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2 years ago

7

The width of a rectangle is the length minus 2 units. The area of the rectangle is 35 units. What is the width, in units, of the

rectangle?

1 answer:

Andrews [41]2 years ago

6 0

Answer: width = 5 units

Step-by-step explanation:

Let L represent the length of the rectangle.

The width of a rectangle is the length minus 2 units. It means that the width of the rectangle is (L - 2) units.

The formula for determining the area of a rectangle is expressed as

Area = length × width

The area of the rectangle is 35 units. It means that

L(L - 2) = 35

L² - 2L = 35

L² - 2L - 35 = 0

L² + 5L - 7L - 35 = 0

L(L + 5) - 7(L + 5) = 0

L - 7 = 0 or L + 5 = 0

L = 7 or L = - 5

Since the length cannot be negative, then

L = 7 units

Width = 7 - 2 = 5 units

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Your utility company charges 13 cents per kilowatt-hour of electricity. What is the daily cost of keeping lit a 75 watt light bu

Tresset [83]

we know that

$1 Kw=1000 w$

The company charges $13$ cents per kilowatt-hour of electricity

$13 cents=0.13 dollars\\\\ 75w=\frac{75}{1000} kw\\\\ 15w=\frac{15}{1000} kw$

Step $1$

Find the cost for the light bulb

a) in one day

$\frac{75}{1000} *12=0.9\ kw\ hour$

$Cost=0.13*0.9$

Cost=$ $0.117$

b) In a year

$\frac{75}{1000} *12=0.9\ kw\ hour$

$0.9*365=328.5\ kw\ hour$

$Cost=0.13*328.5$

Cost=$ $42.71$

Step $2$

Find the cost for the led bulb

a) in one day

$\frac{15}{1000} *12=0.18\ kw\ hour$

$Cost=0.13*0.18$

Cost=$ $0.0234$

b) In a year

$\frac{15}{1000} *12=0.18\ kw\ hour$

$0.18*365=65.7\ kw\ hour$

$Cost=0.13*65.7$

Cost=$ $8.54$

Step $3$

Find the difference in cost

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therefore

the answer is

$34.17 dollars$

3 0

2 years ago

Read 2 more answers

The lifespans of lions in a particular zoo are normally distributed. The average lion lives 12.512.512, point, 5 years; the stan

zaharov [31]

This question was not written properly

Complete Question

The lifespans of lions in a particular zoo are normally distributed. The average lion lives 12.5 years; the standard deviation is 2.4 years. Use the empirical rule (68-95-99.7\%)(68−95−99.7%) to estimate the probability of a lion living between 5.3 to 10. 1 years.

Answer:

Thehe probability of a lion living between 5.3 to 10. 1 years is 0.1585

Step-by-step explanation:

The empirical rule formula states that:

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ.

2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ.

3) 99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ.

Mean is given in the question as: 12.5

Standard deviation : 2.4 years

We start by applying the first rule

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ.

μ - σ

12.5 -2.4

= 10.1

We apply the second rule

2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ.

μ – 2σ

12.5 - 2 × 2.4

12.5 - 4.8

= 7.7

We apply the third rule

3)99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ.

μ - 3σ

= 12.5 - 3(2.4)

= 12.5 - 7.2

= 5.3

From the above calculation , we can see that

5.3 years corresponds to one side of 99.7%

Hence,

100 - 99.7%/2 = 0.3%/2

= 0.15%

And 10.1 years corresponds to one side of 68%

Hence

100 - 68%/2 = 32%/2 = 16%

So,the percentage of a lion living between 5.3 to 10. 1 years is calculated as 16% - 0.15%

= 15.85%

Therefore, the probability of a lion living between 5.3 to 10. 1 years

is converted to decimal =

= 15.85/ 100

= 0.1585

8 0

2 years ago

Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

<u>Ben:</u>

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

<u>Josh:</u>

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

<em><u>Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at </u></em> $h(t)=0$ <em><u>. Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one. </u></em>

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$

Since time can only be positive we reject the $t_{2}$ solution and we keep that Ben's book took $t=4.3276$ seconds to reach the ground.

Similarly solving for Josh we obtain

$t_{1}=3.6794sec\\t_{2}=-0.6794sec$

Thus again we reject the negative and keep the positive solution, so Josh's book took $t=3.6794$ seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

$t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482$

So to answer the original question:

<em>Whose textbook reaches the ground first and by how many seconds?</em>

Josh's textbook reached the ground first
Josh's textbook reached the ground first by a difference of $t=0.6482$

3 0

1 year ago

N(17+x)=34x−r<br> I need to solve for x

Paraphin [41]

Answer:

The value of x is, $x= \frac{17N+r}{34-N}$

Explanation:

Given: $N(17+x)=34x-r$

Distributive Property states that when a number is multiplied by the sum of two numbers, the first number can be distributed to both of those numbers and multiplied by each of them separately.

If $a\cdot(b+c) =a\cdot b + a\cdot c$

Now, using distributive property on left hand side of the given expression as:

$N\cdot 17+N\cdot x = 34x-r$ or $17N+Nx = 34x-r$

Addition Property of equality state that we add the same number from both sides of an equation.

Add r to both sides of an equation:

$17N+Nx+r=34x-r+r$

Simplify:

$17N+Nx+r=34x$

Subtraction Property of equality state that we subtract the same number from both sides of an equation.

Subtract Nx from both sides of an equation;

$17N+Nx+r-Nx=34x-Nx$

Simplify:

$17N+r=34x-Nx$

or

$17N+r=x(34-N)$

Division Property of equality states that we divide the same number from both sides of an equation.

Divide by (34-N) to both sides of an equation;

$\frac{17N+r}{34-N}= \frac{x(34-N)}{34-N}$

On Simplify:

$x= \frac{17N+r}{34-N}$

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