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Ket [755]
3 years ago
6

H e l p i n e e d h e l p

Mathematics
2 answers:
Sedbober [7]3 years ago
6 0
I’m so sorry I don’t know how to answer this is just to ask another question cuz I need help with math I’m so sorry I hope you get a great answer
GarryVolchara [31]3 years ago
5 0

Answer:

OML I have no flipping clue...

Step-by-step explanation:

sorry friend, I just don't know, but thanks for the points☺

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Draw a number line to<br><br> represent the inequality.<br><br> y s 23
Mkey [24]

Answer:

The number lines are shown below.

Step-by-step explanation:

In the given problem the sign of inequality is missing.

We know, that there are 4 signs of inequality ">", "<", "≥" and "≤".

The possible inequalityes are

y>23

y

y\geq 23

y\leq 23

In y>23, all points on the right side of 23 are included in the solution set.

In y, all points on the left side of 23 are included in the solution set.

In y\geq 23, 23 and all points on the right side of 23 are included in the solution set.

In y\leq 23, 23 all points on the left side of 23 are included in the solution set.

4 0
3 years ago
Find the perimeter of a rectangle with a length of 6x+3 and a width of -2x-5.
blsea [12.9K]
The simple way of calculating a perimeter is to add all of the sides.
For a rectangle this would equal 2 x length + 2 x width
P = 2(6x + 3) + 2(-2x - 5)
= 12x + 6 - 4x - 10
= 8x - 4

At a higher level both the length and width must be greater than zero (= zero is a trivial rectangle)
6x + 3 > 0
6x > -3
x > -0.5

-2x - 5 > 0
2x + 5 < 0 (multiplying by -1 reverses the inequality)
2x < -5
x < -2.5

This rectangle cannot exist as x cannot be < -2.5 and > -0.5 at the same time!
8 0
3 years ago
Read 2 more answers
The number line shows an inequality. Describe a real-world situation that the inequality could represent. ONLY ANSWER IF YOU KNO
Orlov [11]
You cant answer unless you have the picture. This question goes to a lot of answers
6 0
3 years ago
Help please jggvnhujff
saveliy_v [14]

Answer:

Step-by-step explanation:

<u><em>8).</em></u>  \left \{ {{4x+3y=1} \atop {x+y=2}} \right.

<em>(2)</em> × [ - 3 ]

4x + 3y = 1 ........ <em>(3)</em>

- 3x - 3y = - 6 .... <em>(4)</em>

<em>(3)</em> + <em>(4)</em>

x = - 5

- 5 + y = 2 ⇒ y = 7

<em>( - 5 , 7 )</em>

<u><em>9).</em></u> \left \{ {{-9x+3y=18} \atop {2x+y=-4}} \right.

<em>(1)</em> ÷ [- 3]

3x - y = - 6  ......... <em>(3)</em>

2x + y = - 4 ........ <em>(4)</em>

<em>(3)</em> + <em>(4)</em>

5x = - 10 ⇒ x = - 2

2(- 2) + y = - 4 ⇒ y = 0

<em>(- 2, 0)</em>

<u><em>10).</em></u> \left \{ {{x-2y=14} \atop {10x-6y=0}} \right.

<em>(2)</em> ÷ 10

x - 0.6y = 0 ⇒ x = 0.6y -----> <em>(1)</em>

0.6y - 2y = 14

- 1.4y = 14

y = - 10

x - 2(- 10) = 14 ⇒ x = - 6

<em>(- 6, - 10)</em>

Now is your turn, you can do it!!

3 0
3 years ago
It's a question from real and complex numbers which I can't solve. so someone PLZ HeLp​
Semmy [17]

Answer:

\frac{1}{5}

Step-by-step explanation:

Using the rules of exponents

a^{m} × a^{n} = a^{(m+n)}, \frac{a^{m} }{a^{n} } = a^{(m-n)}, (a^m)^{n} = a^{mn}

Simplifying the product of the first 2 terms

\frac{a^{p^2+pq} }{a^{pq+q^2} } × \frac{a^{q^2+qr} }{a^{qr+r^2} }

= a^{p^2-q^2} × a^{q^2-r^2}

= a^{p^2-r^2}

Simplifying the third term

5((a^p+r)^{p-r}

= 5a^{(p+r)(p-r)} = 5a^{(p^2-r^2)}

Performing the division, that is

\frac{a^{(p^2-r^2)} }{5a^{(p^2-r^2)} } ← cancel a^{(p^2-r^2)} on numerator/ denominator leaves

= \frac{1}{5}

4 0
3 years ago
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