Question

1. the solution to \\( \frac { 2 } { 3 } ( 3 - 2 x ) = \frac { 3 } { 8 } \\) is

a. \\( - \frac { 11 } { 8 } \\) b. \\( \frac { 3 } { 8 } \\) c. \\( - \frac { 13 } { 16 } \\) d. \\( \frac { 15 } { 16 } \\)

1. corinne planea unas vacaciones en la playa para

julio y está analizando las temperaturas diarias
más altas para sus posibles destinos. le gustaría
elegir un destino con una temperatura media alta
y un pequeño rango intercuartílico. realizó los
diagramas de caja que aparecen a continuación.
¿qué destino tiene una temperatura mediana
superior a los 80 grados y el menor rango
intercuartílico?
a. ocean beach
b. whispering palms
c. serene shores
d. pelican beach

1. ¿qué relación es una función?

a. \\( ( ( 1,3 ), ( 2,1 ), ( 3,1 ), ( 4,7 ) ) \\)
b.
c.
d.

1. una expresión de quinto grado se escribe con un

coeficiente principal de siete y una constante de
sís. ¿qué expresión está escrita correctamente
para estas condiciones?
a. \\( 6 x ^ { 5 } + x ^ { 4 } + 7 \\)
b. \\( 7 x ^ { 6 } - 6 x ^ { 4 } + 5 \\)
c. \\( 6 x ^ { 7 } - x ^ { 5 } + 5 \\)
d. \\( 7 x ^ { 5 } + 2 x ^ { 2 } + 6 \\)

Explanation:

Question 1

Step1: Expand the left - hand side

Multiply \(\frac{3}{4}\) with \((3 - 2x)\) using the distributive property \(a(b - c)=ab - ac\). So, \(\frac{3}{4}(3 - 2x)=\frac{3\times3}{4}-\frac{3\times2x}{4}=\frac{9}{4}-\frac{3x}{2}\)
The equation becomes \(\frac{9}{4}-\frac{3x}{2}=\frac{3}{2}\)

Step2: Move the constant term to the right - hand side

Subtract \(\frac{9}{4}\) from both sides: \(-\frac{3x}{2}=\frac{3}{2}-\frac{9}{4}\)
Find a common denominator for the right - hand side. The common denominator of 2 and 4 is 4. So, \(\frac{3}{2}-\frac{9}{4}=\frac{3\times2}{2\times2}-\frac{9}{4}=\frac{6}{4}-\frac{9}{4}=-\frac{3}{4}\)
The equation is now \(-\frac{3x}{2}=-\frac{3}{4}\)

Step3: Solve for \(x\)

Multiply both sides by \(-\frac{2}{3}\). \(x = (-\frac{3}{4})\times(-\frac{2}{3})\)
\(x=\frac{(-3)\times(-2)}{4\times3}=\frac{6}{12}=\frac{1}{2}\)

A function is a relation where each input (domain value) has exactly one output (range value).

• For option A: Each \(x\) - value (1, 2, 3, 4) has exactly one \(y\) - value.
• For option B: The input \(x = 7\) has two outputs \(y = 3\) and \(y = 5\)
• For option C: A circle (in the \(x - y\) plane) does not represent a function because for most \(x\) values (except at the extreme left and right of the circle), there are two \(y\) values.
• For option D: The input \(x = 6\) has two outputs \(y = 4\) and \(y = 5\)

A fifth - degree polynomial has the form \(a_5x^5+a_4x^4+\cdots+a_0\), where \(a_5
eq0\) and the highest power of \(x\) is 5.

• Option A: The highest degree is 5, but the leading coefficient is 6 (not 7)
• Option B: The highest degree is 6 (not 5)
• Option C: The highest degree is 7 (not 5)
• Option D: The highest degree is 5, the leading coefficient \(a_5 = 7\) and the constant term \(a_0=6\)

Answer:

None of the options A, B, C, D are correct. There might be a mistake in the problem - writing or option - setting. If we assume there was a typo in the original equation, for example, if the equation was \(\frac{3}{4}(3 - 2x)=\frac{3}{16}\)

Step1: Expand the left - hand side

\(\frac{3}{4}(3 - 2x)=\frac{9}{4}-\frac{3x}{2}\)
The equation is \(\frac{9}{4}-\frac{3x}{2}=\frac{3}{16}\)

Step2: Move the constant term to the right - hand side

\(-\frac{3x}{2}=\frac{3}{16}-\frac{9}{4}\)
Common denominator is 16. \(\frac{3}{16}-\frac{9\times4}{4\times4}=\frac{3}{16}-\frac{36}{16}=-\frac{33}{16}\)

Step3: Solve for \(x\)

\(x=(-\frac{33}{16})\times(-\frac{2}{3})=\frac{(-33)\times(-2)}{16\times3}=\frac{66}{48}=\frac{11}{8}\) still not matching. If the equation was \(\frac{4}{3}(3 - 2x)=\frac{3}{4}\)

Step1: Expand the left - hand side

\(\frac{4}{3}(3 - 2x)=4-\frac{8x}{3}\)
The equation is \(4-\frac{8x}{3}=\frac{3}{4}\)

Step2: Move the constant term to the right - hand side

\(-\frac{8x}{3}=\frac{3}{4}-4\)
Common denominator is 4. \(\frac{3}{4}-\frac{16}{4}=-\frac{13}{4}\)

Step3: Solve for \(x\)

\(x = (-\frac{13}{4})\times(-\frac{3}{8})=\frac{39}{32}\)

Question 3