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3 years ago

Help pls 25 points i don’t understand

Mathematics

1 answer:

Brums [2.3K]3 years ago

8 0

Calm Down this is a nasty proof

CB is parallel to ED (Given)

CB is congruent to ED (Given)

Angle CBF is congruent to Angle EDF (Vertical Angles)

Angle DEF is congruent to Angle BCF (Alternate Interior Angles)

Angle EDF is congruent to Angle CBF (Alternate Interior Angles)

Therefore we now have enough information to solve. We have a side angle side triangle here so.....

Triangle CBF is congruent to Triangle EDF because of the Side Angle Side Theorem.

Good Luck and Your Welcome

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Step-by-step explanation:

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3 years ago

Read 2 more answers

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

3 years ago

Simplify the expression (-3)(8)/(-6)

Daniel [21]

Answer:

Step-by-step explanation:

(-3) * (8/-6)

3 * 8/6

3 * (2 * 4)/6

3 * (2 * 4)/(2 * 3)

3 * (4/3)

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3 years ago

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ludmilkaskok [199]

Answer:

$x^{2} + 7x+12=0$

Step-by-step explanation:

Roots of the equation are -4 and -3.

Quadratic equation can be written as

$(x+4)(x+3)=0\\x(x+3)+4(x+3)\\x^{2} +3x+4x+12\\x^{2} + 7x+12=0$

4 0

3 years ago