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Sergeu [11.5K]
3 years ago
7

Plzz HELP me Solve for x. x/2.5<10 x=?

Mathematics
1 answer:
goldfiish [28.3K]3 years ago
7 0

The answer to the inequality would be x<25.

You can solve the inequality by multiplying 2.5 on both sides to isolate x.

\frac{x}{2.5}*2.5<10*2.5

x<25

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Solve for x.<br> 9(x+1)=25+x<br> x=2<br> 0x=3<br> O x=4<br> O x=5
aleksandr82 [10.1K]

Step-by-step explanation:

9(x+1) = 25+x

open the bracket

9x + 9 = 25+x

collect like terms

9x-x = 25-9

8x = 16

x = 2

3 0
2 years ago
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Which student correctly wrote the algebraic expression for the following statement?
Leno4ka [110]
Haileys work is correct so B should be the right answer
3 0
3 years ago
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12.3+9.75=?<br> 10,258÷46=?<br> Show your work
wariber [46]

Answer:

12.3 + 9.75 =22.05    10,258 ÷ 46 =

Step-by-step explanation:

for 10, 258 ÷ 46 i decide to type it

10258 46 = 223

223 = 2230 to the nearest tenth

223 = 223 to the nearest hundredth

223 = 223 to the nearest thousandth

= 0 to the nearest tenth

= 0 to the nearest hundredth

= 0 to the nearest thousandth

Other Divisions Math homework are

10258 divide by half plus 20

Homework answers: (10258/2) + 20 = 5149

10258 divide by half plus 40

Homework answers: (10258/2) + 40 = 5169

10258/46 divided by 2

Answer: (10258/46) ÷ 2 = 111.5

8 0
4 years ago
If y varies inversely asx, and y = 33 when x = 7 what is the constant and what is value of y when x=11?
allsm [11]

Answer:

We have that y = 21 when x = 11.

Step-by-step explanation:

We solve this question by proportions, using a rule of three.

Since they vary inversely, we apply the inverse rule of three, that is, with lateral multiplication instead of diagonal.

33 - 7

y - 11

So

11y = 33 -7

y - 11

Dividing both sides by 11

Y = 3 * 7 = 21

We have that y = 21 when x = 11.

7 0
3 years ago
What is the equation of the quadratic function represented by this table?
MA_775_DIABLO [31]

Answer:

Step-by-step explanation:

I used logic and took the easy way around this as opposed to the long, drawn-out algebraic way.  I noticed right off that at x = -3 and x = -1 the y values were the same.  In the middle of those two x-values is -2, which is the vertex of the parabola with coordinates (-2, 4).  That's the h and k in the formula I'm going to use.  Then I picked a point from the table to use as my x and y in the formula I'm going to use.  I chose (0, 3) because it's easy.  The formula for a quadratic is

y=a(x-h)^2+k

and I have everything I need to solve for a.  Filling in my h, k, x, and y:

3=a(0-(-2))^2+4  and

3=a(2)^2+4  and

-1 = 4a so

a=-\frac{1}{4}

In work/vertex form the equation for the quadratic is

y=-\frac{1}{4}(x+2)^2+4

In standard form it's:

y=-\frac{1}{4}x^2-x+3

8 0
3 years ago
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