Question

3 from unit 6, lesson 13
you know x is a number less than 4. select all the inequalities that must be true.
a ( x < 2 )
b ( x + 6 < 10 )
c ( 5x < 20 )
d ( x - 2 > 2 )
e ( x < 8 )
4 from unit 6, lesson 13
here is an unbalanced hanger diagram.
image of hanger with 3 green triangles on left and 2 red circles on right
a. if you know each circle weighs 4 grams, what do you know about the weight of each triangle? explain your reasoning.
b. if you know each triangle weighs 3 grams, what do you know about the weight of each circle? explain your reasoning.

Explanation:

Problem 3

Step1: Start with given condition

Given $x < 4$

Step2: Test Option A: $x<2$

If $x=3$ (which is $<4$), $3<2$ is false. So A is not always true.

Step3: Test Option B: $x+6<10$

Simplify: $x < 10-6$ → $x < 4$, which matches the given condition. So B is always true.

Step4: Test Option C: $5x<20$

Simplify: $x < \frac{20}{5}$ → $x < 4$, which matches the given condition. So C is always true.

Step5: Test Option D: $x-2>2$

Simplify: $x > 2+2$ → $x > 4$, which contradicts $x<4$. So D is false.

Step6: Test Option E: $x<8$

All numbers less than 4 are also less than 8. So E is always true.

Step1: Define variables and inequality

Let $t$ = weight of 1 triangle, $c=6$ g (weight of 1 circle). The hanger gives $3t > 2c$.

Step2: Substitute circle weight

Substitute $c=6$: $3t > 2\times6$

Step3: Simplify to solve for $t$

$3t > 12$ → $t > \frac{12}{3}$ → $t > 4$

Step1: Define variables and inequality

Let $c$ = weight of 1 circle, $t=3$ g (weight of 1 triangle). The hanger gives $3t > 2c$.

Step2: Substitute triangle weight

Substitute $t=3$: $3\times3 > 2c$

Step3: Simplify to solve for $c$

$9 > 2c$ → $c < \frac{9}{2}$ → $c < 4.5$

Answer:

B. $x + 6 < 10$, C. $5x < 20$, E. $x < 8$

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Problem 4a