x≥6
x≥2.5
x≥7
x≥11
11 answers
To solve the inequality \( 8x + 4 \geq 52 \), follow these steps:
Subtract 4 from both sides: \[ 8x \geq 52 - 4 \] \[ 8x \geq 48 \]
Divide both sides by 8: \[ x \geq \frac{48}{8} \] \[ x \geq 6 \]
So, the solution to the inequality is \( x \geq 6 \).
Among your options:
The final answer is \( x \geq 6 \).
To solve the inequality \( \frac{1}{4}x - \frac{3}{4} > 5 \frac{1}{4} \), follow these steps:
First, convert \( 5 \frac{1}{4} \) to an improper fraction. \[ 5 \frac{1}{4} = 5 + \frac{1}{4} = \frac{20}{4} + \frac{1}{4} = \frac{21}{4} \] So the inequality becomes: \[ \frac{1}{4}x - \frac{3}{4} > \frac{21}{4} \]
Next, add \( \frac{3}{4} \) to both sides: \[ \frac{1}{4}x > \frac{21}{4} + \frac{3}{4} \] \[ \frac{1}{4}x > \frac{21 + 3}{4} \] \[ \frac{1}{4}x > \frac{24}{4} \] \[ \frac{1}{4}x > 6 \]
Now, multiply both sides by 4 to solve for \( x \): \[ x > 6 \times 4 \] \[ x > 24 \]
So the solution to the inequality is \( x > 24 \).
Among your options:
The final answer is \( x > 24 \).
To solve the inequality \( 6x + 9 > 57 \), follow these steps:
Subtract 9 from both sides: \[ 6x > 57 - 9 \] \[ 6x > 48 \]
Divide both sides by 6: \[ x > \frac{48}{6} \] \[ x > 8 \]
So, the solution to the inequality is \( x > 8 \).
To represent this solution graphically:
If you have a set of graphs to choose from, look for the one that shows an open circle at \( 8 \) and shading extending to the right.
To determine which of the inequalities Hessa solved, we can solve each of them to see what the solution looks like.
First inequality: \(25x - 45 \geq 115\)
Add 45 to both sides: \[ 25x \geq 115 + 45 \] \[ 25x \geq 160 \] Divide by 25: \[ x \geq \frac{160}{25} = 6.4 \]
Second inequality: \(37x + 17 \leq 167\)
Subtract 17 from both sides: \[ 37x \leq 167 - 17 \] \[ 37x \leq 150 \] Divide by 37: \[ x \leq \frac{150}{37} \approx 4.05 \]
Third inequality: \(15x - 35 \leq 25\)
Add 35 to both sides: \[ 15x \leq 25 + 35 \] \[ 15x \leq 60 \] Divide by 15: \[ x \leq 4 \]
Fourth inequality: \(35x + 25 \geq 325\)
Subtract 25 from both sides: \[ 35x \geq 325 - 25 \] \[ 35x \geq 300 \] Divide by 35: \[ x \geq \frac{300}{35} = \frac{60}{7} \approx 8.57 \]
If you are trying to match the inequality Hessa solved with the solution she graphed:
Without the specific numeral shown on Hessa's graph, you can choose the appropriate inequality based on whether the solution was a bounded interval or an unbounded one to the right.
If more context about Hessa's graph is provided (like the specific numbers or an open/closed circle), further clarification can be given.
If Hessa's graph shows a closed circle on the number 5 and an arrow pointing left, this indicates that the solution includes 5 and represents all numbers less than or equal to 5.
This means Hessa has solved an inequality of the form:
\[ x \leq 5 \]
Now, let's go back to the inequalities we solved:
None of these exactly lead to \(x \leq 5\). However, it indicates that Hessa's solution should be interpreted slightly differently based on the closest comparison.
Among the given inequalities, none result in \(x \leq 5\) with a closed circle. If the available options don't exactly match the condition shown in the graph, ensure that there wasn't a misinterpretation or potentially another inequality that was not provided in the options.
Thus, based on your description of the graph, the closest valid inequality would ideally have been \(x \leq k\) with \(k\) around or less than 5, but it seems the options provided do not include one that matches this correctly under standard linear inequality forms.
If an extra inequality option like \(x \leq 5\) appears in an updated set, that would be the straightforward answer reflecting Hessa’s graph. If not, this could suggest that perhaps the context of another statement should be revisited or the options reevaluated for suitable adjustments.
Let's solve each inequality step by step to find their solutions.
Convert \( 1 \frac{1}{5} \) to an improper fraction: \[ 1 \frac{1}{5} = 1 + \frac{1}{5} = \frac{5}{5} + \frac{1}{5} = \frac{6}{5} \]
Now, the inequality becomes: \[ \frac{2}{5}x - \frac{4}{5} \geq \frac{6}{5} \]
Add \( \frac{4}{5} \) to both sides: \[ \frac{2}{5}x \geq \frac{6}{5} + \frac{4}{5} \] \[ \frac{2}{5}x \geq \frac{10}{5} \] \[ \frac{2}{5}x \geq 2 \]
Multiply both sides by \( \frac{5}{2} \) (which is positive, so the inequality remains the same): \[ x \geq 2 \cdot \frac{5}{2} \] \[ x \geq 5 \]
Convert \( 1 \frac{6}{7} \) to an improper fraction: \[ 1 \frac{6}{7} = 1 + \frac{6}{7} = \frac{7}{7} + \frac{6}{7} = \frac{13}{7} \]
Now, the inequality becomes: \[ \frac{3}{7}x + \frac{1}{7} \leq \frac{13}{7} \]
Subtract \( \frac{1}{7} \) from both sides: \[ \frac{3}{7}x \leq \frac{13}{7} - \frac{1}{7} \] \[ \frac{3}{7}x \leq \frac{12}{7} \]
Multiply both sides by \( \frac{7}{3} \): \[ x \leq \frac{12}{7} \cdot \frac{7}{3} \] \[ x \leq 4 \]
Add \( \frac{3}{5} \) to both sides: \[ \frac{1}{5}x \leq \frac{2}{5} + \frac{3}{5} \] \[ \frac{1}{5}x \leq \frac{5}{5} \] \[ \frac{1}{5}x \leq 1 \]
Multiply both sides by \( 5 \): \[ x \leq 5 \]
Convert \( 3 \frac{2}{5} \) to an improper fraction: \[ 3 \frac{2}{5} = 3 + \frac{2}{5} = \frac{15}{5} + \frac{2}{5} = \frac{17}{5} \]
Now, the inequality becomes: \[ \frac{3}{5}x + \frac{2}{5} \geq \frac{17}{5} \]
Subtract \( \frac{2}{5} \) from both sides: \[ \frac{3}{5}x \geq \frac{17}{5} - \frac{2}{5} \] \[ \frac{3}{5}x \geq \frac{15}{5} \] \[ \frac{3}{5}x \geq 3 \]
Multiply both sides by \( \frac{5}{3} \): \[ x \geq 3 \cdot \frac{5}{3} \] \[ x \geq 5 \]
