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galben [10]
2 years ago
7

What does p equal?pls help me.

Mathematics
1 answer:
Natalija [7]2 years ago
4 0

\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the Concept of the inequalities.

9p -9 <4p -2

===> 9p - 4p < -2 +9

===> 5p < 7

===> p < 7/5

hence the answer is P < 7/5

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Which of the following ordered pairs make this equation true? Y = 2x - 7 A. (1,9) B. (-3, 4) C. (4,1) D. (-1, 5)​
gregori [183]

Answer:

(4,1)

Step-by-step explanation:

Using substitution you can determine that 2*4-7 is in fact equal to 1

3 0
2 years ago
For what values of x does 25 Superscript x Baseline = 5 Superscript x squared minus 3?
umka2103 [35]

For this case we have the following equation:

25^{x} = 5^{ x^2-3}

Rewriting we have:

5^{2 (x)} = 5^{ x^ 2-3}\\5^{ 2x} = 5^{ x^ 2-3}

For expressions to be equal, exponents must be equal. So:

2x = x ^ 2-3\\x ^ 2-2x-3 = 0

To factor, we look for two numbers that, when multiplied, result in -3 and when added, result in -2. These numbers are -3 and 1.

(x-3) (x + 1) = 0

Thus, the roots are:

x_ {1} = 3\\x_ {2} = - 1

Answer:

x_ {1} = 3\\x_ {2} = - 1

8 0
3 years ago
HELP!! Algebra help!! Will give stars thank u so much &lt;333
Anna35 [415]

Answers:

  • Part a)  \bf{\sqrt{x^2+(x^2-3)^2}
  • Part b)  3
  • Part c)   2.24
  • Part d)  1.58

============================================================

Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

We apply the distance formula to say...

d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying d = \sqrt{x^2 + (x^2-3)^2} is the same as writing d = sqrt(x^2-(x^2-3)^2)

-------------------------------------------

Part (b)

Plug in x = 0 and you should find the following

d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

-------------------------------------------

Part (c)

Repeat for x = 1

d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\

-------------------------------------------

Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.

6 0
2 years ago
Please answer! 5 points BUT brainliest if correct.
arlik [135]

Answer:

it is a because you subtract the ten from both side and that makes the -8 into a -18 then you divide by 3 and your answer is -6

3 0
3 years ago
What is a common factor called
zlopas [31]
  <span>A common factor is a factor of 2 or more numbers. 
the factors of 12 are: 1,2,3,4,6,12 
the factors of 21 are: 1,3,7,21 

1 and 3 are both common factors of 12 and 21.</span>
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