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3 years ago

20 points repost because no one answered. Geomertry question

Mathematics

2 answers:

aniked [119]3 years ago

7 0

Answer:

x=2.25

Step-by-step explanation:

ΔABC and ΔADE are similar.

∴ sides are proportional.

$\frac{5}{1} =\frac{x+9}{x} \\5=\frac{x}{x} +\frac{9}{x} \\5=1+\frac{9}{x}\\\frac{9}{x}=5-1\\\frac{9}{x}=4\\x=\frac{9}{4}=2.25$

Ira Lisetskai [31]3 years ago

4 0

Answer:

x = 9/4

Step-by-step explanation:

We know that the triangles are similar

We can use ratios to solve

x x+9

----- = ----------------

1 5

Using cross products

5x = 1(x+9)

5x = x+9

Subtract x

5x-x = x+9-x

4x= 9

Divide by 4

4x/4 = 9/4

x = 9/4

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2 years ago

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3 years ago

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A drama club made $3,200 from a school play. Seventy-five percent of that amount came from ticket sales. Identify the values of

NISA [10]

Answer:

75%

Step-by-step explanation:

For this case, the first thing we will do is name variables:

a: percentage from ticket sales

b: ticket sales

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a = 75%

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Answer:

a = 75%

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1 year ago

Which is the equation of a line that has a slope of mc005-1.jpg and passes through point (2, –3

goldenfox [79]

Id say B, y=-2/3+3
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3 years ago

Someone please help me with this lol… have no idea what I’m doing

Sholpan [36]

Given:

$\cos \theta =\dfrac{3}{5}$

$\sin \theta$

To find:

The quadrant of the terminal side of $\theta$ and find the value of $\sin\theta$ .

Solution:

We know that,

In Quadrant I, all trigonometric ratios are positive.

In Quadrant II: Only sin and cosec are positive.

In Quadrant III: Only tan and cot are positive.

In Quadrant IV: Only cos and sec are positive.

It is given that,

$\cos \theta =\dfrac{3}{5}$

$\sin \theta$

Here cos is positive and sine is negative. So, $\theta$ must be lies in Quadrant IV.

We know that,

$\sin^2\theta +\cos^2\theta =1$

$\sin^2\theta=1-\cos^2\theta$

$\sin \theta=\pm \sqrt{1-\cos^2\theta}$

It is only negative because $\theta$ lies in Quadrant IV. So,

$\sin \theta=-\sqrt{1-\cos^2\theta}$

After substituting $\cos \theta =\dfrac{3}{5}$ , we get

$\sin \theta=-\sqrt{1-(\dfrac{3}{5})^2}$

$\sin \theta=-\sqrt{1-\dfrac{9}{25}}$

$\sin \theta=-\sqrt{\dfrac{25-9}{25}}$

$\sin \theta=-\sqrt{\dfrac{16}{25}}$

$\sin \theta=-\dfrac{4}{5}$

Therefore, the correct option is B.

6 0

2 years ago