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kobusy [5.1K]
3 years ago
12

What’s a real number between the square root of 13 and the square root of 14

Mathematics
1 answer:
Pie3 years ago
6 0

Real numbers can be both rational and irrational.

sqrt(13) = 3.60555

sqrt(14) = 3.74166

If you're looking for a rational number, there are lots of them to choose from. Easiest is probably 3.7 = 37/10.

An irrational number in this range might be sqrt(27/2).

Hope this helps!

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What is the product of (-3s+2t(4s-t)
galben [10]
Use the FOLD method<span>
(a+b)(c+d)=ac+ad+bc+bd

</span><span>−12<span>s<span><span>​^2 </span><span>​​</span></span></span>+ 3st + 8ts − 2<span>t<span><span>​^2</span><span>​​

</span></span></span></span>Collect like terms
-12{s}^{2}+(3st+8st)-2{t}^{2}

Simplify
<span>-12{s}^{2}+11st-2{t}^{2}<span>−12<span>s<span><span>​2</span><span>​​</span></span></span>+11st−2<span>t<span>​<span>2

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5 0
3 years ago
SUPER DUPER EASY but need help!
Mandarinka [93]

Answer:

D.) 1 + 1 + 5/8 + 2/8

Step-by-step explanation:

1 5/8 + 1 1/4

*set aside ones for a sec*

1/4 x 2/2 = 2/8

5/8 + 2/8

*get the ones back in*

1 + 1 + 5/8 + 2/8

5 0
3 years ago
I need help fast
hram777 [196]

Answer:

On a coordinate plane, a line goes through (0, 3) and (2, 4) and another line goes through (0, 3) and (0.75, 0).

This answer almost coincide with option C. I suppose there was a mistype.

Step-by-step explanation:

The system of equations is formed by:

–x + 2y = 6

4x + y = 3

In the picture attached, the solution set is shown.

The first equation goes through (0, 3) and (2, 4), as can be checked by:

–(0) + 2(3) = 6

–(2) + 2(4) = 6

The second goes through (0, 3) and (0.75, 0), as can be checked by:

4(0) + (3) = 3

4(0.75) + (0) = 3

7 0
2 years ago
Read 2 more answers
-4(d+5)-3d&gt;8 please help i dont need you to explain it only answer please and thank you
Gala2k [10]
Show your work kidddd
8 0
3 years ago
Find+the+positive+value+for+α+if+the+radius+of+the+circle+3x^2+3y-6αx+12y-3α=0+is+4
Alik [6]

Answer:

\alpha=3

Step-by-step explanation:

<u>Equation of a Circle</u>

A circle of radius r and centered on the point (h,k) can be expressed by the equation

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

We are given the equation of a circle as

3x^2+3y^2-6\alpha x+12y-3\alpha=0

Note we have corrected it by adding the square to the y. Simplify by 3

x^2+y^2-2\alpha x+4y-\alpha=0

Complete squares and rearrange:

x^2-2\alpha x+y^2+4y=\alpha

x^2-2\alpha x+\alpha^2+y^2+4y+4=\alpha+\alpha^2+4

(x-\alpha)^2+(y+2)^2=r^2

We can see that, if r=4, then

\alpha+\alpha^2+4=16

Or, equivalently

\alpha^2+\alpha-12=0

There are two solutions for \alpha:

\alpha=-4,\ \alpha=3

Keeping the positive solution, as required:

\boxed{\alpha=3}

8 0
3 years ago
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