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3 years ago

One less than 6 times a number is the same as the number plus 14. find the equation x=

Mathematics

1 answer:

Nataly_w [17]3 years ago

3 0

1-6x=x+14 so that means x=14/-5 i think

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A professor determined the relationship between the time spent studying (in hours) and performance on an exam.

kykrilka [37]

Step 1: Insert 2.6 hours for the variable of time and solve.
70.443 + 4.885(2.6)
70.443+ 12.701
= 83.144
This is Ann's EXPECTED exam score. Because of the concert, Ann scored 16 points lower than this number.
Step 2: Subtract 16 from Ann's expected score.
83.144 - 16
= 67.144

3 0

3 years ago

help it is due today and not links or files of downloads of anything like that just type in your answer plz and thank u :)

pantera1 [17]

The answer should be x=4

4 0

3 years ago

Read 2 more answers

A right circular cone is undergoing a transformation in such a way that the radius of the cone is increasing at a rate of 1/2 in

Ivenika [448]

Answer:

The volume is decreasing at the rate of 1.396 cubic inches per minute

Step-by-step explanation:

Given

Shape: Cone

$\frac{dr}{dt} =\frac{1}{2}$ --- rate of the radius

$\frac{dh}{dt} =-\frac{1}{3}$ --- rate of the height

$r = 2$

$h = \frac{1}{3}$

Required

Determine the rate of change of the cone volume

The volume of a cone is:

$V = \frac{\pi}{3}r^2h$

Differentiate with respect to time (t)

$\frac{dV}{dt} = \frac{\pi}{3}(2rh \frac{dr}{dt} + r^2 \frac{dh}{dt})$

Substitute values for the known variables

$\frac{dV}{dt} = \frac{\pi}{3}(2*2*\frac{1}{3}* \frac{1}{2} - 2^2 *\frac{1}{3})$

$\frac{dV}{dt} = \frac{\pi}{3}(\frac{4}{3}* \frac{1}{2} - \frac{4}{3})$

$\frac{dV}{dt} = \frac{\pi}{3}(\frac{4}{3}(\frac{1}{2} - 1))$

$\frac{dV}{dt} = \frac{\pi}{3}(\frac{4}{3}*- 1)$

$\frac{dV}{dt} = -\frac{\pi}{3}*\frac{4}{3}$

$\frac{dV}{dt} = -\frac{22}{7*3}*\frac{4}{3}$

$\frac{dV}{dt} = -\frac{22}{21}*\frac{4}{3}$

$\frac{dV}{dt} = -\frac{88}{63}$

$\frac{dV}{dt} =-1.396in^3/min$

The volume is decreasing at the rate of 1.396 cubic inches per minute

3 0

3 years ago

A particular isotope has a half-life of 74 days. If you start with 1 kilogram of this isotope, how much will remain after 150

shtirl [24]

$\bf \stackrel{150~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &150\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{150}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{75}{37}}\implies \boxed{A\approx 0.24536}$

$\bf \stackrel{300~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &300\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{300}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{150}{37}}\implies \boxed{A\approx 0.060202}$

6 0

3 years ago

Stephanie took advantage of the interest-only financing from Rooms-To-Go and borrowed $2,000 with a 15 percent annual interest r

AnnyKZ [126]

Answer: $200

Step-by-step explanation

Principal = $2000

Rate = 15%

Time = 8 months = 8/12 = 2/3 years

Interest = Principal × Rate × Time

= 2000 × 15% × 2/3

= 2000 × 15/100 × 2/3

= $200

The interest is $200

8 0

3 years ago