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elena-14-01-66 [18.8K]
3 years ago
13

Triangle ABC is similar to Triangle FGH. What is the value of x in centimeters?

Mathematics
1 answer:
EleoNora [17]3 years ago
8 0

<u>Given</u>:

Given that the triangle ABC is similar to triangle FGH.

We need to determine the value of x.

<u>Value of x:</u>

Since, the triangles are similar, then their sides are proportional.

Thus, we have;

\frac{AC}{FH}=\frac{AB}{GF}=\frac{BC}{GH}

Let us consider the proportion \frac{AB}{GF}=\frac{BC}{GH} to determine the value of x.

Substituting AB = 9 cm, GF = 13.5 cm, BC = 15 cm and GH = x, we get;

\frac{9}{13.5}=\frac{15}{x}

Cross multiplying, we get;

9x=15 \times 13.5

9x=202.5

 x=22.5 \ cm

Thus, the value of x is 22.5 cm

Hence, Option F is the correct answer.

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Which of the following is the domain and range of the ellipse with equation x2 + 4y2 – 2x + 16y – 19 = 0?
Nata [24]

The option second D: [–5, 7]; R: [–5, 1] is correct the domain is D: [–5, 7], and the range is R: [–5, 1]

<h3>What is an ellipse?</h3>

An ellipse is a locus of a point that moves in a plane such that the sum of its distances from the two points called focus adds up to a constant. It is taken from the cone by cutting it at an angle.

We have an ellipse equation:

\rm x^{2}+4y^{2}-2x+16y-19=0

We can write the above equation as:

\rm \dfrac{\left(x-1\right)^{2}}{36}+\dfrac{\left(y+2\right)^{2}}{9}=1

The domain will be:

D: [–5, 7]

The range will be:

R: [–5, 1]

Thus, the option second D: [–5, 7]; R: [–5, 1] is correct the domain is D: [–5, 7] and the range is R: [–5, 1]

Learn more about the ellipse here:

brainly.com/question/19507943

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5 0
2 years ago
Find the components of the vertical force Bold Upper FFequals=left angle 0 comma negative 10 right angle0,−10 in the directions
quester [9]

Solution :

Let $v_0$ be the unit vector in the direction parallel to the plane and let $F_1$ be the component of F in the direction of v_0 and F_2 be the component normal to v_0.

Since, |v_0| = 1,

$(v_0)_x=\cos 60^\circ= \frac{1}{2}$

$(v_0)_y=\sin 60^\circ= \frac{\sqrt 3}{2}$

Therefore, v_0 = \left

From figure,

|F_1|= |F| \cos 30^\circ = 10 \times \frac{\sqrt 3}{2} = 5 \sqrt3

We know that the direction of F_1 is opposite of the direction of v_0, so we have

$F_1 = -5\sqrt3 v_0$

    $=-5\sqrt3 \left$

    $= \left$

The unit vector in the direction normal to the plane, v_1 has components :

$(v_1)_x= \cos 30^\circ = \frac{\sqrt3}{2}$

$(v_1)_y= -\sin 30^\circ =- \frac{1}{2}$

Therefore, $v_1=\left< \frac{\sqrt3}{2}, -\frac{1}{2} \right>$

From figure,

|F_2 | = |F| \sin 30^\circ = 10 \times \frac{1}{2} = 5

∴  F_2 = 5v_1 = 5 \left< \frac{\sqrt3}{2}, - \frac{1}{2} \right>

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Therefore,

$F_1+F_2 = \left< -\frac{5\sqrt3}{2}, -\frac{15}{2} \right> + \left< \frac{5 \sqrt3}{2}, -\frac{5}{2} \right>$

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3 0
2 years ago
One number is 4 times a first number. A third number is 100 more than the first number. If the sum of the three numbers is 538,
larisa [96]

Answer:

73

292

173

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x = first number

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x + 4x + 100 + x = 538

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6x + 100 - 100 = 538 - 100

6x = 438

6x/6 = 438/6

x = 73

Check:

73 + 4(73) + 100 + 73 = 538

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538 = 538

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