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Solve for $x$ : $\newline$ $\begin{array}{l} \frac{x}{-9}=3 \\ x=\square \end{array}$

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Math Problems

Grade 8

Solve multi-step inequalities

Full solution

Q. Solve for

x

\newline

\begin{array}{l} \frac{x}{-9}=3 \\ x=\square \end{array}

Multiply by $-9$ : Multiply both sides by $-9$ to isolate $x$ . $\frac{x}{-9} \times (-9) = 3 \times (-9)$
Simplify equation: Simplify the equation. $x = -27$

More problems from Solve multi-step inequalities

Question

-11 x \geq-33

\newline

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

\newline

Choose

1

answer:

\newline

(A)

x \leq-3

\newline

(B)

x \geq-3

\newline

(C)

x \leq 3

\newline

(D)

x \geq 3

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Posted 4 months ago

Question

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

\newline

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

\newline

Choose

1

answer:

\newline

(A)

q \leq 3

\newline

(B)

q \geq 3

\newline

(C)

q \leq 6

\newline

(D)

q \geq 6

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Posted 4 months ago

Question

Let

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

\newline

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

\newline

Sean's justification:

\newline

g

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

\newline

So, according to the intermediate value theorem,

g(x)=0

must have a solution somewhere between

x=1

and

x=4

\newline

Choose

1

answer:

\newline

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

\newline

(B) No, Sean didn't establish that

0

is between

g(1)

and

g(4)

\newline

g

is continuous.

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Posted 3 months ago

Question

Let

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}

\newline

So, according to the intermediate value theorem,

f(x)=0.1

must have a solution when

x

is between

x=1.5

and

x=2

\newline

Choose

1

answer:

\newline

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

\newline

(B) No, Issac didn't establish that

0

1

is between

f(1.5)

and

f(2)

\newline

f

is continuous.

Get tutor help

Posted 3 months ago

Question

Let

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.

\newline

Choose

1

answer:

\newline

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

\newline

(B) No, Bridget didn't establish that

2

is between

h(0)

and

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

\newline

h

is continuous.

Get tutor help

Posted 3 months ago

Question

Let

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

\newline

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

and

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

, so

10

is between

f(-1)

and

f(4)

\newline

So, according to the intermediate value theorem,

f(x)=10

must have a solution at some point between

x=-1

and

x=4

\newline

Choose

1

answer:

\newline

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

\newline

(B) No, Ibrahim didn't establish that

10

is between

f(-1)

and

f(4)

\newline

f

is continuous.

Get tutor help

Posted 3 months ago

Question

Let

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

\newline

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

\newline

Is Amrita's justification complete?

\newline

If not, why?

\newline

Amrita's justification:

\newline

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

[-1,4]

is within

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

and

x=4

\newline

Choose

1

answer:

\newline

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

\newline

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

g

over

[-1,4]

is equal to

0

4

\newline

g

is differentiable.

Get tutor help

Posted 3 months ago

Question

Let

\newline

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

\newline

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

c

where

\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

such that

h^{\prime}(c)=6

\newline

Choose

1

answer:

\newline

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

\newline

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

h

over

[-2,2]

is equal to

6

\newline

h

is differentiable.

Get tutor help

Posted 3 months ago

Question

Let

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

and

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

Choose

1

answer:

\newline

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

\newline

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

f

over

[0,9]

is equal to

\frac{2}{3}

\newline

f

is differentiable.

Get tutor help

Posted 3 months ago

Question

Let

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]

is within

h

's domain. Also,

h(1)=2

and

h(3)=18

, so

\newline

\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

and

x=3

\newline

Choose

1

answer:

\newline

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

\newline

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

h

over

[1,3]

is equal to

8

\newline

h

is differentiable.

Get tutor help

Posted 3 months ago

View all (7)

Solve for x x x :\newlinex−9=3x=□ \begin{array}{l} \frac{x}{-9}=3 \\ x=\square \end{array} −9x​=3x=□​

Solve for $x$ : $\newline$ $\begin{array}{l} \frac{x}{-9}=3 \\ x=\square \end{array}$