Resourcesdown chevron
Testimonials
Plans
Sign in
Sign up
Resourcesright arrow
Testimonials
Plans
Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

IINDING A SCALE IACTOK Vour vioually impeined friend asked you to enlarge your nomes from chane wo he can study. You touk notes on 8.5 inch by llawes paper. The cnlarged copy has a smaller wide wide a length of 10 inches. What is the sale facter of ains dilation? (See Example 5.)

FINDING A SCALE FACTOR Your visually impaired friend asked you to enlarge your notes from class so he can study. You took notes on 8.58.5 inch by 1111 inches paper. The enlarged copy has a smaller width of 1010 inches. What is the scale factor of this dilation? (See Example 55.)

View step-by-step helpView step-by-step help
Homeright chevron icon
Math Problemsright chevron icon
Grade 6right chevron icon
One-step inequalities: word problemsright chevron icon

Full solution

Q. FINDING A SCALE FACTOR Your visually impaired friend asked you to enlarge your notes from class so he can study. You took notes on 8.58.5 inch by 1111 inches paper. The enlarged copy has a smaller width of 1010 inches. What is the scale factor of this dilation? (See Example 55.)
  1. Identify Widths: Step 11: Identify the original width and the enlarged width.\newlineOriginal width = 8.58.5 inches.\newlineEnlarged width = 1010 inches.
  2. Calculate Scale Factor: Step 22: Calculate the scale factor by dividing the enlarged width by the original width.\newlineScale factor = Enlarged width / Original width = 10 inches8.5 inches=1.1764705882352942\frac{10 \text{ inches}}{8.5 \text{ inches}} = 1.1764705882352942.

More problems from One-step inequalities: word problems

Question
−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
Get tutor helpright-arrow

Posted 3 months ago

Question
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
Get tutor helpright-arrow

Posted 3 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.
Get tutor helpright-arrow

Posted 3 months ago

Question
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.
Get tutor helpright-arrow

Posted 3 months ago

Question
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.
Get tutor helpright-arrow

Posted 3 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.
Get tutor helpright-arrow

Posted 3 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.
Get tutor helpright-arrow

Posted 3 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.
Get tutor helpright-arrow

Posted 3 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.
Get tutor helpright-arrow

Posted 3 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.
Get tutor helpright-arrow

Posted 3 months ago

View all (34)