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2 years ago

Find the area of the parallelogram below.

Mathematics

2 answers:

V125BC [204]2 years ago

8 0

The formula for finding the area of a parallelogram is Base x height

A = Bh

In this case the base (the longest side) is 8 cm and the height (distance between the obtuse angle and the base) is 5 cm

A = 8 * 5

A = 40 cm²

Hope this helps!

~Just a girl in love with Shawn Mendes

VLD [36.1K]2 years ago

6 0

Answer: 40cm²

The formula is Base × Height. We can find the area if we multiply the Base and Height.

A = Bh

As for that, the Base is the upper length, which is 8 and the height is the obtuse angle inside the parallelogram, which is 5.

Since we got our lengths, we can solve.

Remember how I said the formula is A = Bh. (Area = Base × Height)

So, we can multiply 8 by 5 to get the area.

8 times 5 will equal 40.

Hence, 40cm² is the answer.

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Help please!!!!!!!!!

In-s [12.5K]

ANSWER

EXPLANATION

For a matrix A of order n×n, the cofactor $C_{ij}$ of element $a_{ij}$ is defined to be

$C_{ij} = (-1)^{i+j} M_{ij}$

$M_{ij}$ is the minor of element $a_{ij}$ equal to the determinant of the matrix we get by taking matrix A and deleting row i and column j.

Here, we have

$C_{11} = (-1)^{1+1} M_{11} = M_{11}$

M₁₁ is the determinant of the matrix that is matrix A with row 1 and column 1 removed. The bold entries are the row and the column we delete.

$\begin{aligned} A=\begin{bmatrix} \bf 1 & \bf -6 & \bf -4\\ \bf 7 & 0 & -3 \\ \bf -9 & 8 & -8 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 0&-3 \\ 8&-8 \end{bmatrix} \right) \end{aligned}$

Since the determinant of a 2×2 matrix is

$\det\left( \begin{bmatrix} a & b \\ c& d \end{bmatrix} \right) = ad-bc$

it follows that

so $C_{11} = M_{11} = 24$

4 0

3 years ago

137x8 multiply with expanded form please help 10 ponte for your help please help!

Step2247 [10]

Answer: maybe 20

Step-by-step explanation:

or find it out yourself

3 0

2 years ago

ivann1987 [24]

Answer:

AA SIMILARITY statement

SSS similarity statement is also possible.

Step-by-step explanation:

$In \: \triangle CBA \: and \: \triangle TVU \\ \angle B \cong \angle V... (given) \\\angle A \cong \angle U... (given) \\\therefore \triangle CBA \sim \triangle TVU... (By AA\: Postulate) \\$