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3 years ago

Help please, its pre-pre algebra

Mathematics

2 answers:

iren [92.7K]3 years ago

6 0

Answer:

The probability that it lands within the red triangle is 8/81.

Lyrx [107]3 years ago

5 0

8/74 l think it that

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If the ratio of red hair bands to green hair bands is 5 to 9 with a total of 70, how many are green

Eduardwww [97]

$5x-\ red\\9x-\ green\\\\5x+9x=70\\
14x=70\ \ \ | divide\ by\ 14\\\\
x=5\\\\
9x=9*5=45\\\\
There\ are\ 45\ green.$

5 0

3 years ago

A cylinder and its dimensions are shown. Which equation can be used to find V, the volume of the cylinder in cubic centimeters?

irina [24]

Answer:

The volume of a cylinder is given by the equation:

V = π(R)^^2*h

where π = 3.14, R is the radius and h is the height.

If in the image we see that the radius is 7, then the correct option is D.

If in the image we shee that the DIAMETER is 7 (a linea that goes from end to end of the circle and goes trough the middle) then we know that the radus is half of the diameter, 7/2 = 3.5

in this case he correct option is C.

6 0

3 years ago

Select all statements below which are true for all invertible n×n matrices A and B

Mekhanik [1.2K]

Answer:

a. False

b. False

c. True

d. True

e. False

f. False

Step-by-step explanation:

Hi,

We have certain properties for matrices,<em> (where A and B are nxn matrices and I is the identity matrix) </em>:

${(A^{-1})}^{-1} = A$

$(AB)^{-1} = B^{-1}A^{-1}$

$(A')^{-1} = (A^{-1})'$

$(A^{n})^{-1} = (A^{-1})^{n} = A^{-n}$

$AA^{-1} = A^{-1}A = I$

$AI = IA = A$

Using these properties, we verify the provided statements:

A. False.

None of the properties help verify this statement. We ca use an example for counter:

Let $A =\left[\begin{array}{cc}1&2\\2&0\\\end{array}\right]$ and $B = \left[\begin{array}{cc}5&1\\3&2\end{array}\right]$ , we calculate the L.H.S:

$A+B = \left[\begin{array}{cc}1+5&2+1\\2+3&0+2\end{array}\right]\\= \left[\begin{array}{cc}6&3\\5&2\end{array}\right]$

The square of (A+B):

$(A + B)^{2} = \left[\begin{array}{cc}36&9\\25&4\end{array}\right]$

Lets calculate the R.H.S:

$A^{2} =\left[\begin{array}{cc}1&4\\4&0\\\end{array}\right]\\B^{2} = \left[\begin{array}{cc}25&1\\9&4\end{array}\right]\\2AB = \left[\begin{array}{cc} (1 \times 5) + (2 \times 3) &(1 \times 1) + (2 \times 2)\\(2 \times 5) + (0 \times 3)& (2 \times 1) + (0 \times 2)\end{array}\right]\\= \left[\begin{array}{cc} 11 &5\\10& 2\end{array}\right]$

$A^{2} + B^{2} + 2AB = \left[\begin{array}{ccc}1+25+11&4+1+5 \\4+9+10&0+4+2\\\end{array}\right] \\= \left[\begin{array}{ccc}37&10 \\23&6\\\end{array}\right]$

This proves that: L.H.S ≠ R.H.S

Hence, A is false.

B. False

This can only hold when the eigenvalues for A are real.

$trace (A^{2}) > 0, det (A^{2}) > 0 : \\(A + A^{-1}) = ( I + A^{2} ) A^ {- 1} = ( A ( I + A ^{2} )^ {-1})^ {-1}$

C. True

This is a simplification of the distribution property of matrices.

D. True

The property that inverse is possible for any "n" value of the matrix.

E. False

Similar to part A, we can show that this property is invalid for any nxn matrix. Let:

$A = \left[\begin{array}{cc}1&2\\0&1\end{array}\right] \\A^{-1} = \left[\begin{array}{cc}1&-2\\0&1\end{array}\right]$

L.H.S:

$A + A^{-1} = \left[\begin{array}{cc}1+1&2-2\\0+0&1+1\end{array}\right] = \left[\begin{array}{cc}2&0\\0&2\end{array}\right]$

$(A + A^{-1})^{9} = \left[\begin{array}{cc}512&0\\0&512\end{array}\right]$

R.H.S:

$A^{9} = \left[\begin{array}{cc}1&512\\0&1\end{array}\right] \\A^{-9}= \left[\begin{array}{cc}1&-0.001953125\\0&1\end{array}\right]\\\\\\A^{9} + A^{-9} = \left[\begin{array}{cc}2&512\\0&2\end{array}\right] \\$

Since, L.H.S ≠ R.H.S, the statement is false.

F. False

This is a basic matrix rule, that commutative property does not apply on matrices.

3 0

3 years ago

How does a globe represent the fact that there are no parallel lines in elliptical geometry? A. The equator is not parallel to a

Vlada [557]

Answer:

The correct option is;

D. The geodesics connecting the North as South poles intersect at both of the poles

Step-by-step explanation:

As an analog to the straight line, the geodesic of a curved surface is the shortest path between two points on the curved surface

Whereby the Earth is assumed to be spherical, the geodesics around the Earth would then be closed circles and like the longitude and latitude, will meet at the North and South poles

However given that the Earth is an Oblate ellipsoid, Euclid's parallel postulate is not in effect and the geodesics connecting the North and South Poles Intersect at both poles.

7 0

4 years ago

The circumference of the hub cap of a tire is 82.58 centimeters.

Aleksandr-060686 [28]

where is the rest of the question? what do I have to do?

3 0

3 years ago