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4 years ago

The difference of a number and six is seven. Find the number.

Mathematics

1 answer:

NikAS [45]4 years ago

6 0

Answer 13-6=7

Step-by-step explanation:

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The China Company offers a trade-discount rate that is 4 times the complement of the trade-discount rate. If the list price is $

Natali5045456 [20]

Answer: See explanation

Step-by-step explanation:

Since the complement is the difference between 100% and discount rate. Since the China Company offers a trade-discount rate that is 4 times the complement of the trade-discount rate. Therefore, the discount rate will be 80% and the complement 20%.

Therefore, the trade discount will be:

= List price × Trade discount rate

= $58.75 × 80%

= $58.75 × 0.8

= $47

Net price = List price - Trade discount

= $58.75 - $47

= $11.75

7 0

3 years ago

the slopes of a quadrilateral are -1/2, 3/2, -1/2, and 3/2. What conclusion about this shape cam be made.

Talja [164]

The conclusion that can be made given the slopes of the quadrilateral is that: since the slopes of the opposite sides are equal, the opposite sides are parallel, therefore the quadrilateral is a parallelogram. (Option A is correct).

Recall:

The slopes of the opposite sides of a parallelogram are always parallel.

When two lines have a slope that is equal, the lines are said to be parallel to each other.

The slopes given shows are:

-1/2, 3/2, -1/2, and 3/2

This implies that the quadrilateral has two pairs of equal sides that are parallel to each other.

Therefore, the conclusion that can be made given the slopes of the quadrilateral is that: since the slopes of the opposite sides are equal, the opposite sides are parallel, therefore the quadrilateral is a parallelogram. (Option A is correct).

Learn more about parallelogram on:

brainly.com/question/3050890

4 0

2 years ago

Help calculus module 8 DBQ please show work

igor_vitrenko [27]

1. The four subintervals are [0, 2], [2, 3], [3, 7], and [7, 8]. We construct trapezoids with "heights" equal to the lengths of each subinterval - 2, 1, 4, and 1, respectively - and the average of the corresponding "bases" equal to the average of the values of $R(t)$ at the endpoints of each subinterval. The sum is then

$\dfrac{R(0)+R(2)}2(2-0)+\dfrac{R(2)+R(3)}2(3-2)+\dfrac{R(3)+R(7)}2(7-3)+\dfrac{R(7)+R(8)}2(7-8)=\boxed{24.83}$

which is measured in units of gallons, hence representing the amount of water that flows into the tank.

2. Since $R$ is differentiable, the mean value theorem holds on any subinterval of its domain. Then for any interval $[a,b]$ , it guarantees the existence of some $c\in(a,b)$ such that

$\dfrac{R(b)-R(a)}{b-a)=R'(c)$

Computing the difference quotient over each subinterval above gives values of 0.275, 0.3, 0.3, and 0.26. But just because these values are non-zero doesn't guarantee that there is definitely no such $c$ for which $R'(c)=0$ . I would chalk this up to not having enough information.

3. $R(t)$ gives the rate of water flow, and $R(t)\approx W(t)$ , so that the average rate of water flow over [0, 8] is the average value of $W(t)$ , given by the integral

$R_{\rm avg}=\displaystyle\frac1{8-0}\int_0^8\ln(t^2+7)\,\mathrm dt$

If doing this by hand, you can integrate by parts, setting

$u=\ln(t^2+7)\implies\mathrm du=\dfrac{2t}{t^2+7}\,\mathrm dt$

$\mathrm dv=\mathrm dt\implies v=t$

$R_{\rm avg}=\displaystyle\frac18\left(t\ln(t^2+7)\bigg|_{t=0}^{t=8}-\int_0^8\frac{2t^2}{t^2+7}\,\mathrm dt\right)$

For the remaining integral, consider the trigonometric substitution $t=\sqrt 7\tan s$ , so that $\mathrm dt=\sqrt 7\sec^2s\,\mathrm ds$ . Then

$R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}\frac{7\tan^2s}{7\tan^2s+7}\sec^2s\,\mathrm ds$

$R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}\tan^2s\,\mathrm ds$

$R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}(\sec^2s-1)\,\mathrm ds$

$R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\left(\tan s-s\right)\bigg|_{s=0}^{s=\tan^{-1}(8/\sqrt7)}$

$R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\left(\tan\left(\tan^{-1}\frac8{\sqrt7}\right)-\tan^{-1}\frac8{\sqrt7}\right)$

$\boxed{R_{\rm avg}=\displaystyle\ln71-2+\frac{\sqrt7}4\tan^{-1}\frac8{\sqrt7}}$

or approximately 3.0904, measured in gallons per hour (because this is the average value of $R$ ).

4. By the fundamental theorem of calculus,

$g'(x)=f(x)$

and $g(x)$ is increasing whenever $g'(x)=f(x)>0$ . This happens over the interval (-2, 3), since $f(x)=3$ on [-2, 0), and $-x+3>0$ on [0, 3).

5. First, by additivity of the definite integral,

$\displaystyle\int_{-2}^xf(t)\,\mathrm dt=\int_{-2}^0f(t)\,\mathrm dt+\int_0^xf(t)\,\mathrm dt$

Over the interval [-2, 0), we have $f(x)=3$ , and over the interval [0, 6], $f(x)=-x+3$ . So the integral above is

$\displaystyle\int_{-2}^03\,\mathrm dt+\int_0^x(-t+3)\,\mathrm dt=3t\bigg|_{t=-2}^{t=0}+\left(-\dfrac{t^2}2+3t\right)\bigg|_{t=0}^{t=x}=\boxed{6+3x-\dfrac{x^2}2}$

6 0

3 years ago

A health inspector must visit 44 of 1414 restaurants on monday. in how many ways can she pick a first comma second comma third c

ira [324]

Let's make clear the question, because it may be confusing:
"A health inspector must visit 4 of 14 restaurants on Monday. In how many ways can she pick a first, second, third, and fourth restaurant to visit?"

In order to solve the problem, you need to use permutations: you have 14 possible restaurants to choose as first, the second one will be chosen among the 13 restaurants left, the third one among the remaining 12 and the fourth one will be one of the remaining 11.

Therefore, the total number of possibilities is:
14 × 13 × 12 × 11 = 24024

There are 24024 possible ways to choose the four restaurants to visit.

4 0

3 years ago

1 HOUR TIME LIMIT

irina1246 [14]

Answer:

Your answer is 1500 foot pounds.

I hope this helps!

Step-by-step explanation:

Edg 2020

3 0

3 years ago

Read 2 more answers