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One-step inequalities: word problems

Choo Kheng must travel at least 288288 kilometers in a submarine in order to reach her destination.\newlineLet \newlineSS represent the number of hours the submarine can travel on the water's surface and \newlineUU represent the number of hours it can travel underwater in order to reach Choo Kheng's destination.\newline36S+64U≥28836S + 64U \geq 288\newlineThe submarine travels for \newline23\frac{2}{3} hours on the water's surface. What is the least number of hours the submarine must travel underwater in order for Choo Kheng to reach her destination?\newlineChoose 11 answer:\newline(A) The submarine must travel for at least 11 hour.\newline(B) The submarine must travel for at least 22 hours.\newline(C) The submarine must travel for at least 33 hours.\newline(D) The submarine must travel for at least 44 hours.
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A grocery store receives a shipment of bananas. The percent of the shipment that is still suitable for sale is decreasing at a rate of r(t) r(t) percent of the original shipment per day (where t t is the time in days). At t=2 t=2 , the grocery store found 85% 85 \% of the shipment to be suitable for sale.\newlineWhat does 0.85+∫24r(t)dt=0.6 mean?  0.85+\int_{2}^{4} r(t) d t=0.6 \text { mean? } \newlineChoose 11 answer:\newline(A) Between the second and fourth day, another 60% 60 \% of the bananas became unsuitable for sale.\newline(B) Between the second and fourth days, 60% 60 \% of the remaining bananas became unsuitable for sale per day.\newline(C) As of the fourth day, 60% 60 \% of the shipment was still suitable for sale.\newline(D) As of the fourth day, 60% 60 \% of the bananas suitable on the second day are still suitable.
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Consider the following problem:\newlineThe total revenue Addison expects from selling rose bouquets in June changes at a rate of r(x)=8−0.5x r(x)=8-0.5 x thousands of dollars (where x x is the price per bouquet in dollars). When x=30 x=30 , the total expected revenue is $2000 \$ 2000 dollars. By how much does the total expected revenue change between a selling price of $30 \$ 30 and $40 \$ 40 ?\newlineWhich expression can we use to solve the problem?\newlineChoose 11 answer:\newline(A) ∫3040r′(x)dx \int_{30}^{40} r^{\prime}(x) d x \newline(B) 2000+∫3040r(x)dx 2000+\int_{30}^{40} r(x) d x \newline(C) 2000+∫3040r′(x)dx 2000+\int_{30}^{40} r^{\prime}(x) d x \newline(D) ∫3040r(x)dx \int_{30}^{40} r(x) d x
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Consider the following problem:\newlineThe total revenue Addison expects from selling rose bouquets in June changes at a rate of r(x)=8−0.5x r(x)=8-0.5 x thousands of dollars (where x x is the price per bouquet in dollars). When x=30 x=30 , the total expected revenue is $2000 \$ 2000 dollars. By how much does the total expected revenue change between a selling price of $30 \$ 30 and $40 \$ 40 ?\newlineWhich expression can we use to solve the problem?\newlineChoose 11 answer:\newline(A) ∫3040r(x)dx \int_{30}^{40} r(x) d x \newline(B) 2000+∫3040r(x)dx 2000+\int_{30}^{40} r(x) d x \newline(C) 2000+∫3040r′(x)dx 2000+\int_{30}^{40} r^{\prime}(x) d x \newline(D) ∫3040r′(x)dx \int_{30}^{40} r^{\prime}(x) d x
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Let g(x)=2−x3+5x g(x)=2^{-x^{3}+5 x} .\newlineBelow is Lila's attempt to write a formal justification for the fact that the equation g′(x)=−12 g^{\prime}(x)=-12 has a solution in the interval (1,2) (1,2) .\newlineIs Lila's justification complete? If not, why?\newlineLila's justification:\newlineExponential and polynomial functions are differentiable and continuous at all points in their domain, and [1,2] [1,2] is within g′s g^{\prime} s domain.\newlineSo, according to the mean value theorem, g′(x)=−12 g^{\prime}(x)=-12 must have a solution somewhere between x=1 x=1 and x=2 x=2 .\newlineChoose 11 answer:\newline(A) Yes, Lila's justification is complete.\newline(B) No, Lila didn't establish that the average rate of change of g g over [1,2] [1,2] is equal to −12-12 .\newline(C) No, Lila didn't establish that g g is differentiable.
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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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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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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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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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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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