Four more than the quotient of 30 and a number

An inequality has a less than, greater than, less than or equal to, and greater than or equal to. <u>But, there are no equal signs allowed in a inequality. That makes it an expression.</u>

Answer: Option B, 1/4 (x - 5) = 13

7 0

3 years ago

In a basketball game, River scored 15% more points than Paul. If r is the number of points that River scored and p is the number

hoa [83]

Answer:

$r = \dfrac{23}{20} p$

Step-by-step explanation:

Number of points scored by River = $r$

Number of points scored by Paul = $p$

River has scored 15% more points that Paul.

To find:

Equation to represent the given condition.

Solution:

As per the given question statement, Number of points scored by River is 15% more than that of Paul's.

15% of points scored by Paul = 15% of $p$ = $\frac{15}{100}p$

15% more means total points scored by Paul plus 15% of points scored by Paul.

Therefore, we can write the following:

$r = p + 15\%\ of\ p\\\Rightarrow r = p + \dfrac{15}{100} p\\\Rightarrow r = \dfrac{115}{100} p\\\Rightarrow r = \dfrac{23}{20} p$

4 0

3 years ago

Centerville is the headquarters of Greedy Cablevision Inc. The cable company is about to expand service to two nearby towns, Spr

NeX [460]

Answer:

20.2057 Units.

Step-by-step explanation:

First, we determine the length of the cable.

Distance between Centerville (8,0) and point (x,0) is given as:

$\sqrt{(8-x)}^2=8-x$

Distance between point (x,0) and Springfield(0,7) is:

$\sqrt{(7-0)^2+(0-x)^{2}}=\sqrt{(7)^2+(x)^{2}}$

Distance between point (x,0) and Shelvyfield(0,-7) is:

$\sqrt{(-7-0)^2+(0-x)^{2}}=\sqrt{(7)^2+(x)^{2}}$

Therefore the Length of the Cable L(x)

$L(x)=(8-x)+2\sqrt{(7)^2+(x)^{2}}$

To find the critical point, we set the derivative of L(x)=0

$L^{'}(x)=-1+\frac{2x}{\sqrt{\left( 49 - x^{2}\right) }}$

$\frac{-\sqrt{\left( 49 - x^{2}\right)}+2x}{\sqrt{\left( 49 - x^{2}\right)}}=0\\-\sqrt{\left( 49 - x^{2}\right)}+2x=0\\\sqrt{\left( 49 - x^{2}\right)}=2x\\(\sqrt{\left( 49 - x^{2}\right)})^2=(2x)^2\\49 - x^{2}=4x^2\\49=5x^2\\x^2=\frac{49}{5}\\x= 3.1305$

To verify that L(x) has a minimum at this critical number we compute the second derivative L″(x) and find its value at the critical number.

$L^{''}(x)=\frac{\left( 98\right) \,\sqrt{\left( 49 - x^{2}\right) }}{{\left( 7 - x\right) }^{2}\,{\left( 7+x\right) }^{2}}\\At \:x=3.1305, L^{''}=0.3993$

Since $L^{''}(x)$ is positive, the minimum point of L(x) exists.

Next, we find the minimum length by substituting z=3.1305 into L(x)

$L(3.1305)=(8-3.1305)+2\sqrt{(7)^2+(3.1305)^{2}}$

Minimum Length, L=20.2057

The minimum length of the cable is 20.2057 Units.

6 0

2 years ago