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3 years ago

Please help urgent!!! i have attached a picture of the question

Mathematics

2 answers:

rusak2 [61]3 years ago

5 0

The length of EH would be 4.8

joja [24]3 years ago

4 0

Answer:

Step-by-step explanation:

12.5 is 2.5 greater than 5 so, 2 x 2.5 = 5

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Helpppp please ?????

BlackZzzverrR [31]

A) <DAE = 180-126 = 54
b) <EBC = 90 - 48 = 42
c) <BAE = 180-48-54=78

8 0

2 years ago

Brigitte is shorter than 5 feet. (Brigitte's height=h). Help please

KonstantinChe [14]

Brigitte is shorter than 5 feet. (Brigitte's height=h)

We write this in inequality form

Brigitte's height is h

So 'h' is shorter than 5 feet

Shorter means less than so we use < symbol. We should not use = symbol because height is shorter than 5 not equal to 5.

Suppose if it is given ' height is more than 5 feet ' then we use greater than symbol (>)

h is less than 5 feet

h < 5

Inequality is h < 5

3 0

2 years ago

Read 2 more answers

A box was 3/4 full of marbles. The box fell on the floor. Thirty marbles fell out. This was 20% of the marbles in the box. How m

Hunter-Best [27]

Answer: -5.7

Step-by-step explanation:

3/4- 30= -29.25 and that was only 20% of the box that was dropped out of the box. -29.25 of 20% equals -5.85. 3/4 of 20% equals 0.15 subtract the 2 decimals and you will get -5.7

7 0

2 years ago

Write an expression that represents 5 squared.

Korvikt [17]

5 squared is 5^2.

Squared means the power of two, so exponent two.

4 0

2 years ago

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