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taurus [48]
3 years ago
10

Find the value of x. Round to the nearest tenth.

Mathematics
1 answer:
Aliun [14]3 years ago
7 0

Answer:

Step-by-step explanation:

Triangle XYZ is a right angle triangle.

From the given right angle triangle

XZ represents the hypotenuse of the right angle triangle.

With ∠34 as the reference angle,

YZ represents the adjacent side of the right angle triangle.

XY represents the opposite side of the right angle triangle.

To determine x, we would apply the Cosine trigonometric ratio

Cos θ = opposite side/hypotenuse. Therefore,

Cos 34 = x/28

x = 28Cos34 = 28 × 0.8290

x = 23.212

To the nearest tenth, it becomes

23.2

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Which number is equivalent to 5√8 x 3√4<br><br>A. 15√2<br><br>B. 60√2<br><br>C. 30√3<br><br>D. 60√3​
Elena L [17]

Answer:

B

Step-by-step explanation:

5×sqrt(8)×3×sqrt(4) = 15×sqrt(8)×sqrt(4) = 15×sqrt(8)×2 =

= 30×sqrt(8) = 30×sqrt(4×2) =

= 30×2×sqrt(2) = 60×sqrt(2)

4 0
2 years ago
PLS HELP BRAINLEIST AND 25 POINTS
kramer

Given:

The vertices of a triangle ABC are A( 0, 2), B (2, 5), and C (−1, 7).

To find:

Part A: The length of each side of the triangle.

Part B: The slope of each side of the triangle.

Part C: Classify the triangle.

Solution:

Part A: The vertices of a triangle ABC are A( 0, 2), B (2, 5), and C (−1, 7).

Distance formula:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Using the distance formula, we get

AB=\sqrt{(2-0)^2+(5-2)^2}

AB=\sqrt{2^2+3^2}

AB=\sqrt{4+9}

AB=\sqrt{13}

Similarly,

BC=\sqrt{\left(-1-2\right)^2+\left(7-5\right)^2}

BC=\sqrt{13}

And,

AC=\sqrt{\left(-1-0\right)^2+\left(7-2\right)^2}

AC=\sqrt{26}

Therefore, the length of each side of the triangle are AB=\sqrt{13},BC=\sqrt{13},AC=\sqrt{26}.

Part B:

Slope formula:

m=\dfrac{y_2-y_1}{x_2-x_1}

Slope of side AB is:

m_{AB}=\dfrac{5-2}{2-0}

m_{AB}=\dfrac{3}{2}

Slope of side BC is:

m_{BC}=\dfrac{7-5}{-1-2}

m_{BC}=\dfrac{2}{-3}

m_{BC}=-\dfrac{2}{3}

Slope of side AC is:

m_{AC}=\dfrac{7-2}{-1-0}

m_{AC}=\dfrac{5}{-1}

m_{AC}=-5

Therefore, the slopes of sides AB, BC, AC are \dfrac{3}{2},-\dfrac{2}{3},-5 respectively.

Part C:

AB=BC=\sqrt{13}

Product of slopes of AB and BC is:

\dfrac{3}{2}\times \dfrac{-2}{3}=-1

So, side AB and BC are perpendicular to each other.

Two sides of triangle ABC are equal and one angle is a right angle.

Therefore, the triangle ABC is an isosceles right triangle.

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3 years ago
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