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 inequality represents this interval?\newline[−5,1)[-5, 1)\newlineChoices:\newline(A) −5≤x≤1-5 \leq x \leq 1\newline(B) −5≤x<1-5 \leq x < 1\newline(C) −5<x<1-5 < x < 1\newline(D) −5<x≤1-5 < x \leq 1

View step-by-step helpView step-by-step help
Homeright chevron icon
Math Problemsright chevron icon
Algebra 1right chevron icon
Interval notationright chevron icon

Full solution

Q. What inequality represents this interval?\newline[−5,1)[-5, 1)\newlineChoices:\newline(A) −5≤x≤1-5 \leq x \leq 1\newline(B) −5≤x<1-5 \leq x < 1\newline(C) −5<x<1-5 < x < 1\newline(D) −5<x≤1-5 < x \leq 1
  1. Identify Endpoints: Identify the endpoints of the interval and whether they are included in the interval. The interval is [−5,1)[-5, 1), which means −5-5 is included (as indicated by the square bracket) and 11 is not included (as indicated by the parenthesis).
  2. Translate to Inequality: Translate the interval notation into an inequality. Since −5-5 is included, the inequality must use ＂less than or equal to＂ for −5-5. Since 11 is not included, the inequality must use ＂less than＂ for 11.
  3. Write Correct Inequality: Write the inequality using the appropriate symbols. The correct inequality is −5≤x<1-5 \leq x < 1, which means xx is greater than or equal to −5-5 and less than 11.

More problems from Interval notation

Question
z=−24−12.7i z=-24-12.7 i \newlineWhat are the real and imaginary parts of z z ?\newlineChoose 11 answer:\newline(A)\newlineRe⁡(z)=−24 and Im⁡(z)=−12.7 \begin{array}{l} \operatorname{Re}(z)=-24 \text { and } \\ \operatorname{Im}(z)=-12.7 \end{array} \newline(B)\newlineRe⁡(z)=−24 and Im⁡(z)=−12.7i \begin{array}{l} \operatorname{Re}(z)=-24 \text { and } \\ \operatorname{Im}(z)=-12.7 i \end{array} \newline(C)\newlineRe⁡(z)=−12.7 and Im⁡(z)=−24 \begin{array}{l} \operatorname{Re}(z)=-12.7 \text { and } \\ \operatorname{Im}(z)=-24 \end{array} \newline(D)\newlineRe⁡(z)=−12.7i and Im⁡(z)=−24 \begin{array}{l} \operatorname{Re}(z)=-12.7 i \text { and } \\ \operatorname{Im}(z)=-24 \end{array}
Get tutor helpright-arrow

Posted 4 months ago

Question
z=77−27.8i z=77-27.8 i \newlineWhat are the real and imaginary parts of z z ?\newlineChoose 11 answer:\newline(A)\newlineRe⁡(z)=77 and Im⁡(z)=−27.8 \begin{array}{l} \operatorname{Re}(z)=77 \text { and } \\ \operatorname{Im}(z)=-27.8 \end{array} \newline(B)\newlineRe⁡(z)=−27.8i and Im⁡(z)=77 \begin{array}{l} \operatorname{Re}(z)=-27.8 i \text { and } \\ \operatorname{Im}(z)=77 \end{array} \newline(C)\newlineRe⁡(z)=−27.8 and Im⁡(z)=77 \begin{array}{l} \operatorname{Re}(z)=-27.8 \text { and } \\ \operatorname{Im}(z)=77 \end{array} \newline(D)\newlineRe⁡(z)=77 and Im⁡(z)=−27.8i \begin{array}{l} \operatorname{Re}(z)=77 \text { and } \\ \operatorname{Im}(z)=-27.8 i \end{array}
Get tutor helpright-arrow

Posted 4 months ago

Question
z=−16i−92.3 z=-16 i-92.3 \newlineWhat are the real and imaginary parts of z z ?\newlineChoose 11 answer:\newline(A)\newlineRe⁡(z)=−92.3 and Im⁡(z)=−16i \begin{array}{l} \operatorname{Re}(z)=-92.3 \text { and } \\ \operatorname{Im}(z)=-16 i \end{array} \newline(B)\newlineRe⁡(z)=−92.3 and Im⁡(z)=−16 \begin{array}{l} \operatorname{Re}(z)=-92.3 \text { and } \\ \operatorname{Im}(z)=-16 \end{array} \newline(C)\newlineRe⁡(z)=−16i and Im⁡(z)=−92.3 \begin{array}{l} \operatorname{Re}(z)=-16 i \text { and } \\ \operatorname{Im}(z)=-92.3 \end{array} \newline(D)\newlineRe⁡(z)=−16 and Im⁡(z)=−92.3 \begin{array}{l} \operatorname{Re}(z)=-16 \text { and } \\ \operatorname{Im}(z)=-92.3 \end{array}
Get tutor helpright-arrow

Posted 4 months ago

Question
x2+8x+6=0 x^{2}+8 x+6=0 \newlineWhat are the solutions to the given equation?\newlineChoose 11 answer:\newline(A)\newlinex=−8−10 and x=−8+10 \begin{array}{l} x=-8-\sqrt{10} \text { and } \\ x=-8+\sqrt{10} \end{array} \newline(B)\newlinex=−8−7 and x=−8+7 \begin{array}{l} x=-8-\sqrt{7} \text { and } \\ x=-8+\sqrt{7} \end{array} \newline(C)\newlinex=−4−10 and x=−4+10 \begin{array}{l} x=-4-\sqrt{10} \text { and } \\ x=-4+\sqrt{10} \end{array} \newline(D)\newlinex=−4−7 and x=−4+7 \begin{array}{l} x=-4-\sqrt{7} \text { and } \\ x=-4+\sqrt{7} \end{array}
Get tutor helpright-arrow

Posted 4 months ago

Question
If x2+6x−5=0 x^{2}+6 x-5=0 , what are the values of x x ?\newlineChoose 11 answer:\newline(A) −3−14 -3-\sqrt{14} and −3+14 -3+\sqrt{14} \newline(B) 3−14 3-\sqrt{14} and 3+14 3+\sqrt{14} \newline(C) −1-1 and −5-5\newline(D) 11 and 55
Get tutor helpright-arrow

Posted 4 months ago

Question
Consider the expression 4(8x+5) 4(8 x+5) .\newlineComplete 22 descriptions of the parts of the expression.\newline11. The entire expression is the product of □\square. \newline22. On its own, (8x+5) (8 x+5) is a sum with □\square.
Get tutor helpright-arrow

Posted 3 months ago

Question
Evaluate −a+(−b) -a+(-b) where a=6.05 a=6.05 and b=3.611 b=3.611 .
Get tutor helpright-arrow

Posted 3 months ago

Question
E=[−1−2−2−1] E=\left[\begin{array}{ll}-1 & -2 \\ -2 & -1\end{array}\right] and A=[202−1]. A=\left[\begin{array}{rr} 2 & 0 \\ 2 & -1 \end{array}\right] . \newlineLet H=EA \mathrm{H}=\mathrm{EA} . Find H \mathrm{H} .\newlineH= \mathbf{H}=
Get tutor helpright-arrow

Posted 3 months ago

Question
B=[5−2−15] \mathbf{B}=\left[\begin{array}{rr}5 & -2 \\ -1 & 5\end{array}\right] and A=[2002] \mathrm{A}=\left[\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right] .\newlineLet H=BA \mathrm{H}=\mathrm{BA} . Find H \mathrm{H} .\newlineH= \mathbf{H}=
Get tutor helpright-arrow

Posted 3 months ago

Question
B=[0−12] \mathrm{B}=\left[\begin{array}{r}0 \\ -1 \\ 2\end{array}\right] and A=[12] \mathrm{A}=\left[\begin{array}{ll}1 & 2\end{array}\right] \newlineLet H=BA \mathrm{H}=\mathrm{BA} . Find H \mathrm{H} .\newlineH= \mathbf{H}=
Get tutor helpright-arrow

Posted 3 months ago

View all (8)