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3 years ago

True or false

Mathematics

2 answers:

faust18 [17]3 years ago

7 0

True
Hope this helped

seropon [69]3 years ago

4 0

It depends. Some functions can have inverses that are also functions, such as the function y = x. However, a function like y = x^2 would not have an inverse that is also a function.

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What multiplies to -26 and adds to 5

Kryger [21]

The two numbers are (5+ √(129))/2 and (5-√(129))/2.

(5+ √(129))/2+ (5-√(129))/2= 5
(5+ √(129))/2* (5-√(129))/2= -26

Hope this helps~

6 0

3 years ago

Consider two vectors A and B A=14i and B= -4i+8j

DaniilM [7]

Assuming $\mathbf a,\mathbf b\in\mathbb R^3$ , you have

$\mathbf a\cdot\mathbf b=(14)(-4)+(0)(8)=-56$

so

$((\mathbf a\cdot\mathbf b)\,\mathbf i)\cdot\mathbf a=(-56\,\mathbf i)\cdot(14\,\mathbf i)=(-56)(14)=-784$

Next,

$\mathbf a+\mathbf b=(14\,\mathbf i)+(-4\,\mathbf i+8\,\mathbf j)=10\,\mathbf i+8\,\mathbf j$

Then

$(\mathbf a+\mathbf b)\times\mathbf b=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\10&8&0\\-4&8&0\end{vmatrix}=112\,\mathbf k$

$((\mathbf a+\mathbf b)\times\mathbf b)\times\mathbf k=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\0&0&112\\0&0&1\end{vmatrix}=\mathbf 0$

4 0

3 years ago

Can someone help me pls

Simora [160]

Sorry I can not help u right now

4 0

3 years ago

Read 2 more answers

Plsss help me !!!!

Yuri [45]

Answer:

The enrollment after 5 years is 10,724

Step-by-step explanation:

Generally, we can have the depreciation formula written as follows;

A = P(1 - r)^t

A is the number of enrollment in after a certain number of years t

P is the initial population which is 13,500

r is the rate of depreciation which is 4.5% = 4.5/100 = 0.045

t = 5 years

Substituting these values, we have it that;

A = 13,500(1-0.045)^5

A = 10,723.84

4 0

3 years ago

Which functions are symmetric with respect to the y axis<br>

Inessa05 [86]

Answer:

A, B, and D

Step-by-step explanation:

Only the functions that have x by itself between the absolute value signs (A, B, and D) are symmetric with respect to the y-axis .

Placing a constant outside the absolute value signs moves the function up or down the y-axis but retains the symmetry.

Adding a constant inside the absolute value signs (as in C and E) moves the axis of symmetry to the left or right of the y-axis.

In the diagram, both A and B are symmetric with respect to the y-axis, but C has been shifted three units to the left.

4 0

3 years ago