Question

name: abbygail weekly math review - q1:8 date: oct 2
oct 1 wednesday
in a problem from 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.\ abe found the part for only $60 instead of $130. how will this affect break - even point?
solve the equation for x: 6(x + 3)-9 = 27
solve the equation for x: $\frac{x}{8}-\frac{3}{4}=\frac{5}{8}$
solve the equation for y: $\frac{8 + 2y}{3}=x$
what kind of transformation is depicted in the pictures below?
solve for x in the figure below. x + 42, 60°
solve for x in the figure below. 9x + 12, 7x + 6, 22°
oct 2 thursday
the average of mannys three tests is an 84. what must he get on a 4th test to raise his average to a 87?
solve the equation for x: 7(x - 2)=7x + 14
solve the equation for x: $\frac{x}{9}+\frac{2}{3}=\frac{4}{9}$
set up an equation for the perimeter, p, of the rectangle below. solve for l. l, w
draw △hij after a 90° rotation about the origin. h(-3,3), i(1,2), j(-2,1)
give the vertices of a triangle that dilated twice as big as △abc if a(0, 0), b(8, 2), and c(12,5) (centered at the origin).
solve for x in the figure below. 5x, 3x + 2, 130°
solve for x in the figure below. x + 66, x + 86, 50°

Explanation:

Solve $6(x + 3)-9 = 27$ for $x$:

Step1: Expand the left - hand side

$6(x + 3)-9=6x+18 - 9=6x + 9$
So the equation becomes $6x+9 = 27$.

Step2: Isolate the term with $x$

Subtract 9 from both sides: $6x=27 - 9=18$.

Step3: Solve for $x$

Divide both sides by 6: $x=\frac{18}{6}=3$.

Solve $7(x - 2)=7x + 14$ for $x$:

Step1: Expand the left - hand side

$7(x - 2)=7x-14$.
So the equation is $7x-14 = 7x + 14$.

Step2: Try to isolate $x$

Subtract $7x$ from both sides: $7x-7x-14=7x-7x + 14$, which simplifies to $-14 = 14$. This is a contradiction, so there is no solution.

Solve $\frac{x}{8}-\frac{3}{4}=\frac{5}{8}$ for $x$:

Step1: Get a common denominator

The common denominator of 8 and 4 is 8. Rewrite $\frac{3}{4}=\frac{6}{8}$.
The equation becomes $\frac{x}{8}-\frac{6}{8}=\frac{5}{8}$.

Step2: Combine the fractions on the left - hand side

$\frac{x - 6}{8}=\frac{5}{8}$.

Step3: Solve for $x$

Cross - multiply: $x-6 = 5$, then $x=5 + 6=11$.

Solve $\frac{x}{9}+\frac{2}{3}=\frac{4}{9}$ for $x$:

Step1: Get a common denominator

Rewrite $\frac{2}{3}=\frac{6}{9}$.
The equation is $\frac{x}{9}+\frac{6}{9}=\frac{4}{9}$.

Step2: Combine the fractions on the left - hand side

$\frac{x + 6}{9}=\frac{4}{9}$.

Step3: Solve for $x$

Cross - multiply: $x+6 = 4$, then $x=4 - 6=-2$.

Solve $\frac{8 + 2y}{3}=x$ for $y$:

Step1: Multiply both sides by 3

$8 + 2y=3x$.

Step2: Isolate the term with $y$

Subtract 8 from both sides: $2y=3x - 8$.

Step3: Solve for $y$

Divide both sides by 2: $y=\frac{3x - 8}{2}$.

For the rectangle with perimeter $P$, the perimeter formula is $P = 2(L + W)$. Solve for $L$:

Step1: Expand the formula

$P=2L+2W$.

Step2: Isolate the term with $L$

Subtract $2W$ from both sides: $2L=P - 2W$.

Step3: Solve for $L$

Divide both sides by 2: $L=\frac{P - 2W}{2}$.

For the rotation of $\triangle HIJ$ with $H(-3,3)$, $I(1,2)$, $J(-2,1)$ 90° about the origin, the transformation rule for a 90° counter - clockwise rotation about the origin is $(x,y)\to(-y,x)$.

$H(-3,3)\to H'(-3,-3)$
$I(1,2)\to I'(-2,1)$
$J(-2,1)\to J'(-1,-2)$

For the dilation of $\triangle ABC$ with $A(0,0)$, $B(8,2)$, $C(12,5)$ by a scale factor of 2 centered at the origin, the rule is $(x,y)\to(2x,2y)$.

$A(0,0)\to A'(0,0)$
$B(8,2)\to B'(16,4)$
$C(12,5)\to C'(24,10)$

For the right - triangle with an angle of 60° and side $x + 42$:

Using the fact that $\sin60^{\circ}=\frac{\text{opposite}}{\text{hypotenuse}}$ (assuming $x + 42$ is the hypotenuse and if we consider the relationship with the right - angle). If we assume it's a non - hypotenuse side, and using the tangent function. Since it's a right - triangle and one angle is 60°, the other non - right angle is 30°. If $x + 42$ is the side opposite the 60° angle and the adjacent side to the 60° angle is $a$, $\tan60^{\circ}=\sqrt{3}=\frac{x + 42}{a}$. But if we assume the side adjacent to 60° is 1, then $x+42=\sqrt{3}$, $x=\sqrt{3}-42$. If we use the cosine function $\cos60^{\circ}=\frac{1}{2}=\frac{\text{adjacent}}{\text{hypotenuse}}$. Assuming the adjacent side to 60° is $a$ and hypotenuse is $x + 42$, $a=\frac{x + 42}{2}$. Without more information about the sides, if we assume it's a 30 - 60 - 90 triangle and the side opposite 30° is 1, the side opposite 60° is $\sqrt{3}$ and the hypotenuse is 2. If $x + 42$ is the hypotenuse, and the side opposite 30° is 1, then $x+42 = 2$, $x=-40$.

For the triangle with angles 130°, and sides $5x$ and $3x + 2$:

Using the angle - sum property of a triangle (the sum of interior angles of a triangle is 180°), the third angle is $180-(130)=50^{\circ}$. Then, using the Law of Sines $\frac{5x}{\sin130^{\circ}}=\frac{3x + 2}{\sin50^{\circ}}$.
$5x\sin50^{\circ}=(3x + 2)\sin130^{\circ}$. Since $\sin130^{\circ}=\sin(180 - 50)^{\circ}=\sin50^{\circ}$, we have $5x\sin50^{\circ}=(3x + 2)\sin50^{\circ}$. Divide both sides by $\sin50^{\circ}$ (since $\sin50^{\circ}
eq0$), $5x=3x + 2$. Subtract $3x$ from both sides: $2x=2$, $x = 1$.

For the triangle with angles $9x+12$, $7x + 6$ and $22^{\circ}$:

Using the angle - sum property of a triangle: $(9x+12)+(7x + 6)+22=180$.
Combine like terms: $16x+40 = 180$.
Subtract 40 from both sides: $16x=140$.
Divide both sides by 16: $x=\frac{140}{16}=\frac{35}{4}=8.75$.

For the triangle with angles $x + 66$, $x + 86$ and $50^{\circ}$:

Using the angle - sum property of a triangle: $(x + 66)+(x + 86)+50=180$.
Combine like terms: $2x+202 = 180$.
Subtract 202 from both sides: $2x=-22$.
Divide both sides by 2: $x=-11$.

Answer:

$x = 3$ for $6(x + 3)-9 = 27$
No solution for $7(x - 2)=7x + 14$
$x = 11$ for $\frac{x}{8}-\frac{3}{4}=\frac{5}{8}$
$x=-2$ for $\frac{x}{9}+\frac{2}{3}=\frac{4}{9}$
$y=\frac{3x - 8}{2}$ for $\frac{8 + 2y}{3}=x$
$L=\frac{P - 2W}{2}$ for $P = 2(L + W)$
$H'(-3,-3)$, $I'(-2,1)$, $J'(-1,-2)$ for the rotation of $\triangle HIJ$
$A'(0,0)$, $B'(16,4)$, $C'(24,10)$ for the dilation of $\triangle ABC$
$x=\sqrt{3}-42$ (with assumptions) for the right - triangle with 60° angle
$x = 1$ for the triangle with 130° angle
$x = 8.75$ for the triangle with angles $9x+12$, $7x + 6$ and $22^{\circ}$
$x=-11$ for the triangle with angles $x + 66$, $x + 86$ and $50^{\circ}$

Answer:

$x = 3$ for $6(x + 3)-9 = 27$
No solution for $7(x - 2)=7x + 14$
$x = 11$ for $\frac{x}{8}-\frac{3}{4}=\frac{5}{8}$
$x=-2$ for $\frac{x}{9}+\frac{2}{3}=\frac{4}{9}$
$y=\frac{3x - 8}{2}$ for $\frac{8 + 2y}{3}=x$
$L=\frac{P - 2W}{2}$ for $P = 2(L + W)$
$H'(-3,-3)$, $I'(-2,1)$, $J'(-1,-2)$ for the rotation of $\triangle HIJ$
$A'(0,0)$, $B'(16,4)$, $C'(24,10)$ for the dilation of $\triangle ABC$
$x=\sqrt{3}-42$ (with assumptions) for the right - triangle with 60° angle
$x = 1$ for the triangle with 130° angle
$x = 8.75$ for the triangle with angles $9x+12$, $7x + 6$ and $22^{\circ}$
$x=-11$ for the triangle with angles $x + 66$, $x + 86$ and $50^{\circ}$