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3 years ago

Which rectangle has an area of 30 square feet and a perimeter of 26 ft? 2ft by 15 ft 3ft by 10 ft 5ft by 6 ft 2ft by 15 ft

Mathematics

1 answer:

Kay [80]3 years ago

6 0

Answer:

Ok Im kinda of confused on how to wrote this out, I can understand the question, but i dont understand the options you've given me...

Step-by-step explanation:

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Special precision bolts are made by another machine. These are for jobs that require more exact measurements. The bolts must be

zalisa [80]

The absolute value of a number $x$ is the positive version of that number. So, if $x$ is positive, its absolute value is $x$ itself. If $x$ is negative, its absolute value is $-x$ , so that it will be positive again.

So, if the absolute value of a number is smaller than a certain quantity, i.e. $|x|$ it means that the positive version of $x$ is less than $a$ .

So, if $x$ is positive, the inequality becomes $x$ . If $x$ is negative, the inequality becomes $-x-a$ .

So, as a compound inequality, we have

$|x|$

So, in your case, we have

$|x-8|$

Add 8 to all sides in this inequality:

$8-0.01 < x < 8+0.01 \iff 7.79 < x$

So, on a number line, you must higlight all numbers between 7.79 and 8.01, endpoints excluded.

Conceptually, this means that the bolt measurement should be exactly 8mm, but you can accept bolts that are 0.01 millimeters shorter of larger.

7 0

3 years ago

A-b)3+(b-c)3+(c-a)3:(a-b)(b-c)(c-a)

Ivenika [448]

The answer to the above equation is 3

Step-by-step explanation:

(a-b)³+(b-c)³+(c-a)³: (a-b)(b-c)(c-a)

Let us consider (a−b)= x, (b−c)= y and (c−a)= z.

Hence, It is obvious that:

x+y+z =0 ∵all the terms gets cancelled out

⇒We must remember the algebraic formula

x³+y³+z³−3xyz= (x+y+z) (x²+y²+z²-xy-xz-yz)

Since x+y+z=0 ⇒Whole “(x+y+z) (x²+y²+z²-xy-xz-yz) ” term becomes 0

x³+y³+z³−3xyz =0

Alternatively, x³+y³+z³= 3xyz

Now putting the value of x, y, z in the original equation

(a-b)³+(b-c)³+(c-a)³ can be written as 3(a-b)(b-c)(c-a) since (a−b)= x, (b−c)= y and (c−a)= z.

3(a-b)(b-c)(c-a): (a-b)(b-c)(c-a)

= 3 ∵Common factor (a-b)(b-c)(c-a) gets cancelled out

Answer to the above question is 3

7 0

3 years ago

Write the definite integral for the summation: the limit as n goes to infinity of the summation from k equals 1 to n of the prod

zlopas [31]

Sounds like you have

$\displaystyle\lim_{n\to\infty}\sum_{k=1}^n\left(1+\frac kn\right)^2\frac1n$

which translates to the sum of the areas of $n$ rectangles with dimensions $\left(1+\dfrac kn\right)^2$ (height) and $\dfrac1n$ (width). This is the right-endpoint Riemann sum for approximating the area under $x^2$ over the interval [1, 2].