All categories

3 years ago

Pre college need help

Mathematics

2 answers:

DanielleElmas [232]3 years ago

8 0

27) Function.

reasoning: it only touches the curve/line once. that’s a trick my old teachers taught me.

28) f(-8) =169

I hoped I helped. #28 is confusing so I sincerely apologize if I’m incorrect.

Goshia [24]3 years ago

3 0

Answer:

f(-8) = 169

Step-by-step explanation:

You might be interested in

A skyscraper is 1,150 feet above ground level. It extends 35 feet below ground level. What is the total length of the skyscraper

Ne4ueva [31]

1,150ft + 35ft = 1,185ft

The reason you wouldnt subtract would be that you cant have negative 35 ft. You add the two heights together to produce your answer.

6 0

3 years ago

SOMEONE PLEASE HELP ME ASAP PLEASE!!!

Strike441 [17]

Answer:

105.12 ft^2

Step-by-step explanation:

Area of a rectangle: bh

In this case 8*10.... so area of the rectangle is 80

Area of a circle: pir^2

Half it for a semicircle.

so 1/2 pi r^2

radius is 4 cuz its half of 8.

so 1/2(3.14)(4^2)=(0.5)(3.14)(16)=25.12

Now add up 80+25.12

Total is 105.12

Hope I helped :)

5 0

3 years ago

Triangle N M L is shown. Angle N M L is 90 degrees. The length of M L is 15, the length of N L is 25, and the length of N M is 2

lys-0071 [83]

Answer: sin(L) = 4/5, cos(L) = 3/5, tan(L) = 4/3

Step-by-step explanation:

i got it wrong so here's the correct answers for people who are trying to pass

6 0

3 years ago

Read 2 more answers

43 Points AND BRAINIEST NEED HELP ASAP What are the coordinates of point p?

Llana [10]

Answer:

taking a vector approach to the question

−−−→

−−→

⇒

(

−

−−

)

(

−

−−

)

⇒

−

−−

−

−−

⇒

−

−−

⇒

−

(

−

)

(

)

⇒

−

(

−

)

(

)

(

)

⇒

−

(

)

(

)

⇒

(

)

Sorry the format dident turn out how i wanted it to be, I hope this helps.

3 0

4 years ago

Read 2 more answers

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, (where A and B are nxn matrices and I is the identity matrix) :

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

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

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

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

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

$AI = IA = A$

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