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Solve the problem. -The locations of three receiving stations and the distances to the epicenter of an earthquake are contained in the following three equations: (x+2) 2+(y8) 2=16,(x+7) 2+(y4) 2=25( x + 2 ) ^ { 2 } + ( y - 8 ) ^ { 2 } = 16 , ( x + 7 ) ^ { 2 } + ( y - 4 ) ^ { 2 } = 25 , (x4) 2+(y+4) 2=100( x - 4 ) ^ { 2 } + ( y + 4 ) ^ { 2 } = 100 . Determine the location of the epicenter.


A) at (3,6) ( - 3,6 )
B) at (1,3) ( - 1,3 )
C) at (1,4) ( - 1,4 )
D) at (2,4) ( - 2,4 )

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For the pair of functions, find the indicated sum, difference, product, or quotient. - f(x) =64x,g(x) =9x2+4f ( x ) = 6 - 4 x , g ( x ) = - 9 x ^ { 2 } + 4 Find (f+g) (x) ( f + g ) ( x ) .


A) 13x24x+10- 13 x ^ { 2 } - 4 x + 10
B) 9x2+6- 9 x ^ { 2 } + 6
C) 9x24x+10- 9 x ^ { 2 } - 4 x + 10
D) 13x+10- 13 x + 10

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Match the equation with the correct graph. - y=14x+1y = - \frac { 1 } { 4 } x + 1


A)
 Match the equation with the correct graph. - y = - \frac { 1 } { 4 } x + 1  A)    B)    C)    D)
B)
 Match the equation with the correct graph. - y = - \frac { 1 } { 4 } x + 1  A)    B)    C)    D)
C)
 Match the equation with the correct graph. - y = - \frac { 1 } { 4 } x + 1  A)    B)    C)    D)
D)
 Match the equation with the correct graph. - y = - \frac { 1 } { 4 } x + 1  A)    B)    C)    D)

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Consider the function h as defined. Find functions f and g so that (f ∘ g) (x) = h(x) . - h(x) =10x2+7h ( x ) = \frac { 10 } { x ^ { 2 } } + 7


A) f(x) =1x,g(x) =10x+7f ( x ) = \frac { 1 } { x } , g ( x ) = \frac { 10 } { x } + 7
B) f(x) =x+7,g(x) =10x2f ( x ) = x + 7 , g ( x ) = \frac { 10 } { x ^ { 2 } }
C) f(x) =10x2,g(x) =7 f ( x ) = \frac { 10 } { x ^ { 2 } } , g ( x ) = 7
D) f(x) =x,g(x) =10x+7f ( x ) = x , g ( x ) = \frac { 10 } { x } + 7

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Graph the function. -Graph the function. -    A)    B)    C)    D)     Graph the function. -    A)    B)    C)    D)


A)
Graph the function. -    A)    B)    C)    D)
B)
Graph the function. -    A)    B)    C)    D)
C)
Graph the function. -    A)    B)    C)    D)
D)
Graph the function. -    A)    B)    C)    D)

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Graph the function. - f(x) =7x2f(x) =7 x^{2}  Graph the function. - f(x) =7 x^{2}     A)    B)    C)    D)


A)
 Graph the function. - f(x) =7 x^{2}     A)    B)    C)    D)
B)
 Graph the function. - f(x) =7 x^{2}     A)    B)    C)    D)
C)
 Graph the function. - f(x) =7 x^{2}     A)    B)    C)    D)
D)
 Graph the function. - f(x) =7 x^{2}     A)    B)    C)    D)

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Consider the function h as defined. Find functions f and g so that (f ∘ g) (x) = h(x) . - h(x) =1x27h ( x ) = \frac { 1 } { x ^ { 2 } - 7 }


A) f(x) =1x2,g(x) =17f ( x ) = \frac { 1 } { x ^ { 2 } } , g ( x ) = - \frac { 1 } { 7 }
B) f(x) =1x2,g(x) =x7f ( x ) = \frac { 1 } { x ^ { 2 } } , g ( x ) = x - 7
C) f(x) =1x,g(x) =x27f ( x ) = \frac { 1 } { x } , g ( x ) = x ^ { 2 } - 7
D) f(x) =17,g(x) =x27f ( x ) = \frac { 1 } { 7 } , g ( x ) = x ^ { 2 } - 7

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Use a graphing calculator to solve the linear equation. - 4(2z2) =7(z+4) 4 ( 2 z - 2 ) = 7 ( z + 4 )


A) {20}\{ - 20 \}
B) {20}\{ 20 \}
C) {36}\{ 36 \}
D) {24}\{ 24 \}

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Find the slope of the line and sketch the graph. - 2x3y=82 x-3 y=-8  Find the slope of the line and sketch the graph. - 2 x-3 y=-8    A)   \mathrm { m } = \frac { 3 } { 2 }    B)   \mathrm { m } = - \frac { 3 } { 2 }    C)   \mathrm { m } = \frac { 2 } { 3 }    D)   \mathrm { m } = - \frac { 2 } { 3 }


A) m=32\mathrm { m } = \frac { 3 } { 2 }
 Find the slope of the line and sketch the graph. - 2 x-3 y=-8    A)   \mathrm { m } = \frac { 3 } { 2 }    B)   \mathrm { m } = - \frac { 3 } { 2 }    C)   \mathrm { m } = \frac { 2 } { 3 }    D)   \mathrm { m } = - \frac { 2 } { 3 }
B) m=32\mathrm { m } = - \frac { 3 } { 2 }
 Find the slope of the line and sketch the graph. - 2 x-3 y=-8    A)   \mathrm { m } = \frac { 3 } { 2 }    B)   \mathrm { m } = - \frac { 3 } { 2 }    C)   \mathrm { m } = \frac { 2 } { 3 }    D)   \mathrm { m } = - \frac { 2 } { 3 }
C) m=23\mathrm { m } = \frac { 2 } { 3 }
 Find the slope of the line and sketch the graph. - 2 x-3 y=-8    A)   \mathrm { m } = \frac { 3 } { 2 }    B)   \mathrm { m } = - \frac { 3 } { 2 }    C)   \mathrm { m } = \frac { 2 } { 3 }    D)   \mathrm { m } = - \frac { 2 } { 3 }
D) m=23\mathrm { m } = - \frac { 2 } { 3 }
 Find the slope of the line and sketch the graph. - 2 x-3 y=-8    A)   \mathrm { m } = \frac { 3 } { 2 }    B)   \mathrm { m } = - \frac { 3 } { 2 }    C)   \mathrm { m } = \frac { 2 } { 3 }    D)   \mathrm { m } = - \frac { 2 } { 3 }

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Give the domain and range of the relation. - y=8x4y = \frac { - 8 } { x - 4 }


A) domain: (,4) (4,) ( - \infty , - 4 ) \cup ( - 4 , \infty ) ; range: (,0) (0,) ( - \infty , 0 ) \cup ( 0 , \infty )
B) domain: (,4) (4,) ( - \infty , 4 ) \cup ( 4 , \infty ) ; range: (,) ( - \infty , \infty )
C) domain: (,4) (4,) ( - \infty , - 4 ) \cup ( 4 , \infty ) ; range: (,) ( - \infty , \infty )
D) domain: (,4) (4,) ( - \infty , 4 ) \cup ( 4 , \infty ) ; range: (,0) (0,) ( - \infty , 0 ) \cup ( 0 , \infty )

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Solve the problem. -In Country X, the average hourly wage in dollars from 1960 to 2010 can be modeled by f(x) ={0.078(x1960) +0.32 if 1960x<19950.184(x1995) +3.04 if 1995x2010f ( x ) = \left\{ \begin{array} { l l } 0.078 ( x - 1960 ) + 0.32 & \text { if } 1960 \leq x < 1995 \\0.184 ( x - 1995 ) + 3.04 & \text { if } 1995 \leq x \leq 2010\end{array} \right. Use ff to estimate the average hourly wages in 1965,1985 , and 2005.2005 .


A) $3.43,$0.32,$6.72\$ 3.43 , \$ 0.32 , \$ 6.72
B) $0.71,$3.04,$6.72\$ 0.71 , \$ 3.04 , \$ 6.72
C) $0.71,$2.27,$6.72\$ 0.71 , \$ 2.27 , \$ 6.72

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Graph the line described. -  through (0,4) ;m=14\text { through }(0,4) ; \mathrm{m}=-\frac{1}{4}  Graph the line described. - \text { through }(0,4)  ; \mathrm{m}=-\frac{1}{4}    A)    B)    C)    D)


A)
 Graph the line described. - \text { through }(0,4)  ; \mathrm{m}=-\frac{1}{4}    A)    B)    C)    D)
B)
 Graph the line described. - \text { through }(0,4)  ; \mathrm{m}=-\frac{1}{4}    A)    B)    C)    D)
C)
 Graph the line described. - \text { through }(0,4)  ; \mathrm{m}=-\frac{1}{4}    A)    B)    C)    D)
D)
 Graph the line described. - \text { through }(0,4)  ; \mathrm{m}=-\frac{1}{4}    A)    B)    C)    D)

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The graph of a linear function f is shown. Write the equation that defines f. Write the equation in slope -intercept form. - The graph of a linear function f is shown. Write the equation that defines f. Write the equation in slope -intercept form. -  A)   y = \frac { x } { - 4 }  B)   y = \frac { x } { 4 }  C)   y = - 4 x  D)   y = 4 x


A) y=x4y = \frac { x } { - 4 }
B) y=x4y = \frac { x } { 4 }
C) y=4xy = - 4 x
D) y=4xy = 4 x

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Give the domain and range of the relation. - Give the domain and range of the relation. -  A)  domain:  ( - \infty , 0 )  \cup ( 0 , \infty )  ; range:  ( - \infty , 0 )  \cup ( 0 , \infty )   B)  domain:  ( - \infty , 0 )  ; range:  ( - \infty , 0 )   C)  domain:  ( 0 , \infty )  ; range:  [ 3 , \infty )   D)  domain:  ( - \infty , \infty )  ; range:  [ 6 , \infty )


A) domain: (,0) (0,) ( - \infty , 0 ) \cup ( 0 , \infty ) ; range: (,0) (0,) ( - \infty , 0 ) \cup ( 0 , \infty )
B) domain: (,0) ( - \infty , 0 ) ; range: (,0) ( - \infty , 0 )
C) domain: (0,) ( 0 , \infty ) ; range: [3,) [ 3 , \infty )
D) domain: (,) ( - \infty , \infty ) ; range: [6,) [ 6 , \infty )

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Graph the point symmetric to the given point. -Plot the point (-2, -1) , then plot the point that is symmetric to (-2, -1) with respect to the x-axis. Graph the point symmetric to the given point. -Plot the point (-2, -1) , then plot the point that is symmetric to (-2, -1)  with respect to the x-axis.    A)    B)    C)    D)


A)
Graph the point symmetric to the given point. -Plot the point (-2, -1) , then plot the point that is symmetric to (-2, -1)  with respect to the x-axis.    A)    B)    C)    D)
B)
Graph the point symmetric to the given point. -Plot the point (-2, -1) , then plot the point that is symmetric to (-2, -1)  with respect to the x-axis.    A)    B)    C)    D)
C)
Graph the point symmetric to the given point. -Plot the point (-2, -1) , then plot the point that is symmetric to (-2, -1)  with respect to the x-axis.    A)    B)    C)    D)
D)
Graph the point symmetric to the given point. -Plot the point (-2, -1) , then plot the point that is symmetric to (-2, -1)  with respect to the x-axis.    A)    B)    C)    D)

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Decide whether the relation defines a function. - x=y2x = y ^ { 2 }


A) Function
B) Not a function

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Determine whether the equation has a graph that is symmetric with respect to the y -axis, the x-axis, the origin, or none of these. -y = (x - 6) (x + 3)


A) y-axis only
B) none of these
C) x-axis only
D) x-axis, y-axis, origin

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Graph the equation by plotting points. - y=x2+1y=-x^{2}+1  Graph the equation by plotting points. - y=-x^{2}+1     A)    B)    C)    D)


A)
 Graph the equation by plotting points. - y=-x^{2}+1     A)    B)    C)    D)
B)
 Graph the equation by plotting points. - y=-x^{2}+1     A)    B)    C)    D)
C)
 Graph the equation by plotting points. - y=-x^{2}+1     A)    B)    C)    D)
D)
 Graph the equation by plotting points. - y=-x^{2}+1     A)    B)    C)    D)

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Write an equation for the line described. Give your answer in standard form. -through (1,5) ( 1,5 ) , undefined slope


A) x=5x = 5
B) y=1y = 1
C) x=1x = 1
D) y=5\mathrm { y } = 5

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Describe the transformations and give the equation for the graph. - Describe the transformations and give the equation for the graph. -  A)  It is the graph of  f ( x )  = | x |  shrunken vertically by a factor of 8 and translated 9 units down. The equation is  y = 8 | x | + 9  B)  It is the graph of  f ( x )  = | x |  stretched vertically by a factor of 8 and translated 9 units down. The equation is  y = 8 | x | - 9  C)  It is the graph of  f ( x )  = | x |  stretched vertically by a factor of 8 and translated 9 units down. The equation is  y = \frac { 1 } { 8 } | x | + 9   D)  It is the graph of  f ( x )  = | x |  shrunken vertically by a factor of 8 and translated 9 units down. The equation is  \mathrm { y } = \frac { 1 } { 8 } | \mathrm { x } | - 9


A) It is the graph of f(x) =xf ( x ) = | x | shrunken vertically by a factor of 8 and translated 9 units down. The equation is y=8x+9y = 8 | x | + 9
B) It is the graph of f(x) =xf ( x ) = | x | stretched vertically by a factor of 8 and translated 9 units down. The equation is y=8x9y = 8 | x | - 9
C) It is the graph of f(x) =xf ( x ) = | x | stretched vertically by a factor of 8 and translated 9 units down. The equation is y=18x+9y = \frac { 1 } { 8 } | x | + 9

D) It is the graph of f(x) =xf ( x ) = | x | shrunken vertically by a factor of 8 and translated 9 units down. The equation is y=18x9\mathrm { y } = \frac { 1 } { 8 } | \mathrm { x } | - 9

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