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jasenka [17]
1 year ago
14

GEOMETRY - FIND X ON THIS TRIANGLE

Mathematics
1 answer:
m_a_m_a [10]1 year ago
8 0

Answer: \frac{7\sqrt{3}}{2}

Step-by-step explanation:

\sin 60^{\circ}=\frac{x}{7}\\\\\frac{\sqrt{3}}{2}=\frac{x}{7}\\\\x=\frac{7\sqrt{3}}{2}

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Use the graph of f "(x) below to state x-coordinates of the inflection points for the graph of f(x).
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If the area of a rectangle is 24a2b and the length is 8ab2, what would be the width of the rectangle, given that width is found
Goshia [24]
(24ba^2)/(8ab^2)=3a/b=width
5 0
3 years ago
Write an expression that is equivalont to 4(5+2)-3h,​
zhuklara [117]

Answer:

28 - 3h

Step-by-step explanation:

We must do the work seen inside parentheses first.

Thus, we havve 4(7) - 3h, which simplifies to

28 - 3h

8 0
3 years ago
Triangle PQR is transformed to triangle P'Q'R'. Triangle PQR has vertices P(4, 0), Q(0, −4), and R(−8, −4). Triangle P'Q'R' has
Whitepunk [10]

Answer:

Please find attached the required plot accomplished with an online tool

Part A:

1/4

Part B:

P''(-1, 0),  Q''(0, -1), and R''(2, -1)

Part C:

Triangle PQR is similar to triangle P''Q''R'' but they are not congruent

Step-by-step explanation:

Part A:

Triangle ΔPQR has vertices P(4, 0), Q(0, -4), R(-8, -4)

Triangle ΔP'Q'R' has vertices P'(1, 0), Q'(0, -1), R'(-2, -1)

The dimensions of the sides of the triangle are given by the relation;

l = \sqrt{\left (y_{2}-y_{1}  \right )^{2}+\left (x_{2}-x_{1}  \right )^{2}}

Where;

(x₁, y₁) and (x₂, y₂) are the coordinates on the ends of the segment

For segment PQ, we place (x₁, y₁) = (4, 0) and (x₂, y₂) = (0, -4);

By substitution into the length equation, we get;

The length of segment PQ = 4·√2

The length of segment PR = 4·√10

The length of segment RQ = 8  

The length of segment P'Q' = √2

The length of segment P'R' = √10

The length of segment R'Q' = 2

Therefore, the scale factor of the dilation of ΔPQR to ΔP'Q'R' is 1/4

Part B:

Reflection of (x, y) across the y-axis gives;

(x, y) image after reflection across the y-axis = (-x, y)

The coordinates after reflection of P'(1, 0), Q'(0, -1), R'(-2, -1) across the y-axis is given as follows;

P'(1, 0) image after reflection across the y-axis = P''(-1, 0)

Q'(0, -1) image after reflection across the y-axis = Q''(0, -1)

R'(-2, -1) image after reflection across the y-axis = R''(2, -1)

Part C:

Triangle PQR is similar to triangle P''Q''R'' but they are not congruent as the dimensions of the sides of triangle PQR and P''Q''R'' are not the same.

6 0
3 years ago
Can someone please help me with this? Mhanifa? I will mark brainliest
den301095 [7]

Answer:

Option C 20

GOOD LUCK FOR THE FUTURE! :)

8 0
2 years ago
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