Question

monday
the average of johnnys two tests is a 92. what must he get on a 3rd test to raise his average to a 94?
how many solutions does the equation have? -10(x - 1)=10 - 10x
solve the equation for x: 16 = 2(x + 1)
solve the equation for x: $\frac{x}{14}-\frac{1}{2}=\frac{1}{14}$
solve the equation for x: $\frac{x}{12}+\frac{1}{3}=\frac{1}{12}$
solve the equation for y: y = $\frac{x + 3}{2}$, x = 2y - 3
reflect △uvw over the y - axis
if △xyz is dilated by a factor of 3, what would be the area of the △xyz?
find the missing angle.
find the missing angle.
find the missing angle.
tuesday
a car needs fixing and abe can fix it for $70 per hour with a $130 part, but gabe can fix it for $80 an hour with a $40 part. how long will it take for both abe and gabe to cost the same amount?
solve the equation for w: 4 - $\frac{2}{7}$w = 18
point f(-17,8) is rotated 180° about the origin. what are the coordinates of image f?
find the missing angle.
find the missing angle.
find the missing angle.

Answer:

Monday

1. The average of Johnny's two tests is a 92. What must he get on a 3rd test to raise his average to a 94?

• Explanation:
• Step1: Find the total score of the first two - tests
• The formula for the average $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$. Given $n = 2$ and $\bar{x}=92$, the total score of the first two tests is $2\times92 = 184$.
• Step2: Find the total score needed for three tests to have an average of 94
• For $n = 3$ and $\bar{x}=94$, the total score needed is $3\times94=282$.
• Step3: Find the score on the third test
• Let the score on the third test be $x$. Then $184 + x=282$, so $x=282 - 184=98$.
• Answer: 98

1. How many solutions does the equation $-10(x - 1)=10-10x$ have?

• Explanation:
• Step1: Expand the left - hand side of the equation
• Using the distributive property $a(b - c)=ab - ac$, we have $-10(x - 1)=-10x+10$.
• Step2: Compare the two sides of the equation
• The equation becomes $-10x + 10=10-10x$. Since the left - hand side and the right - hand side are identical for all real values of $x$, the equation has infinitely many solutions.
• Answer: Infinitely many

1. Solve the equation for $x$: $\frac{x}{14}-\frac{1}{2}=\frac{1}{14}$

• Explanation:
• Step1: Add $\frac{1}{2}$ to both sides of the equation
• $\frac{x}{14}=\frac{1}{14}+\frac{1}{2}$.
• Step2: Find a common denominator for the right - hand side
• The common denominator of 14 and 2 is 14. So $\frac{1}{14}+\frac{1}{2}=\frac{1}{14}+\frac{7}{14}=\frac{1 + 7}{14}=\frac{8}{14}=\frac{4}{7}$.
• Step3: Solve for $x$
• Multiply both sides of $\frac{x}{14}=\frac{4}{7}$ by 14, we get $x=\frac{4}{7}\times14 = 8$.
• Answer: 8

1. Solve the equation for $y$ given $x = 2y-3$

• Explanation:
• Step1: Add 3 to both sides of the equation
• $x + 3=2y$.
• Step2: Divide both sides by 2
• $y=\frac{x + 3}{2}$.
• Answer: $y=\frac{x + 3}{2}$

1. Reflect $\triangle UVW$ over the $y$ - axis. Given $U(-2,1)$, $V(-1,4)$, $W(1,2)$

• Explanation:
• Step1: Apply the rule for reflecting over the $y$ - axis
• The rule for reflecting a point $(x,y)$ over the $y$ - axis is $(-x,y)$.
• For point $U(-2,1)$, its image $U'(2,1)$.
• For point $V(-1,4)$, its image $V'(1,4)$.
• For point $W(1,2)$, its image $W'(-1,2)$.
• Answer: $U'(2,1)$, $V'(1,4)$, $W'(-1,2)$

1. If $\triangle XYZ$ is dilated by a factor of 3, what would be the area of the $\triangle X'Y'Z'$?

• Explanation:
• Step1: Recall the relationship between the areas of similar triangles
• If the scale factor of dilation is $k$, the ratio of the areas of two similar triangles is $k^{2}$. Here $k = 3$.
• Let the area of $\triangle XYZ$ be $A$. Then the area of $\triangle X'Y'Z'$ is $k^{2}A=9A$. Since the area of the original triangle is not given, we can only say that the area of the dilated triangle is 9 times the area of the original triangle.
• Answer: 9 times the area of the original $\triangle XYZ$

1. Find the missing angle in a triangle with angles $60^{\circ}$ and $80^{\circ}$

• Explanation:
• Step1: Recall the sum of angles in a triangle
• The sum of the interior angles of a triangle is $180^{\circ}$. Let the missing angle be $x$.
• Then $x+60^{\circ}+80^{\circ}=180^{\circ}$.
• $x=180^{\circ}-(60^{\circ}+80^{\circ})=40^{\circ}$.
• Answer: $40^{\circ}$

Tuesday

1. A car needs fixing and Abe can fix it for $\$70$ per hour with a $\$130$ part, but Gabe can fix it for $\$80$ per hour with a $\$40$ part. How long will it take for both Abe and Gabe to cost the same amount?

• Explanation:
• Step1: Set up equations for Abe's and Gabe's costs
• Let $h$ be the number of hours of work. Abe's cost $C_{A}=70h + 130$. Gabe's cost $C_{G}=80h+40$.
• Step2: Set the two costs equal to each other
• $70h + 130=80h+40$.
• Step3: Solve for $h$
• Subtract $70h$ from both sides: $130 = 10h+40$.
• Subtract 40 from both sides: $90 = 10h$.
• Divide both sides by 10: $h = 9$.
• Answer: 9 hours

1. Solve the equation for $x$: $16 = 2(x + 1)$

• Explanation:
• Step1: Divide both sides of the equation by 2
• $\frac{16}{2}=x + 1$, so $8=x + 1$.
• Step2: Subtract 1 from both sides
• $x=8 - 1=7$.
• Answer: 7

1. Solve the equation for $x$: $\frac{x}{12}+\frac{1}{3}=\frac{1}{12}$

• Explanation:
• Step1: Subtract $\frac{1}{3}$ from both sides of the equation
• $\frac{x}{12}=\frac{1}{12}-\frac{1}{3}$.
• Step2: Find a common denominator for the right - hand side
• The common denominator of 12 and 3 is 12. So $\frac{1}{12}-\frac{1}{3}=\frac{1}{12}-\frac{4}{12}=-\frac{3}{12}=-\frac{1}{4}$.
• Step3: Solve for $x$
• Multiply both sides of $\frac{x}{12}=-\frac{1}{4}$ by 12, we get $x=-\frac{1}{4}\times12=-3$.
• Answer: - 3

1. Solve the equation for $w$: $4-\frac{2}{7}w = 18$

• Explanation:
• Step1: Subtract 4 from both sides of the equation
• $-\frac{2}{7}w=18 - 4=14$.
• Step2: Solve for $w$
• Multiply both sides by $-\frac{7}{2}$. So $w=14\times(-\frac{7}{2})=-49$.
• Answer: - 49

1. Point $F(-17,8)$ is rotated $180^{\circ}$ about the origin. What are the coordinates of image $F'$?

• Explanation:
• Step1: Apply the rule for rotating a point $180^{\circ}$ about the origin
• The rule for rotating a point $(x,y)$ $180^{\circ}$ about the origin is $(-x,-y)$.
• For point $F(-17,8)$, its image $F'(17,-8)$.
• Answer: $(17,-8)$

1. Find the missing angle in a triangle with angles $45^{\circ}$ and $75^{\circ}$

• Explanation:
• Step1: Recall the sum of angles in a triangle
• The sum of the interior angles of a triangle is $180^{\circ}$. Let the missing angle be $x$.
• Then $x + 45^{\circ}+75^{\circ}=180^{\circ}$.
• $x=180^{\circ}-(45^{\circ}+75^{\circ})=60^{\circ}$.
• Answer: $60^{\circ}$

1. Find the missing angle in a right - triangle with one non - right angle of $35^{\circ}$

• Explanation:
• Step1: Recall the sum of angles in a triangle
• The sum of the interior angles of a triangle is $180^{\circ}$, and in a right - triangle one angle is $90^{\circ}$. Let the missing angle be $x$.
• Then $x+90^{\circ}+35^{\circ}=180^{\circ}$.
• $x=180^{\circ}-(90^{\circ}+35^{\circ})=55^{\circ}$.
• Answer: $55^{\circ}$