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Ulleksa [173]
3 years ago
10

What digit is in the hundredth place of 6.589

Mathematics
2 answers:
yuradex [85]3 years ago
7 0
8 is the digit in the hundredth place.  Note: it goes decimal tenth hundredth thousandth.
Nimfa-mama [501]3 years ago
6 0
The 8 is in the hundredth place.
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PLEASE ANSWER ASAP Which function has a constant additive rate of change of –1/4?
Hatshy [7]

Answer:

the first graph

Step-by-step explanation:

The first one  - y = 1/4x + 3/2.

sorry im a bit late but hopefully this helps

6 0
3 years ago
Someone Please Help Me On This
luda_lava [24]
1) 18
2) 4
3) 27
4) 9
5) x=8
6) x=6
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Similar:<br> Similarity Statement:<br> Scale Factor:<br> Please help, NO LINKS
nalin [4]

Answer:

No similarity and no scale factor (I could be wrong)

Step-by-step explanation:

Don't worry, no links :)

You would see if they are similar if they have similar sides. so if there is an equal ratio, to both, they are similar. Sometimes they may not look similar until you rotate them. So for the following, you can see that if E were on the bottom, it would look like the triangle with N and M on the bottom you can ensure this to look at the ratios of each side. To find the scale factor, it depends on which way you are going. are you going from GEF to MNL or MNL to GEF? To me, it doesn't look like there is a scale factor, but I could be wrong.

8 0
3 years ago
Find the value of k for x^2 - 2kx+7k -12=0 so that they have two equal roots
Alisiya [41]

Step-by-step explanation:

x^2 -2kx +7k -12 = 0

two equal roots -->Δ = b^2 -4ac = 4k^2 - 4(7k-12)=0 = 4(k^2 - 7k +12)=4(k-4)(k-3)=0

so k =3 and k =4

5 0
3 years ago
For waht values of x do the vectors -1,0,-1), (2,1,2), (1,1, x) form a basis for R3?
DaniilM [7]
<h2>Answer:</h2>

The values of x for which the given vectors are basis for R³ is:

                        x\neq 1

<h2>Step-by-step explanation:</h2>

We know that for a set of vectors are linearly independent if the matrix formed by these set of vectors is non-singular i.e. the determinant of the matrix formed by these vectors is non-zero.

We are given three vectors as:

(-1,0,-1), (2,1,2), (1,1, x)

The matrix formed by these vectors is:

\left[\begin{array}{ccc}-1&2&1\\0&1&1\\-1&2&x\end{array}\right]

Now, the determinant of this matrix is:

\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-1(x-2)-2(1)+1\\\\\\\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-x+2-2+1\\\\\\\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-x+1

Hence,

-x+1\neq 0\\\\\\i.e.\\\\\\x\neq 1

4 0
3 years ago
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