# Recent questions tagged calculus

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Given $y=\sqrt{x^{7}}-\frac{5}{x^{3}}$ Which of the following is equivalent to $y$ ?
Given $y=\sqrt{x^{7}}-\frac{5}{x^{3}}$ Which of the following is equivalent to $y$ ?Given \ y=\sqrt{x^{7}}-\frac{5}{x^{3}} \ Which of the following is equivalent to $y$ ? \begin{align} \begin{array}{|l|l|} \hline \text { A } &amp ...
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Evaluate $\frac{d}{d x} f(x)$ using the rules for differentiation: $f(x)=-\frac{1}{2} x^{4}-2 x^{3}+\frac{1}{2} \sqrt{x^{5}}$
Evaluate $\frac{d}{d x} f(x)$ using the rules for differentiation: $f(x)=-\frac{1}{2} x^{4}-2 x^{3}+\frac{1}{2} \sqrt{x^{5}}$Evaluate $\frac{d}{d x} f(x)$ using the rules for differentiation: \ f(x)=-\frac{1}{2} x^{4}-2 x^{3}+\frac{1}{2} \sqrt{x^{5}} \ ...
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Find $h^{\prime}(x)$ using the rules for differentiation: $h(x)=2 x^{2}+4 x-1$
Find $h^{\prime}(x)$ using the rules for differentiation: $h(x)=2 x^{2}+4 x-1$Find $h^{\prime}(x)$ using the rules for differentiation: \ h(x)=2 x^{2}+4 x-1 \ ...
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Given: $a=\frac{1}{2} y^{3}+5 y+2$ Determine $\frac{d}{d y} a$ using the rules for differentiation.
Given: $a=\frac{1}{2} y^{3}+5 y+2$ Determine $\frac{d}{d y} a$ using the rules for differentiation.Given: \ a=\frac{1}{2} y^{3}+5 y+2 \ Determine $\frac{d}{d y} a$ using the rules for differentiation. ...
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Calculate $h^{\prime}(y)$ for: $h(y)=-8 y^{3}-7 y^{2}+y^{1,5}+3 y^{0,8}$
Calculate $h^{\prime}(y)$ for: $h(y)=-8 y^{3}-7 y^{2}+y^{1,5}+3 y^{0,8}$Calculate $h^{\prime}(y)$ for: \ h(y)=-8 y^{3}-7 y^{2}+y^{1,5}+3 y^{0,8} \ ...
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Find $D_{x} h(x)$ using the rules for differentiation: $h(x)=6 x^{4}-6 x^{3}-5 x^{2}$
Find $D_{x} h(x)$ using the rules for differentiation: $h(x)=6 x^{4}-6 x^{3}-5 x^{2}$Find $D_{x} h(x)$ using the rules for differentiation: \ h(x)=6 x^{4}-6 x^{3}-5 x^{2} \ ...
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Evaluate $g^{\prime}(x)$ using the rules for differentiation: $g(x)=-\frac{1}{2} x^{\frac{7}{2}}+\frac{3}{4} \sqrt[4]{x^{3}}-\frac{4}{3} x^{\frac{1}{2}}$
Evaluate $g^{\prime}(x)$ using the rules for differentiation: $g(x)=-\frac{1}{2} x^{\frac{7}{2}}+\frac{3}{4} \sqrt[4]{x^{3}}-\frac{4}{3} x^{\frac{1}{2}}$Evaluate $g^{\prime}(x)$ using the rules for differentiation: \ g(x)=-\frac{1}{2} x^{\frac{7}{2}}+\frac{3}{4} \sqrt{x^{3}}-\frac{4}{3} x^{\frac{ ...
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Find $\frac{d}{d x} f(x)$ using the rules for differentiation: $f(x)=-3 x^{2}-6 x+6$
Find $\frac{d}{d x} f(x)$ using the rules for differentiation: $f(x)=-3 x^{2}-6 x+6$Find $\frac{d}{d x} f(x)$ using the rules for differentiation: \ f(x)=-3 x^{2}-6 x+6 \ ...
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What are the rules of differentiation?
What are the rules of differentiation?What are the rules of differentiation? ...
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$$\text { Consider the system } A X=B, \text { where } X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \text {. }$$
$$\text { Consider the system } A X=B, \text { where } X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \text {. }$$ \begin{aligned} &amp;\text { Consider the system } A X=B \text {, where } X=\left\begin{array}{l} x \\ y \\ z \end{array}\right \te ...
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If $f$ has an inverse, $f$ is differentiable at $0, f(0)=-2$ and $f^{\prime}(0)=-3$, then
If $f$ has an inverse, $f$ is differentiable at $0, f(0)=-2$ and $f^{\prime}(0)=-3$, thenIf $f$ has an inverse, $f$ is differentiable at $0, f(0)=-2$ and $f^{\prime}(0)=-3$, then &nbsp; A. $\quad f(-2)=0$ B. \$\quad\left(f^{-1}\right)^{\ ...
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State Rolle's Theorem.
State Rolle's Theorem.State Rolle's Theorem. ...
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Is algebra easier than calculus?
Is algebra easier than calculus?Is algebra easier than calculus? ...
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Prove that $x^2+2xy+y^2$ cannot be negative
Prove that $x^2+2xy+y^2$ cannot be negativeProve that &nbsp;$x^2+2xy+y^2$ cannot be negative ...
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State the fundamental theorem of calculus
State the fundamental theorem of calculusState the fundamental theorem of calculus ...
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What is the definition of innovation?
What is the definition of innovation?What is the definition of innovation? ...
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Prove that $\ln x \leq x-1$ for $x>0$
Prove that $\ln x \leq x-1$ for $x>0$Prove that \ \ln x \leq x-1 \ for $x&gt;0$ ...
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Proof: Let $f: A \rightarrow B$ and $g: B \rightarrow C$ be functions. If $g \circ f$ is bijective, then $f$ is injective and $g$ is surjective.
Proof: Let $f: A \rightarrow B$ and $g: B \rightarrow C$ be functions. If $g \circ f$ is bijective, then $f$ is injective and $g$ is surjective.Proof: Let $f: A \rightarrow B$ and $g: B \rightarrow C$ be functions. If $g \circ f$ is bijective, then $f$ is injective and $g$ is surject ...
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Proof: $\frac{n^{2}+5}{n+1} \rightarrow \infty$ as $n \rightarrow \infty$
Proof: $\frac{n^{2}+5}{n+1} \rightarrow \infty$ as $n \rightarrow \infty$Proof: $\frac{n^{2}+5}{n+1} \rightarrow \infty$ as $n \rightarrow \infty$ ...
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If a 12 inch diameter pizza requires 10 ounces of dough, how much dough is needed for a 16 inch pizza?
If a 12 inch diameter pizza requires 10 ounces of dough, how much dough is needed for a 16 inch pizza?If a 12 inch diameter pizza requires 10 ounces of dough, how much dough is needed for a 16 inch pizza? ...
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Solve the proportion $\dfrac{5}{3}=\dfrac{x}{6}$ for the unknown value $x$.
Solve the proportion $\dfrac{5}{3}=\dfrac{x}{6}$ for the unknown value $x$.Solve the proportion $\dfrac{5}{3}=\dfrac{x}{6}$ for the unknown value $x$. ...
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Write each as a percent: a) $\frac{1}{4}$ b) $0.02$ c) $2.35$
Write each as a percent: a) $\frac{1}{4}$ b) $0.02$ c) $2.35$Write each as a percent: a) $\frac{1}{4}$ b) $0.02$ c) $2.35$ ...
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If $\frac{3 \pi}{4}<\alpha<\pi$ then $\sqrt{2 \cot \alpha+\frac{1}{\sin ^{2} \alpha}}$ is equal to
If $\frac{3 \pi}{4}<\alpha<\pi$ then $\sqrt{2 \cot \alpha+\frac{1}{\sin ^{2} \alpha}}$ is equal toIf $\frac{3 \pi}{4}&lt;\alpha&lt;\pi$ then $\sqrt{2 \cot \alpha+\frac{1}{\sin ^{2} \alpha}}$ is equal to ...
The set of all possible values of $\alpha \operatorname{in}[-\pi, \pi] \operatorname{such}$ that $\sqrt{\frac{1-\sin \alpha}{1+\sin \alpha}}$ is equal to $\sec \alpha-\tan \alpha$ isThe set of all possible values of $\alpha \operatorname{in}-\pi, \pi \operatorname{such}$ that $\sqrt{\frac{1-\sin \alpha}{1+\sin \alpha}}$ is e ...
If $\pi / 2<\alpha<\pi$, then the expression $\left|\sqrt{\frac{1-\sin \alpha}{1+\sin \alpha}}+\sqrt{\frac{1+\sin \alpha}{1-\sin \alpha}}\right|$ is equal to
If $\pi / 2<\alpha<\pi$, then the expression $\left|\sqrt{\frac{1-\sin \alpha}{1+\sin \alpha}}+\sqrt{\frac{1+\sin \alpha}{1-\sin \alpha}}\right|$ is equal toIf $\pi / 2&lt;\alpha&lt;\pi$, then the expression \(\left|\sqrt{\frac{1-\sin \alpha}{1+\sin \alpha}}+\sqrt{\frac{1+\sin \alpha}{1-\sin \alpha}}\rig ...