2x + 1500 = 3x$ → $x=1500$, dinner=1500-750=750 (incorrect)
Option 3: 500 (incorrect)
Option 4: $\frac{2}{3}(x) = 400 + 300$ (typo, 350, but if 400+350=750, $\frac{2}{3}x=750$ → x=1125, dinner=375 (incorrect)
Option 5: 1000 (incorrect)
Option 6: $\frac{2}{3}x(400+300)=x$ (invalid, units don't make sense)

Wait, the correct model is: If $\frac{2}{3}$ of daily calories are dinner, then $dinner = \frac{2}{3}x$, so $400 + 350 = x - \frac{2}{3}x = \frac{1}{3}x$ → $x=2250$, dinner=1500. The only matching option is:

1. Leena consumed 1500 calories at dinner

Wait, but no equation matches? Wait the second option: $\frac{2}{3}(x + 400 + 350) = x$ is wrong. Wait maybe the problem says she consumes $\frac{2}{3}$ of daily calories at breakfast + lunch, so dinner is $\frac{1}{3}x$. Then $400+350=\frac{2}{3}x$ → $x=1125$, dinner=375 (no option). So the only valid pair is:

Answer:

• Leena consumed 1 500 calories at dinner
• (Note: No equation option is correctly formatted, but if we assume a typo, the intended correct equation would be $\frac{2}{3}x = x - (400 + 350)$, but among given options, only the first calorie count is valid if $\frac{2}{3}$ of daily calories are dinner.)

Wait, re-express the correct equation for dinner being $\frac{2}{3}x$:
$400 + 350 + \frac{2}{3}x = x$
$750 = x - \frac{2}{3}x$
$750 = \frac{1}{3}x$
$x=2250$
Dinner = $\frac{2}{3} \times 2250 = 1500$. This matches the first option. No given equation correctly models this, but the first calorie count is correct.

Final valid answers:
A. Leena consumed 1 500 calories at dinner