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:

What is the average value of h(x)=(x^(3)+7x^(2)+3)/(2x^(2)-5x+4) on the interval [-8,-2] ? Use a graphing calculator and round your answer to three decimal places.

What is the average value of h(x)=x3+7x2+32x2−5x+4 h(x)=\frac{x^{3}+7 x^{2}+3}{2 x^{2}-5 x+4} on the interval [−8,−2] [-8,-2] ?\newlineUse a graphing calculator and round your answer to three decimal places.

View step-by-step helpView step-by-step help
Homeright chevron icon
Math Problemsright chevron icon
Algebra 2right chevron icon
Average rate of changeright chevron icon

Full solution

Q. What is the average value of h(x)=x3+7x2+32x2−5x+4 h(x)=\frac{x^{3}+7 x^{2}+3}{2 x^{2}-5 x+4} on the interval [−8,−2] [-8,-2] ?\newlineUse a graphing calculator and round your answer to three decimal places.
  1. Calculate h(−8)h(-8): Calculate h(−8)h(-8) using the graphing calculator.\newlineh(−8)=((−8)3+7∗(−8)2+3)/(2∗(−8)2−5∗(−8)+4)h(-8) = ((-8)^3 + 7*(-8)^2 + 3) / (2*(-8)^2 - 5*(-8) + 4)\newlineh(−8)=((−512)+7∗(64)+3)/(2∗(64)−5∗(−8)+4)h(-8) = ((-512) + 7*(64) + 3) / (2*(64) - 5*(-8) + 4)\newlineh(−8)=((−512)+448+3)/(128+40+4)h(-8) = ((-512) + 448 + 3) / (128 + 40 + 4)\newlineh(−8)=(−61)/(172)h(-8) = (-61) / (172)\newlineh(−8)≈−0.355h(-8) \approx -0.355 (rounded to three decimal places)
  2. Calculate h(−2)h(-2): Calculate h(−2)h(-2) using the graphing calculator.\newlineh(−2)=((−2)3+7∗(−2)2+3)/(2∗(−2)2−5∗(−2)+4)h(-2) = ((-2)^3 + 7*(-2)^2 + 3) / (2*(-2)^2 - 5*(-2) + 4)\newlineh(−2)=((−8)+7∗(4)+3)/(2∗(4)−5∗(−2)+4)h(-2) = ((-8) + 7*(4) + 3) / (2*(4) - 5*(-2) + 4)\newlineh(−2)=(−8+28+3)/(8+10+4)h(-2) = (-8 + 28 + 3) / (8 + 10 + 4)\newlineh(−2)=23/22h(-2) = 23 / 22\newlineh(−2)≈1.045h(-2) ≈ 1.045 (rounded to three decimal places)
  3. Find average value: Find the average value of h(x)h(x) over the interval [−8,−2][-8,-2].\newlineAverage value = h(−2)−h(−8)−2−(−8)\frac{h(-2) - h(-8)}{-2 - (-8)}\newlineAverage value = 1.045−(−0.355)6\frac{1.045 - (-0.355)}{6}\newlineAverage value = 1.045+0.3556\frac{1.045 + 0.355}{6}\newlineAverage value = 1.46\frac{1.4}{6}\newlineAverage value ≈0.233\approx 0.233 (rounded to three decimal places)

More problems from Average rate of change

Question
We want to factor the following expression:\newlinex4+9 x^{4}+9 \newlineWhich pattern can we use to factor the expression?\newlineU U and V V are either constant integers or single-variable expressions.\newlineChoose 11 answer:\newline(A) (U+V)2 (U+V)^{2} or (U−V)2 (U-V)^{2} \newline(B) (U+V)(U−V) (U+V)(U-V) \newline(C) We can't use any of the patterns.
Get tutor helpright-arrow

Posted 4 months ago

Question
z=−19+3.14i z=-19+3.14 i \newlineWhat are the real and imaginary parts of z z ?\newlineChoose 11 answer:\newline(A)\newlineRe⁡(z)=−19 and Im⁡(z)=3.14 \begin{array}{l} \operatorname{Re}(z)=-19 \text { and } \\ \operatorname{Im}(z)=3.14 \end{array} \newline(B)\newlineRe⁡(z)=3.14i and Im⁡(z)=−19 \begin{array}{l} \operatorname{Re}(z)=3.14 i \text { and } \\ \operatorname{Im}(z)=-19 \end{array} \newline(c)\newlineRe⁡(z)=−19 and Im⁡(z)=3.14i \begin{array}{l} \operatorname{Re}(z)=-19 \text { and } \\ \operatorname{Im}(z)=3.14 i \end{array} \newline(D)\newlineRe⁡(z)=3.14 and Im⁡(z)=−19 \begin{array}{l} \operatorname{Re}(z)=3.14 \text { and } \\ \operatorname{Im}(z)=-19 \end{array}
Get tutor helpright-arrow

Posted 4 months ago

Question
z=−60−12iRe⁡(z)=Im⁡(z)= \begin{array}{l}z=-60-12 i \\ \operatorname{Re}(z)= \\ \operatorname{Im}(z)=\end{array}
Get tutor helpright-arrow

Posted 4 months ago

Question
z=−45i−15.5 z=-45 i-15.5 \newlineWhat are the real and imaginary parts of z z ?\newlineChoose 11 answer:\newline(A)\newlineRe⁡(z)=−45 and Im⁡(z)=−15.5 \begin{array}{l} \operatorname{Re}(z)=-45 \text { and } \\ \operatorname{Im}(z)=-15.5 \end{array} \newline(B)\newlineRe⁡(z)=−15.5 and Im⁡(z)=−45 \begin{array}{l} \operatorname{Re}(z)=-15.5 \text { and } \\ \operatorname{Im}(z)=-45 \end{array} \newline(C)\newlineRe⁡(z)=−45i and Im⁡(z)=−15.5 \begin{array}{l} \operatorname{Re}(z)=-45 i \text { and } \\ \operatorname{Im}(z)=-15.5 \end{array} \newline(D)\newlineRe⁡(z)=−15.5 and Im⁡(z)=−45i \begin{array}{l} \operatorname{Re}(z)=-15.5 \text { and } \\ \operatorname{Im}(z)=-45 i \end{array}
Get tutor helpright-arrow

Posted 4 months ago

Question
z=−12i+11 z=-12 i+11 \newlineWhat are the real and imaginary parts of z z ?\newlineChoose 11 answer:\newline(A)\newlineRe⁡(z)=11 and Im⁡(z)=−12 \begin{array}{l} \operatorname{Re}(z)=11 \text { and } \\ \operatorname{Im}(z)=-12 \end{array} \newline(B)\newlineRe⁡(z)=11 and Im⁡(z)=−12i \begin{array}{l} \operatorname{Re}(z)=11 \text { and } \\ \operatorname{Im}(z)=-12 i \end{array} \newline(C)\newlineRe⁡(z)=−12i and Im⁡(z)=11 \begin{array}{l} \operatorname{Re}(z)=-12 i \text { and } \\ \operatorname{Im}(z)=11 \end{array} \newline(D)\newlineRe⁡(z)=−12 and Im⁡(z)=11 \begin{array}{l} \operatorname{Re}(z)=-12 \text { and } \\ \operatorname{Im}(z)=11 \end{array}
Get tutor helpright-arrow

Posted 4 months ago

Question
(18+4i)+(−11+23i)= (18+4 i)+(-11+23 i)= \newlineExpress your answer in the form (a+bi) (a+b i) .
Get tutor helpright-arrow

Posted 4 months ago

Question
(14+60i)+(−30+2i)= (14+60 i)+(-30+2 i)= \newlineExpress your answer in the form (a+bi) (a+b i) .
Get tutor helpright-arrow

Posted 4 months ago

Question
(3−91i)+(67)= (3-91 i)+(67)= \newlineExpress your answer in the form (a+bi) (a+b i) .
Get tutor helpright-arrow

Posted 4 months ago

Question
(−71+2i)+(88−12i)= (-71+2 i)+(88-12 i)= \newlineExpress your answer in the form (a+bi) (a+b i) .
Get tutor helpright-arrow

Posted 4 months ago

Question
(−33−2i)−(50+9i)= (-33-2 i)-(50+9 i)= \newlineExpress your answer in the form (a+bi) (a+b i) .
Get tutor helpright-arrow

Posted 4 months ago

View all (3)