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Solve for $x$ . $2x +10 = 22$

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Solve two-step inequalities

Full solution

Q. Solve for

x

.

2x +10 = 22

Subtract $10$ : Subtract $10$ from both sides of the equation to isolate the term with the variable $x$ . $\newline$ $2x + 10 - 10 = 22 - 10$ $\newline$ This simplifies to: $\newline$ $2x = 12$
Isolate variable $x$ : Divide both sides of the equation by $2$ to solve for $x$ . $\frac{2x}{2} = \frac{12}{2}$ This simplifies to: $x = 6$

More problems from Solve two-step inequalities

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-11 x \geq-33

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Which of the following best describes the solutions to the inequality shown?

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1

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x \leq-3

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x \geq-3

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x \leq 3

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x \geq 3

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Question

\frac{3}{2} q \leq \frac{9}{2} q-18

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Which of the following best describes the solutions to the inequality shown?

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q \leq 3

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q \geq 3

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q \leq 6

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q \geq 6

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Question

g(x)=5-e^{x}

\newline

Below is Sean's attempt to write a formal justification for the fact that the equation

g(x)=0

has a solution where

1 \leq x \leq 4

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Is Sean's justification complete? If not, why?

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Sean's justification:

\newline

g

is defined for all real numbers, and exponential functions are continuous at all points in their domains.

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So, according to the intermediate value theorem,

g(x)=0

must have a solution somewhere between

x=1

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(A) Yes, Sean's justification is complete.

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(B) No, Sean didn't establish that

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g(4)

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(C) No, Sean didn't establish that

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Posted 9 months ago

Question

f(x)=\sqrt[3]{\cos (\pi \cdot x)}

\newline

Below is Issac's attempt to write a formal justification for the fact that the equation

f(x)=0.1

has a solution where

1.5 \leq x \leq 2

\newline

Is Issac's justification complete? If not, why?

\newline

Issac's justification:

\newline

\begin{array}{l} f(1.5)=0 \text { and } \\ f(2)=1 \text {, so } \\ f(1.5)<0.1<f(2) . \end{array}

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So, according to the intermediate value theorem,

f(x)=0.1

must have a solution when

x

x=1.5

x=2

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1

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(A) Yes, Issac's justification is complete.

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(B) No, Issac didn't establish that

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f(1.5)

f(2)

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(C) No, Issac didn't establish that

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Posted 9 months ago

Question

h(x)=5 \cdot \tan (x)

\newline

Below is Bridget's attempt to write a formal justification for the fact that the equation

h(x)=2

has a solution where

0 \leq x \leq \frac{\pi}{4}

\newline

Is Bridget's justification complete? If not, why?

\newline

Bridget's justification:

\newline

h

is defined over the entire interval

\left[0, \frac{\pi}{4}\right]

, and trigonometric functions are continuous at all points in their domains.

\newline

So, according to the intermediate value theorem,

h(x)=2

must have a solution at some point in that interval.

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1

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(A) Yes, Bridget's justification is complete.

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(B) No, Bridget didn't establish that

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h(0)

h\left(\frac{\pi}{4}\right)

\newline

(C) No, Bridget didn't establish that

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Posted 9 months ago

Question

f(x)=2^{5-2 x}

\newline

Below is Ibrahim's attempt to write a formal justification for the fact that the equation

f(x)=10

has a solution where

-1 \leq x \leq 4

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Is Ibrahim's justification complete? If not, why?

\newline

Ibrahim's justification:

\newline

f

is defined for all real numbers, and exponential functions are continuous at all points in their domains. Also,

f(-1)=128

f(4)=\frac{1}{8}

10

f(-1)

f(4)

\newline

So, according to the intermediate value theorem,

f(x)=10

must have a solution at some point between

x=-1

x=4

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1

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(A) Yes, Ibrahim's justification is complete.

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(B) No, Ibrahim didn't establish that

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(c) No, Ibrahim didn't establish that

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Question

g(x)=\cos \left(\pi x^{2}\right)

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Below is Amrita's attempt to write a formal justification for the fact that the equation

g^{\prime}(x)=0.4

has a solution where

-1<x<4

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Is Amrita's justification complete?

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\newline

Amrita's justification:

\newline

Polynomial and trigonometric functions are differentiable and continuous at all points in their domain, and

[-1,4]

g^{\prime}

s domain. Also,

\newline

\begin{array}{l} g(-1)=-1 \text { and } \\ g(4)=1 \text {, so } \\ \frac{g(4)-g(-1)}{4-(-1)}=0.4 . \end{array}

\newline

So, according to the mean value theorem,

g^{\prime}(x)=0.4

must have a solution somewhere between

x=-1

x=4

\newline

1

\newline

(A) Yes, Amrita's justification is complete.

\newline

(B) No, Amrita didn't establish that the average rate of change of

g

[-1,4]

0

4

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(C) No, Amrita didn't establish that

g

is differentiable.

Posted 9 months ago

Question

\newline

h(x)=x^{5}-2 x^{4}+10 x^{2}-10 x \text {. }

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Below is Omar's attempt to write a formal justification for the fact that there exists a value

c

\newline

-2<c<2 \text { such that } h^{\prime}(c)=6 \text {. }

\newline

Is Omar's justification complete? If not, why?

\newline

Omar's justification:

\newline

\begin{array}{l} h(-2)=-4 \text { and } \\ h(2)=20 \text {, so } \\ \frac{h(2)-h(-2)}{2-(-2)}=6 . \end{array}

\newline

So, according to the mean value theorem, there exists a value

c

somewhere in the interval

-2<c<2

h^{\prime}(c)=6

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1

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(A) Yes, Omar's justification is complete.

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(B) No, Omar didn't establish that the average rate of change of

h

[-2,2]

6

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(C) No, Omar didn't establish that

h

is differentiable.

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Question

f(x)=\sqrt{x+\sqrt{x^{3}}}

\newline

Below is Lena's attempt to write a formal justification for the fact that the equation

f^{\prime}(x)=\frac{2}{3}

has a solution where

0<x<9

\newline

Is Lena's justification complete? If not, why?

\newline

Lena's justification:

\newline

f(0)=0

f(9)=6

, so the average rate of change of

f

over the interval

0 \leq x \leq 9

\frac{2}{3}

\newline

So, according to the mean value theorem,

f^{\prime}(x)=\frac{2}{3}

must have a solution somewhere in the interval

0<x<9

\newline

1

\newline

(A) Yes, Lena's justification is complete.

\newline

(B) No, Lena didn't establish that the average rate of change of

f

[0,9]

\frac{2}{3}

\newline

(C) No, Lena didn't establish that

f

is differentiable.

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Question

h(x)=3^{x}-x^{2}

\newline

Below is Uriah's attempt to write a formal justification for the fact that the equation

h^{\prime}(x)=8

has a solution where

1<x<3

\newline

Is Uriah's justification complete? If not, why?

\newline

Uriah's justification:

\newline

Polynomial and exponential functions are differentiable and continuous at all points in their domain, and

[1,3]

h

's domain. Also,

h(1)=2

h(3)=18

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\frac{h(3)-h(1)}{3-1}=8 \text {. }

\newline

So, according to the mean value theorem,

h^{\prime}(x)=8

must have a solution somewhere between

x=1

x=3

\newline

1

\newline

(A) Yes, Uriah's justification is complete.

\newline

(B) No, Uriah didn't establish that the average rate of change of

h

[1,3]

8

\newline

(C) No, Uriah didn't establish that

h

is differentiable.

Posted 9 months ago

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