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2 years ago

Which is true about the solution to the system of inequalities shown

Mathematics

2 answers:

rosijanka [135]2 years ago

7 0

Pls provide the problem

Rainbow [258]2 years ago

5 0

Answer:

look below for question

Step-by-step explanation:

Which is true about the solution to the system of inequalities shown?

y > 3x + 1

y < 3x – 3

Only values that satisfy y > 3x + 1 are solutions.

Only values that satisfy y < 3x – 3 are solutions.

Values that satisfy either y > 3x + 1 or y < 3x – 3 are solutions.

There are no solutions.

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There are four nickels and seven dimes in your pocket. One of the nickels and one of the dimes are Canadian. The others are US c

natima [27]

Answer:

8/11

Step-by-step explanation:

Only 3 of the nickels are neither dimes nor Canadian. The other 8 of 11 coins are dimes or Canadian. The probability of choosing one of them at random is 8/11.

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2 years ago

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The pizza has an area of 78.5 in². What is the minimum width of a placemat that can be placed underneath it so that the pizza do

xz_007 [3.2K]

Answer:

The minimum width of the placemat is 10 inches.

Step-by-step explanation:

Let suppose that placemat has a square form, whose width must be at least equal to the diameter of the pizza, so that pizza does not touch the table. Hence, the following relationship is obtained:

$w = D$

Where:

$w$ - Width of the placemat, measured in inches.

$D$ - Diameter of pizza, measured in inches.

The area of the pizza, measured in square inches, is determined by this formula:

$A = \frac{\pi}{4} \cdot D^{2}$

The diameter is cleared afterwards:

$D = \sqrt{\frac{4\cdot A}{\pi} }$

If $A = 78.5\,in^{2}$ and $\pi = 3.14$ , then:

$D = \sqrt{\frac{4\cdot (78.5\,in^{2})}{3.14} }$

$D = 10\,in$

The minimum width of the placemat is 10 inches.

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2 years ago

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Please help me. The original blueprint of the Moreno’s’ living room has a scale of 2 inches= 5 feet. The family wants to use a n

Svetradugi [14.3K]

Answer:

Part a) The scale of the new blueprint is $\frac{5}{8} \frac{in}{ft}$

Part b) The width of the living room in the new blueprint is $9.4\ in$

Step-by-step explanation:

we know that

The scale of the original blueprint is

$\frac{2}{5}\frac{in}{ft}$

and

the width of the living room on the original blueprint is 6 inches

Find the actual width of the living room, using proportion

$\frac{2}{5}\frac{in}{ft}=\frac{6}{x}\frac{in}{ft}\\ \\x=5*6/2\\ \\x=15\ ft$

Find the actual length of the living room, using proportion

$\frac{2}{5}\frac{in}{ft}=\frac{9.6}{x}\frac{in}{ft}\\ \\x=5*9.6/2\\ \\x=24\ ft$

Find the scale of the new blueprint, divide the length of the living room on the new blueprint by the actual length of the living room

$\frac{15}{24} \frac{in}{ft}$

simplify

$\frac{5}{8} \frac{in}{ft}$

Find the width of the living room in the new blueprint, using proportion

$\frac{5}{8}\frac{in}{ft}=\frac{x}{15}\frac{in}{ft}\\ \\x=15*5/8\\ \\x=9.4\ in$

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2 years ago

To make a profit, a company’s revenue must be greater than its operating costs. The company’s revenue is modeled by the expressi

myrzilka [38]

Revenue = 7.5x - 100
Operation Costs = 5.8x + 79.86

To break even, operation cost = Revenue
⇒ 7.5x - 100 = 5.8x + 79.86
7.5x = 5.8x + 179.86 (Add 100 to both sides)

7.5x - 5.8x = 179.86
1.7x = 179.86
x = 105.8

This implies that the company will need to sell at least 106 items to make a profit.

The inequality that will determine the number of items at need to be sold to make a profit is x ≥ 106

The solution to the inequality is as follows

Revenue = 7.5x - 100
if x =106
Revenue = 7.5(106) - 100
Revenue = 695

Operational Cost = 5.8x + 79.86
if x = 106
Operational Cost = 5.8(106) + 79.86
Operational Cost = 694.66

Profit ≥ (695 - 694.66)
Profit ≥ 0.34

The company must sell at least 106 items to make a profit.

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2 years ago

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. For each of these intervals, list all its elements or explain why it is empty. a) [a, a] b) [a, a) c) (a, a] d) (a, a) e) (a,

Eva8 [605]

Answer:

Elements are of the form

(i) $[a,a]=\{[x,y] : a\leq x\leq a, a\leq y\leq a; a\in \mathbb R\}$

(ii) $[a,b)=\{[x,y) :a\leq x$

(iii) $(a,a]=\{(x,y] :a$

(iv) $(a,a)=\{(x,y): a$

(v) (a,b) where a>b= $\{(x,y) : a>x>b,a>y>b;a>b,a,b \in \mathbb R\}$

(vi) [a,b] where a>b= $\{[x,y] : a\geq x\geq b,a\geq y\geq b;a>b,a,b \in \mathbb R\}$

Step-by-step explanation:

Given intervals are,

(i) [a,a] (ii) [a,a) (iii) (a,a] (iv) (a,a) (v) (a,b) where a>b (vi) [a,b] where a>b.

To show all its elements,

(i) [a,a]

Imply the set including aa from left as well as right side.

Its elements are of the form.

$\{[a,a] : a\in \mathbb R\}=\{[0,0],[1, 1],[-1,-1],[2,2],[-2,-2],[3,3],[-3,-3],........\}$

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set [a,a] represents singleton sets, and singleton sets are empty so is [a,a].

(ii) [a,a)

This means given interval containing a by left and exclude a by right.

Its elements are of the form.

$[ 1, 1),[-1,-1),[2,2),[-2,-2),[3,3),[-3,-3),........$

Since there is a singleton element a of real numbers withis the set, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set [a,a) represents singleton sets, and singleton sets are empty so is [a.a).

(iii) (a,a]

It means the interval not taking a by left and include a by right.

Its elements are of the form.

$( 1, 1],(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........$

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set (a,a] represents singleton sets, and singleton sets are empty so is (a,a].

(iv) (a,a)

Means given set excluding a by left as well as right.

Since there is a singleton element a of real numbers, this set is empty.

Its elements are of the form.

$( 1, 1),(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........$

Because there is no increment so if $a\in \mathbb R$ then the set (a,a) represents singleton sets, and singleton sets are empty, so is (a,a).

(v) (a,b) where a>b.

Which indicate the interval containing a, b such that increment of x is always greater than increment of y which not take x and y by any side of the interval.

That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

$(a,b)=\{(5,0),(5,1),(5,2).....\}$ e.t.c

So this set is connected and we know singletons are connected in $\mathbb R$ . Hence given set is empty.

(vi) [a,b] where $a\leq b$ .

Which indicate the interval containing a, b such that increment of x is always greater than increment of y which include both x and y.

That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

$[a,b]=\{[5,0],[5,1],[5,2].....\}$ e.t.c

So this set is connected and we know singletons are connected in $\mathbb R$ . Hence given set is empty.

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2 years ago