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2 years ago

Solve for k: 31 - 3k = 10

Mathematics

2 answers:

mash [69]2 years ago

8 0

Answer:

k=-7

Step-by-step explanation:

31-3k=10

31-31=0 10-31=-21

-3k=-21 -3/3=0

-21/3=-7

k=-7

Crazy boy [7]2 years ago

5 0

Answer:

k = 7

Step-by-step explanation:

Step 1: Write equation

31 - 3k = 10

Step 2: Solve for <em>k</em>

Subtract 31 on both sides: -3k = -21
Divide both sides by -3: k = 7

Step 3: Check

<em>Plug in k to verify it's a solution.</em>

31 - 3(7) = 10

31 - 21 = 10

10 = 10

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Just a shot in the dark: 9*30 and 9*8?

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2 years ago

If tan theta = startfraction 11 over 60 endfraction, what is the value of cot theta?

Dennis_Churaev [7]

The value of $cot\theta$ when $tan\theta =\frac{11}{60}$ comes to be $\frac{60}{11}$ .

Given that trigonometric ratio:

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<h3>What is the tangent of an angle?</h3>

The tangent of an angle is the ratio of the opposite side(to that angle) to the adjacent side(to that angle).

So, for the given problem

Opposite side to $\theta$ = 11

Adjacent side to $\theta$ = 60

So, $cot \theta =\frac{AdjacentSide }{OppositeSide}$

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6 0

1 year ago

In the 1990s the demand for personal computers in the home went up with household income. For a given community in the 1990s, th

WITCHER [35]

Answer:

a) 0.5198 computers per household

b) 0.01153 computers

Step-by-step explanation:

Given:

number of computers in a home,

q = 0.3458 ln x - 3.045 ; 10,000 ≤ x ≤ 125,000

here x is mean household income

mean income = $30,000

increasing rate, $\frac{dx}{dt}$ = $1,000

Now,

a) computers per household are

since,

mean income of $30,000 lies in the range of 10,000 ≤ x ≤ 125,000

thus,

q = 0.3458 ln(30,000) - 3.045

q = 0.5198 computers per household

b) Rate of increase in computers i.e $\frac{dq}{dt}$

$\frac{dq}{dt}$ = $\frac{d(0.3458 ln x - 3.045)}{dt}$

$\frac{dq}{dt}=0.3458\times(\frac{1}{x})\frac{dx}{dt} - 0$

on substituting the values, we get

$\frac{dq}{dt}=0.3458\times(\frac{1}{30,000})\times1,000$

= 0.01153 computers

6 0

2 years ago

.please answer correctly and show steps

kupik [55]

Answer:

6/5m

1 1/5m

Step-by-step explanation:

5 - 3 4/5 = x

5/1 - 19/5 = x

5/1(5/5) - 19/5 = x

25/5 - 19/5 = x

25/5 - 19/5 = 6/5

x = 6/5 or 1 1/5

8 0

2 years ago

For the cost function c equals 0.1 q squared plus 2.1 q plus 8, how fast does c change with respect to q when q equals 11? Det

Soloha48 [4]

Answer:

Rate of change of c with respect to q is 4.3

Percentage rate of change c with respect to q is 9.95%

Step-by-step explanation:

Cost function is given as, $c=0.1\:q^{2}+2.1\:q+8$

Given that c changes with respect to q that is, $\dfrac{dc}{dq}$ . So differentiating given function,

$\dfrac{dc}{dq}=\dfrac{d}{dq}\left (0.1\:q^{2}+2.1\:q+8 \right)$

Applying sum rule of derivative,

$\dfrac{dc}{dq}=\dfrac{d}{dq}\left(0.1\:q^{2}\right)+\dfrac{d}{dq}\left(2.1\:q\right)+\dfrac{d}{dq}\left(8\right)$

Applying power rule and constant rule of derivative,

$\dfrac{dc}{dt}=0.1\left(2\:q^{2-1}\right)+2.1\left(1\right)+0$

$\dfrac{dc}{dt}=0.1\left(2\:q\right)+2.1$

$\dfrac{dc}{dt}=0.2\left(q\right)+2.1$

Substituting the value of $q=11$ ,

$\dfrac{dc}{dt}=0.2\left(11\right)+21.$

$\dfrac{dc}{dt}=2.2+2.1$

$\dfrac{dc}{dt}=4.3$

Rate of change of c with respect to q is 4.3

Formula for percentage rate of change is given as,

$Percentage\:rate\:of\:change=\dfrac{Q'\left(x\right)}{Q\left(x\right)}\times 100$

Rewriting in terms of cost C,

$Percentage\:rate\:of\:change=\dfrac{C'\left(q\right)}{C\left(q\right)}\times 100$

Calculating value of $C\left(q \right)$

$C\left(q\right)=0.1\:q^{2}+2.1\:q+8$

Substituting the value of $q=11$ ,

$C\left(q\right)=0.1\left(11\right)^{2}+2.1\left(11\right)+8$

$C\left(q\right)=0.1\left(121\right)+23.1+8$

$C\left(q\right)=12.1+23.1+8$

$C\left(q\right)=43.2$

Now using the formula for percentage,

$Percentage\:rate\:of\:change=\dfrac{4.3}{43.2}\times 100$

$Percentage\:rate\:of\:change=0.0995\times 100$

$Percentage\:rate\:of\:change=9.95%$

Percentage rate of change of c with respect to q is 9.95%

7 0

2 years ago

Solve for k: 31 - 3k = 10​

Solve for k: 31 - 3k = 10