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2 years ago

BRAINILEST IF ANSWEED QUICKLY AND CORRECT

Mathematics

2 answers:

Anna [14]2 years ago

6 0

The answer is B. It is one-half the area of a square of side length 4 units.

r-ruslan [8.4K]2 years ago

4 0

Answer:

The answer is B.It is one-half the area of a square of side length 4 units.

Step-by-step explanation:

Given a coordinate grid shown with triangle of measurements with base 4 units and height also 4 units.

We have to find the area of triangle which can be calculated as

⇒

we have to choose the statement describes the area of the triangle evaluate above.

Option A: Not matched ∵equals to [area of rectangle of length 4 units width 2 units]=

Option B: matched : given one-half the area of a square of side length 4 units.

⇒

Option C : Not matched : It is twice the area of a rectangle of length 4 units and width 2 units

⇒

Option 4 : Not matched : It is twice the area of a square of side length 4 units.

⇒

So, the correct option is B. It is one-half the area of a square of side length 4 units.

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How do you turn this into a simplified fraction?

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-0.0103

Step-by-step explanation:

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2 years ago

Find the length of the third side. If necessary, write in simplest radical form. 1 2

tigry1 [53]

Answer:

√3

Step-by-step explanation:

Using the pythagorean theorem, we know that a^2 + b^2 = c^2

In this case, c = 2, which means that a^2 + b^2 = 4

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Therefore, b^2 = 3

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7 0

2 years ago

Read 2 more answers

Find Sin (BAC+30) 80POINTS

Ann [662]

Answer:

$\dfrac{2}{5}\sqrt{3}+\dfrac{3}{10}$

Step-by-step explanation:

Trigonometric Identities

$\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B$

Trigonometric ratios

$\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}$

where:

$\theta$ is the angle
O is the side opposite the angle
A is the side adjacent the angle
H is the hypotenuse (the side opposite the right angle)

Using the trig ratio formulas for cosine and sine:

$\cos (\angle BAC)=\dfrac{12}{20}$

$\sin (\angle BAC)=\dfrac{16}{20}$

Angles

$\sin (30^{\circ})=\dfrac{1}{2}$

$\cos (30^{\circ})=\dfrac{\sqrt{3}}{2}$

Therefore, using the trig identities and ratios:

$\begin{aligned}\sin (\angle BAC + 30^{\circ}) & = \sin (\angle BAC) \cos (30^{\circ})+\cos (\angle BAC) \sin (30^{\circ})\\\\& = \dfrac{16}{20} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{12}{20} \cdot \dfrac{1}{2}\\\\& = \dfrac{16}{40}\sqrt{3}+\dfrac{12}{40}\\\\& = \dfrac{2}{5}\sqrt{3}+\dfrac{3}{10}\end{aligned}$