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anyanavicka [17]
3 years ago
15

How do you round 9.246 to the nearest tenth?

Mathematics
1 answer:
Flauer [41]3 years ago
3 0
You look at the hundreths place (4) and if it is 5 or more you round the tenths place up. so the number is rounded to 9.2
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Which type of triangle will always have exactly 1-fold reflectional symmetry? right triangle obtuse triangle equilateral triangl
ivolga24 [154]
Isosceles triangle ig
8 0
3 years ago
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A. The slope of the line will change from positive to negative
Novay_Z [31]

Answer:

The slope of the line will change from negative to positive

Step-by-step explanation:

the equation is in the form of y=mx+c and m is negative. By multiplying m with another negative number it turns positive, thus slope changes from negative to positive

3 0
3 years ago
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HELP ME PLEASE............................................................
mestny [16]

Answer:

2√6

Step-by-step explanation:

find the hypotenuse of the triangle thats 3 and 7

pythagorean

3 squared + 7 squared

16 + 49 = 57

√57 is the hypotenuse

√57 is also the base of the triangle that has side thats 9

pythagorean theorem

x squared + √57 squared = 9 squared

x^2 + 57 = 81

x^2 = 81 - 57

x^2 = 24

x = √24

x = √4 * √6

x = 2√6

7 0
2 years ago
The product of twice a number and six is the same as the difference of eleven times the number and 6/5 . Find the number
Rom4ik [11]

Answer:

-6/5

Step-by-step explanation:

solution;

let,the number be x,

product of <u>twice</u><u> </u><u>of</u><u> </u><u>x</u> and <u>six</u> = the <u>difference </u>

<u>difference of eleven times x and 6/5</u>

or, (2x×6)=(11×x)-(6/5)

or, 12x = 11x - 6/5

or, 12x - 11x = -6/5

therefore,x =-6/5

Therefore, the number required is (-6/5) whose product of twice a number and six is the same as the difference of eleven times the number and 6/5 .

3 0
4 years ago
Determine whether the set of vectors <img src="https://tex.z-dn.net/?f=%20v_%7B1%3D%283%2C2%2C1%29%2C%20v_%7B2%7D%20%3D%28-1%2C-
Korolek [52]
Since each vector is a member of \mathbb R^3, the vectors will span \mathbb R^3 if they form a basis for \mathbb R^3, which requires that they be linearly independent of one another.

To show this, you have to establish that the only linear combination of the three vectors c_1\mathbf v_1+c_2\mathbf v_2+c_3\mathbf v_3 that gives the zero vector \mathbf0 occurs for scalars c_1=c_2=c_3=0.

c_1\begin{bmatrix}3\\2\\1\end{bmatrix}+c_2\begin{bmatrix}-1\\-2\\-4\end{bmatrix}+c_3\begin{bmatrix}1\\1\\-1\end{bmatrix}=(0,0,0)\iff\begin{bmatrix}3&-1&1\\2&-2&1\\1&-4&-1\end{bmatrix}\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}

Solving this, you'll find that c_1=c_2=c_3=0, so the vectors are indeed linearly independent, thus forming a basis for \mathbb R^3 and therefore they must span \mathbb R^3.
4 0
3 years ago
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