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3 years ago

every three days Seth goes to the grocery store every brain goes to the store every 5 days what day will they both go

Mathematics

2 answers:

Brrunno [24]3 years ago

8 0

Just multiply 3x5 which is 15

Rom4ik [11]3 years ago

6 0

If seth goes every three days and Brian im assuming is the name.Goes every five days then they will both go on day 15

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What is 2/5 v + 3/4 = 2/3x² as a sentence

Bond [772]

The representation of the equation given in the task content; 2/5 v + 3/4 = 2/3x² as a s sentence is; The sum of Two-fifths of a number, v and quarters is equal to two-thirds of the square of the number x.

<h3>What is the representation of the equation as given in the task content?</h3>

It follows from the convention that the following fractions can be named as follows;

2/5 means two-fifths.
3/4 means three-quarters.
2/3 means two-thirds.

Hence, it follows from the iterations above and how they relate to the equation given in the task content that the representation is; The sum of Two-fifths of a number, v and quarters is equal to two-thirds of the square of the number x.

Read more on equations as words;

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2 years ago

SIMPLIFY THE EXPRESSION (m − 1)(-4)

anastassius [24]

Answer:

-4m + 4

Step-by-step explanation:

you distribute the -4 to the m making -4m and then distribute -4 to the -1 which makes +4

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3 years ago

Let FS R be a finite set. (a) Prove that F is closed (b) Prove that every point in F is an isolated point.

Lorico [155]

Answer:

a) Suppose that F is ordered in ascending order: $F = \{x_1,\ldots, x_n\}$ . Then, the complement of F can be written as

$F^c = (-\infty,x_1)\cup (x_1,x_2)\cup (x_2,x_3)\cup \cdots \cup (x_{n-1}, x_{n})\cup (x_n,+\infty)$

which is the union of a finite number of open intervals, then $F^c$ is an open set. Thus, F is a closed subset of the real numbers.

b) Take an arbitrary element of F, let us say $x_k$ . Now, choose a real number $\epsilon$ such that

$0$ there are not other element of F, because $\epsilon$ is less that the minimum distance between $x_k$ and its neighbors.

In case that $k=1$ we only consider $0$ , and if $k=n$ we only consider $0$ .

Then, all points of F are isolated.

Step-by-step explanation:

8 0

3 years ago

A Rectangle is removed from a right triangle to create the shaded region below. Find the area of the shaded region be sure to in

Scorpion4ik [409]

The given shapes area is found to be 4 square yards.

Step-by-step explanation:

Step 1; To determine the area of an indefinite shape we must divide it into the shapes that we know. The shape that is given can be split into a triangle with a rectangle in it.

Step 2; Any given triangles area is half the multiplication of its base value and its height value. In the given diagram, the base of the triangle measures a length of 5 yards and the length from base to top is 4 yards

Area of the triangle = $\frac{1}{2}$ * base * height = $\frac{1}{2}$ * 5 * 4 = 10 square yards.

The given rectangle measures a length of 3 yards and a width of 2 yards. The area of any given rectangle is the multiplication of its length and width.

Area of the rectangle = Length * Width = 3 yards * 2 yards = 6 square yards.

Step 3; To find the total area of the shape we must subtract the the rectangle's area from the triangle's area.

Area of given shape = Area of the triangle - Area of the rectangle.

Area of given shape = 10 square yards - 6 square yards = 4 square yards.

7 0

3 years 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

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3 years ago