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2 years ago

Find the area of a rectangle that has a length of 4/5 of a meter and a width of 3/5

Mathematics

1 answer:

Helen [10]2 years ago

6 0

Answer:

Step-by-step explanation:

Givens

L = 4/5 m

w = 3/5 m

Formula

A = L * w

Solution

Area = 4/5 * 3/5

Area = (4 *3 ) / (5 * 5)

Area = 12/ 25 m^2

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Identify the value of b. PLEASE HELP!!

Degger [83]

Answer:

b = 18

Step-by-step explanation:

From an external point, the products of distances to the near circle intercept and the far circle intercept are the same. For a tangent, such as AC, point A is both the near and far intercept point, so that product is the square of the length of AC.

(AC)² = (CG)(CV)

b² = 12·27 = 324 . . . . substitute known values

b = √324 . . . . . . . . . . take the square root

b = 18

7 0

3 years ago

State the property that is demonstrated: (3)(8) = (8)(3).

bagirrra123 [75]

(3)(8) = (8)(3)

answer is A. Commutative

4 0

3 years ago

Evaluate the expression. x2−4x for x=−1 no links or files pls.

qaws [65]

Answer:

${x}^{2} - 4x \\ {( - 1)}^{2} - 4( - 1) \\ 1 + 4 = 5$

I hope I helped you^_^

7 0

2 years ago

For which pair of triangles would you use AAS to prove the congruence of the 2 triangles

DerKrebs [107]

Answer:

The correct option is B.

Step-by-step explanation:

According to AAS congruence rule, two triangles are congruent if two angles and a non included side are congruent to corresponding angles and side of another triangle.

We need two angles and a non included side, to use AAS postulate.

In option A, two sides and their inclined angle are congruent, therefore these triangles are congruent by SAS postulate and option A is incorrect.

In option B, two angles and a non included side are congruent, therefore these triangles are congruent by AAS postulate and option B is correct.

In option C, two angles and their included side are congruent, therefore these triangles are congruent by ASA postulate and option C is incorrect.

In option D, all sides are congruent, therefore these triangles are congruent by SSS postulate and option D is incorrect.

8 0

3 years ago

Read 2 more answers

Given the graph of a function f. Identify the function by name. Then Graph, state domain & range in set notation:A) f(x) +2B

WARRIOR [948]

The function in the graph has the name of square function.

The domain of a function is all values of x the function can have. The domain of this function is all real numbers:

$\mleft\lbrace x\in\R\mright\rbrace$

The range of a function is all values of y the function can have. The range of this function is all positive numbers, including zero:

$\mleft\lbrace y\in\R\mright|y\ge0\}$

In order to graph f(x) + 2, we just need to translate the graph 2 units up. To find the new points, we need to increase all y-coordinates by 2:

(-2, 6), (-1, 3), (0, 2), (1, 3), (2, 6)

Domain: {x ∈ ℝ}

Range: {y ∈ ℝ | y ≥ 2}

Then, in order to graph f(x) - 2, we just need to translate the graph 2 units down. To find the new points, we need to decrease all y-coordinates by 2:

(-2, 2), (-1, -1), (0, -2), (1, -1), (2, 2)

Domain: {x ∈ ℝ}

Range: {y ∈ ℝ | y ≥ -2}

4 0

1 year ago