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2 years ago

When a base is not indicated, what is the base equal to?

Mathematics

2 answers:

Damm [24]2 years ago

8 0

Answer:

Step-by-step explanation:

if it is in log then 10

Bezzdna [24]2 years ago

7 0

Answer:

Dude its me again, most of your questions have too little information, and this one does not have a picture.

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Mariana wears her winter coat when the temperature is colder than -4 degrees Celsius

zhuklara [117]

Answer:

It is very cool that Mariana wears her winter coat when the temperature is colder than -4 degrees Celsius

that means when it is 10 degrees Celsius, she won't wear her winter coat

4 0

2 years ago

Enter the equivalent distance in km in the box.

devlian [24]

100 cm = 1m

To find 60 000cm, multiply by 600 :
60 000 cm = 600m

To convert from m to km, divide by 1000 :
600m ÷ 1000 = 0.6km

3 0

2 years ago

Solve the equation <br><br>9y^3+8y=4y+18y^2

kari74 [83]

9y^3+8y=4y+18y^2
9y^3-18y^2+4y=0

*Divide the whole thing by y

9y^2-18y+4=0

*use quadratic formula to solve

3 0

2 years ago

Resolve the factors ( xy+z)^2 (y-xz)^2

skelet666 [1.2K]

Answer:

=x4y2z2−2x3y3z+2x3yz3+x2y4−4x2y2z2+x2z4+2xy3z−2xyz3+y2z2

Make me brainliest

5 0

2 years ago

Read 2 more answers

Find the inverse of the given matrix, if it exists.Aequals=left bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 0 3

BabaBlast [244]

Answer:

$A^{-1}=\left[ \begin{array}{ccc} \frac{1}{9} & \frac{4}{27} & - \frac{2}{27} \\\\ \frac{8}{9} & \frac{5}{27} & \frac{11}{27} \\\\ - \frac{4}{9} & \frac{2}{27} & - \frac{1}{27} \end{array} \right]$

Step-by-step explanation:

We want to find the inverse of $A=\left[ \begin{array}{ccc} 1 & 0 & -2 \\\\ 4 & 1 & 3 \\\\ -4 & 2 & 3 \end{array} \right]$

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 4&1&3&0&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]$

Subtract row 1 multiplied by 4 from row 2

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]$

Add row 1 multiplied by 4 to row 3

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&2&-5&4&0&1\end{array}\right]$

Subtract row 2 multiplied by 2 from row 3

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&-27&12&-2&1\end{array}\right]$

Divide row 3 by −27

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

Add row 3 multiplied by 2 to row 1

$\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

Subtract row 3 multiplied by 11 from row 2

$\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&0&\frac{8}{9}&\frac{5}{27}&\frac{11}{27} \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

As can be seen, we have obtained the identity matrix to the left. So, we are done.

6 0

2 years ago