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3 years ago

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What are the inequality solutions for 3 ≤ 3 x − 4 < 2 x + 1

1 answer:

krek1111 [17]3 years ago

3 0

Answer:

7/3 ≤ x < 5.

Step-by-step explanation:

3 ≤ 3 x − 4 < 2 x + 1

Split this up into 2 inequalities:

3x - 4 ≥ 3

3x ≥ 7

x ≥ 7/3.

and

3x - 4 < 2x + 1

3x - 2x < 1 + 4

x < 5.

So it's 7/3 ≤ x < 5.

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ehidna [41]

Answer:

A. the distance between points W and X

Step-by-step explanation:

<u><em>Looking for the distance of 5 units</em></u>

A. the distance between points W and X

-3 - (-8) = -3 + 8 = 5
Correct

B. the distance between points X and Y

1 - (-3) = 1 + 3 = 4 < 5
Incorrect

C. the distance between points X and Z

5 - (-3) = 5 + 3 = 8 > 5
Incorrect

D. the distance between points Y and Z

5 - 1 = 4 < 5
Incorrect

7 0

3 years ago

Please help me on this TIA!!

tamaranim1 [39]

50.24 hope it helps:)

8 0

3 years ago

A cash box contains $74 made up of quarters, half-dollars, and one-dollar

Strike441 [17]

Answer:

25 one-dollar coins, 16 half-dollar coins, and 164 quarters

Step-by-step explanation:

First, set up equations based on the information given:

$0.25q+0.50h+1.00d=74$

$\displaystyle{h=\frac{3}{5}d+1}$

$q=4(d+h)$

Then, substitute <em>q</em> in the first equation with the expression from the third equation:

$0.25[4(d+h)]+0.50h+1.00d=74\\1d+1h+0.50h+1.00d=74\\2d+1.5h=74$

Next, substitute <em>h</em> in that equation with the expression from the second equation:

$\displaystyle{2d+1.5(\frac{3}{5}d+1)=74}$

$2d+0.9d+1.5=74\\2.9d+1.5=74$

Solve for <em>d</em>, the number of one-dollar coins:

$2.9d+1.5=74\\2.9d=72.5\\d=25$

Substitute 25 for <em>d</em> in the second equation to find <em>h</em>, the number of half-dollar coins:

$\displaystyle{h=\frac{3}{5}d+1}$

$\displaystyle{h=\frac{3}{5}(25)+1}$

$h=15+1\\h=16$

Substitute 25 for <em>d</em> and 16 for <em>h</em> in the third equation to find <em>q</em>, the number of quarters:

$q=4(d+h)\\q=4(25+16)\\q=4(41)\\q=164$

Then, verify that the coins total $74:

$0.25(164)+0.50(16)+1.00(25)=74\\41+8+25=74\\74=74\\\text{Check.}$

Next, verify that the number of half-dollar coins is one more than three-fifths of the number of one-dollar coins:

$\displaystyle{h=\frac{3}{5}d+1}$

$\displaystyle{16=\frac{3}{5}(25)+1}$

$16 = 15 + 1\\16 = 16\\\text{Check.}$

Finally, verify that the number of quarters is four times the number one-dollar and half-dollar coins together:

$q=4(d+h)\\164=4(25+16)\\164=4(41)\\164=164\\\text{Check.}$

6 0

3 years ago

ALOT OF POINTS IF YOU HELP ME PLEASE EXPLAIN

Svetllana [295]

Answer:

Helena gave the answer as (7y²z + 6yz²- 5 - 3yz² + 2) which is equivalent to (7y²z + 3yz² - 3).

Step-by-step explanation:

Misha's group was asked to write an expression equivalent to

7y²z + 3yz² - 3

When Mr. Chen checked their answers, he found only one to be correct.

And she was Helena.

Helena gave the answer as (7y²z + 6yz²- 5 - 3yz² + 2) which is equivalent to (7y²z + 3yz² - 3).

Because, (7y²z + 6yz²- 5 - 3yz² + 2)

= 7y²z + (6yz² - 3yz²) - (5 - 2)

= 7y²z + 3yz² - 3 (Answer)

4 0

3 years ago

Please help! i dont understand

kvv77 [185]

QUESTION 1

The given inequality is

$y\leq x-3$ and $y\geq -x-2$ .

If (3,-2) is a solution; then it must satisfy both inequalities.

We put x=3 and y=-2 in to both inequalities.

$-2\leq 3-3$ and $-2\geq -3-2$ .

$-2\leq 0$ :True and $-2\geq -5$ :True

Both inequalities are satisfied, hence (3,-2) is a solution to the given system of inequality.

QUESTION 2

The given inequality is

$y\:>\:-3x+3$ and $y\:>\: x+2$ .

If (1,4) is a solution; then it must satisfy both inequalities.

We put x=1 and y=4 in to both inequalities.

$4\:>\:-3(1)+3$ and $4\:>\: 1+2$ .

$4\:>\:0$ :True and $4\:>\: 3$ :True

Both inequalities are satisfied, hence (1,4) is a solution to the given system of inequality.

Ans: True

QUESTION 3

The given inequality is

$y\leq 3x-6$ and $y\:>\: -4x+2$ .

If (0,-2) is a solution; then it must satisfy both inequalities.

We put x=0 and y=-2 in to both inequalities.

$-2\leq 3(0)-6$ and $-2\:>\: -4(0)+2$ .

$-2\leq -6$ :False and $-2\:>\:2$ :False

Both inequalities are not satisfied, hence (0,-2) is a solution to the given system of inequality.

Ans:False

QUESTION 4

The given inequality is

$2x-y\:$ and $x+y\:>\:-1$ .

If (0,3) is a solution; then it must satisfy both inequalities.

We put x=0 and y=3 in to both inequalities.

$2(0)-3\:$ and $0+3\:>\:-1$ .

$-3\:$ :True and $3\:>\:-1$ : True

Both inequalities are satisfied, hence (0,3) is a solution to the given system of inequality.

Ans:True

QUESTION 5

The given system of inequality is

$y\:>\:2x-3$ and $y\:$ .

If (-3,0) is a solution; then it must satisfy both inequalities.

We put x=-3 and y=0 in to both inequalities.

$0\:>\:2(-3)-3$ and $0\:$ .

$0\:>\:-9$ ;True and $0\:$ :True

Both inequalities are satisfied, hence (-3,0) is a solution to the given system of inequality.

Ans:True

3 0

3 years ago