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The Jacksons and the Simpsons were competing in the final leg of the Amazing Race, which was 240 kilometers long.
In their race to the finish, the Jacksons immediately took off traveling at a average speed of
v kilometers per hour. The Simpsons' start was delayed by an hour. When they eventually took off, they traveled at an average speed that was 40 kilometers per hour faster than the Jacksons' speed. Sadly for them, that didn't help, and the Jacksons won.
Write an inequality in terms of
v that models the situation.
◻

The Jacksons and the Simpsons were competing in the final leg of the Amazing Race, which was $240$ kilometers long. $\newline$ In their race to the finish, the Jacksons immediately took off traveling at a average speed of $v$ kilometers per hour. The Simpsons' start was delayed by an hour. When they eventually took off, they traveled at an average speed that was $40$ kilometers per hour faster than the Jacksons' speed. Sadly for them, that didn't help, and the Jacksons won. $\newline$ Write an inequality in terms of $v$ that models the situation. $\square$

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Math Problems

Grade 6

One-step inequalities: word problems

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More problems from One-step inequalities: word problems

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 10 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 10 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 9 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.

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Posted 9 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.

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Posted 9 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.

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Posted 9 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.

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Posted 9 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.

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Posted 9 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.

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Posted 9 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.

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Posted 9 months ago

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