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Roselyn is driving to visit her family, who live 150150 kilometers away. Her average speed is 6060 kilometers per hour. The car's tank has 2020 liters of fuel at the beginning of the drive, and its fuel efficiency is 66 kilometers per liter. Fuel costs 0.600.60 dollars per liter. How long can Roselyn drive before she runs out of fuel? hours

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One-step inequalities: word problemsright chevron icon

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−11x≥−33 -11 x \geq-33 \newlineWhich of the following best describes the solutions to the inequality shown?\newlineChoose 11 answer:\newline(A) x≤−3 x \leq-3 \newline(B) x≥−3 x \geq-3 \newline(C) x≤3 x \leq 3 \newline(D) x≥3 x \geq 3
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Posted 10 months ago

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32q≤92q−18 \frac{3}{2} q \leq \frac{9}{2} q-18 \newlineWhich of the following best describes the solutions to the inequality shown?\newlineChoose 11 answer:\newline(A) q≤3 q \leq 3 \newline(B) q≥3 q \geq 3 \newline(C) q≤6 q \leq 6 \newline(D) q≥6 q \geq 6
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Posted 10 months ago

Question
Let g(x)=5−ex g(x)=5-e^{x} .\newlineBelow is Sean's attempt to write a formal justification for the fact that the equation g(x)=0 g(x)=0 has a solution where 1≤x≤4 1 \leq x \leq 4 .\newlineIs Sean's justification complete? If not, why?\newlineSean's justification:\newlineg g is defined for all real numbers, and exponential functions are continuous at all points in their domains.\newlineSo, according to the intermediate value theorem, g(x)=0 g(x)=0 must have a solution somewhere between x=1 x=1 and x=4 x=4 .\newlineChoose 11 answer:\newline(A) Yes, Sean's justification is complete.\newline(B) No, Sean didn't establish that 00 is between g(1) g(1) and g(4) g(4) .\newline(C) No, Sean didn't establish that g g is continuous.
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Posted 9 months ago

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Let f(x)=cos⁡(π⋅x)3 f(x)=\sqrt[3]{\cos (\pi \cdot x)} .\newlineBelow is Issac's attempt to write a formal justification for the fact that the equation f(x)=0.1 f(x)=0.1 has a solution where 1.5≤x≤2 1.5 \leq x \leq 2 .\newlineIs Issac's justification complete? If not, why?\newlineIssac's justification:\newlinef(1.5)=0 and f(2)=1, so f(1.5)<0.1<f(2). \begin{array}{l} f(1.5)=0 \text { and } \\ f(2)=1 \text {, so } \\ f(1.5)<0.1<f(2) . \end{array} \newlineSo, according to the intermediate value theorem, f(x)=0.1 f(x)=0.1 must have a solution when x x is between x=1.5 x=1.5 and x=2 x=2 .\newlineChoose 11 answer:\newline(A) Yes, Issac's justification is complete.\newline(B) No, Issac didn't establish that 00.11 is between f(1.5) f(1.5) and f(2) f(2) .\newline(C) No, Issac didn't establish that f f is continuous.
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Posted 9 months ago

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Let h(x)=5⋅tan⁡(x) h(x)=5 \cdot \tan (x) .\newlineBelow is Bridget's attempt to write a formal justification for the fact that the equation h(x)=2 h(x)=2 has a solution where 0≤x≤π4 0 \leq x \leq \frac{\pi}{4} .\newlineIs Bridget's justification complete? If not, why?\newlineBridget's justification:\newlineh h is defined over the entire interval [0,π4] \left[0, \frac{\pi}{4}\right] , and trigonometric functions are continuous at all points in their domains.\newlineSo, according to the intermediate value theorem, h(x)=2 h(x)=2 must have a solution at some point in that interval.\newlineChoose 11 answer:\newline(A) Yes, Bridget's justification is complete.\newline(B) No, Bridget didn't establish that 22 is between h(0) h(0) and h(π4) h\left(\frac{\pi}{4}\right) .\newline(C) No, Bridget didn't establish that h h is continuous.
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Posted 9 months ago

Question
Let f(x)=25−2x f(x)=2^{5-2 x} .\newlineBelow is Ibrahim's attempt to write a formal justification for the fact that the equation f(x)=10 f(x)=10 has a solution where −1≤x≤4 -1 \leq x \leq 4 .\newlineIs Ibrahim's justification complete? If not, why?\newlineIbrahim's justification:\newlinef f is defined for all real numbers, and exponential functions are continuous at all points in their domains. Also, f(−1)=128 f(-1)=128 and f(4)=18 f(4)=\frac{1}{8} , so 1010 is between f(−1) f(-1) and f(4) f(4) .\newlineSo, according to the intermediate value theorem, f(x)=10 f(x)=10 must have a solution at some point between x=−1 x=-1 and x=4x=4 .\newlineChoose 11 answer:\newline(A) Yes, Ibrahim's justification is complete.\newline(B) No, Ibrahim didn't establish that 1010 is between f(−1) f(-1) and f(4) f(4) .\newline(c) No, Ibrahim didn't establish that f f is continuous.
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Posted 9 months ago

Question
Let g(x)=cos⁡(πx2) g(x)=\cos \left(\pi x^{2}\right) .\newlineBelow is Amrita's attempt to write a formal justification for the fact that the equation g′(x)=0.4 g^{\prime}(x)=0.4 has a solution where −1<x<4 -1<x<4 .\newlineIs Amrita's justification complete?\newlineIf not, why?\newlineAmrita's justification:\newlinePolynomial and trigonometric functions are differentiable and continuous at all points in their domain, and [−1,4] [-1,4] is within g′ g^{\prime} s domain. Also,\newlineg(−1)=−1 and g(4)=1, so g(4)−g(−1)4−(−1)=0.4. \begin{array}{l} g(-1)=-1 \text { and } \\ g(4)=1 \text {, so } \\ \frac{g(4)-g(-1)}{4-(-1)}=0.4 . \end{array} \newlineSo, according to the mean value theorem, g′(x)=0.4 g^{\prime}(x)=0.4 must have a solution somewhere between x=−1 x=-1 and x=4 x=4 .\newlineChoose 11 answer:\newline(A) Yes, Amrita's justification is complete.\newline(B) No, Amrita didn't establish that the average rate of change of g g over [−1,4] [-1,4] is equal to 00.44 .\newline(C) No, Amrita didn't establish that g g is differentiable.
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Posted 9 months ago

Question
Let\newlineh(x)=x5−2x4+10x2−10x.  h(x)=x^{5}-2 x^{4}+10 x^{2}-10 x \text {. } \newlineBelow is Omar's attempt to write a formal justification for the fact that there exists a value c c where\newline−2<c<2 such that h′(c)=6.  -2<c<2 \text { such that } h^{\prime}(c)=6 \text {. } \newlineIs Omar's justification complete? If not, why?\newlineOmar's justification:\newlineh(−2)=−4 and h(2)=20, so h(2)−h(−2)2−(−2)=6. \begin{array}{l} h(-2)=-4 \text { and } \\ h(2)=20 \text {, so } \\ \frac{h(2)-h(-2)}{2-(-2)}=6 . \end{array} \newlineSo, according to the mean value theorem, there exists a value c c somewhere in the interval −2<c<2 -2<c<2 such that h′(c)=6 h^{\prime}(c)=6 .\newlineChoose 11 answer:\newline(A) Yes, Omar's justification is complete.\newline(B) No, Omar didn't establish that the average rate of change of h h over [−2,2] [-2,2] is equal to 66 .\newline(C) No, Omar didn't establish that h h is differentiable.
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Posted 9 months ago

Question
Let f(x)=x+x3 f(x)=\sqrt{x+\sqrt{x^{3}}} .\newlineBelow is Lena's attempt to write a formal justification for the fact that the equation f′(x)=23 f^{\prime}(x)=\frac{2}{3} has a solution where 0<x<9 0<x<9 .\newlineIs Lena's justification complete? If not, why?\newlineLena's justification:\newlinef(0)=0 f(0)=0 and f(9)=6 f(9)=6 , so the average rate of change of f f over the interval 0≤x≤9 0 \leq x \leq 9 is 23 \frac{2}{3} .\newlineSo, according to the mean value theorem, f′(x)=23 f^{\prime}(x)=\frac{2}{3} must have a solution somewhere in the interval 0<x<9 0<x<9 .\newlineChoose 11 answer:\newline(A) Yes, Lena's justification is complete.\newline(B) No, Lena didn't establish that the average rate of change of f f over[0,9] [0,9] is equal to 23 \frac{2}{3} .\newline(C) No, Lena didn't establish that f f is differentiable.
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Posted 9 months ago

Question
Let h(x)=3x−x2 h(x)=3^{x}-x^{2} .\newlineBelow is Uriah's attempt to write a formal justification for the fact that the equation h′(x)=8 h^{\prime}(x)=8 has a solution where 1<x<3 1<x<3 .\newlineIs Uriah's justification complete? If not, why?\newlineUriah's justification:\newlinePolynomial and exponential functions are differentiable and continuous at all points in their domain, and [1,3] [1,3] is within h h 's domain. Also, h(1)=2 h(1)=2 and h(3)=18 h(3)=18 , so\newlineh(3)−h(1)3−1=8.  \frac{h(3)-h(1)}{3-1}=8 \text {. } \newlineSo, according to the mean value theorem, h′(x)=8 h^{\prime}(x)=8 must have a solution somewhere between x=1 x=1 and x=3 x=3 .\newlineChoose 11 answer:\newline(A) Yes, Uriah's justification is complete.\newline(B) No, Uriah didn't establish that the average rate of change of h h over [1,3] [1,3] is equal to 88 .\newline(C) No, Uriah didn't establish that h h is differentiable.
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Posted 9 months ago

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