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3 years ago

What is the area of the kite?

Mathematics

2 answers:

Elena-2011 [213]3 years ago

6 0

the most reasonable answer would be c

Anastaziya [24]3 years ago

4 0

Answer:

Step-by-step explanation: 6*6=36×1/2=18

6*6=36×1/2=18

6*10=60×1/2=30

30+30+18+18

60+36=96

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The isosceles triangle has a perimeter of 7.5 m.

shepuryov [24]

D) 2.1+2x=7.5

All the sides added together will equal the perimeter

7 0

3 years ago

Read 2 more answers

Kinda need some help

aksik [14]

Answer:

1) x=10

2) x=6

3) x=13

Step-by-step explanation:

1) 15x-17+4x+7=180

19x-10=180

19x=190

x=10

2) 13x+5=16x-13

5+13=16x-13x

18=3x

x=6

3) 9x+7=11x-19

7+19=11x-9x

26=2x

x=13

6 0

3 years ago

Write down mrs Chauke balance on the 25 march 2021 in words

Yanka [14]

Answer:

A line passes through the points (-4,1) and (-3,3). This line can be modeled by the equation y

3 0

2 years ago

If ADB = 70°,

Delvig [45]

Diameter of the circle is BD as shown, as indicated and passed through center where BD passes, triangle ABD is a right triangle with angle ABD as the right angle. It means ADB and ABD are complementary which means
AB = 140 degree
AD=40 degree

7 0

3 years ago

[10] In the following given system, determine a matrix A and vector b so that the system can be represented as a matrix equation

irina1246 [14]

Answer:

$y=-\frac{158}{579}$

Step-by-step explanation:

To find the matrix A, took all the numeric coefficient of the variables, the first column is for x, the second column for y, the third column for z and the last column for w:

$A=\left[\begin{array}{cccc}1&1&2&2\\-7&-3&5&-8\\4&1&1&1\\3&7&-1&1\end{array}\right]$

And the vector B is formed with the solution of each equation of the system: $b=\left[\begin{array}{c}3\\-3\\6\\1\end{array}\right]$

To apply the Cramer's rule, take the matrix A and replace the column assigned to the variable that you need to solve with the vector b, in this case, that would be the second column. This new matrix is going to be called $A_{2}$ .

$A_{2}=\left[\begin{array}{cccc}1&3&2&2\\-7&-3&5&-8\\4&6&1&1\\3&1&-1&1\end{array}\right]$

The value of y using Cramer's rule is:

$y=\frac{det(A_{2}) }{det(A)}$

Find the value of the determinant of each matrix, and divide:

$y==\frac{\left|\begin{array}{cccc}1&3&2&2\\-7&-3&5&-8\\4&6&1&1\\3&1&-1&1\end{array}\right|}{\left|\begin{array}{cccc}1&1&2&2\\-7&-3&5&-8\\4&1&1&1\\3&7&-1&1\end{array}\right|} =\frac{158}{-579}$

$y=-\frac{158}{579}$

7 0

3 years ago