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2 years ago

13 If CDEF is a rhombus, find m FED.

Mathematics

1 answer:

Tatiana [17]2 years ago

8 0

Answer:

Step-by-step explanation:

8x-20=5x+1

-5 -5

3x=-20+1

+20 +20

3x=21

3 3

x=7

8(7)-20

FED=36

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The value pi/4 is a solution for the equation 3 sqrt 2 cos theta+2=-1

natulia [17]

Answer:

<h2>FALSE</h2>

Step-by-step explanation:

$3\sqrt2\cos\theta+2=-1\\\\\text{Method 1}\\\\\text{Put}\ \theta=\dfrac{\pi}{4}\ \text{to the equation and check the equality:}\\\\\cos\dfrac{\pi}{4}=\dfrac{\sqrt2}{2}\\\\L_s=3\sqrt2\cos\dfrac{\pi}{4}+2=3\sqrt2\left(\dfrac{\sqrt2}{2}\right)+2=\dfrac{(3\sqrt2)(\sqrt2)}{2}+2\\\\=\dfrac{(3)(2)}{2}+2=3+2=5\\\\R_s=-1\\\\L_s\neq R_s\\\\\boxed{FALSE}$

$\text{Method 2}\\\\\text{Solve the equation:}\\\\3\sqrt2\cos\theta+2=-1\qquad\text{subtract 2 from both sides}\\\\3\sqrt2\cos\theta=-3\qquad\text{divide both sides by}\ 3\sqrt2\\\\\cos\theta=-\dfrac{3}{3\sqrt2}\\\\\cos\theta=-\dfrac{1}{\sqrt2}\cdot\dfrac{\sqrt2}{\sqrt2}\\\\\cos\theta=-\dfrac{\sqrt2}{2}\to\theta=\dfrac{3\pi}{4}+2k\pi\ \vee\ \theta=-\dfrac{3\pi}{4}+2k\pi\ \text{for}\ k\in\mathbb{Z}\\\\\text{It's not equal to}\ \dfrac{\pi}{4}\ \text{for any value of }\ k.$

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2 years ago

3x + (x-8) = 23 how do I solve it

Ostrovityanka [42]

<span>3x + (x-8) = 23

Combine like terms first. You can combine 3x and x because they have the same variable.

4x - 8 = 23
Now add 8 to both sides.
4x = 31

Last step is to divide both sides by 4 to isolate the x.
4/4x = 31/4

x = 7.75 is the answer.</span>

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3 years ago

The amount of paper that the copy shop used this month was a 25% increase of the amount of paper they used last month. What perc

likoan [24]

Answer:

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If last month's consumption is represented by 4 equal shapes, then the current month's consumption will be 5 shapes.

If last month's consumption is represented by 8 equal shapes, then the current month's consumption will be 10 shapes.

this pattern keeps repeating.

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3 years ago

What context can be represented with the numbers 4 and -14?

DanielleElmas [232]

Is there anything else that goes with this question like a picture or something

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2 years ago

Consider ΔWXY and ΔBCD with ∠X ≅∠C, WX ≅ BC, and WY ≅ BD.

LenKa [72]

Answer:- B. No, because the corresponding congruent angles listed are not the included angles.

Explanation:-

Given:- ΔWXY and ΔBCD with ∠X ≅∠C, WX ≅ BC, and WY ≅ BD.

Now, look at the attachment

We can see that ∠X and ∠C are not included angles by the corresponding equal sides.

⇒ We cannot use SAS postulate to show ΔWXY ≅ ΔBCD .

⇒ B is the right option.

SAS postulate tells the if two sides of a triangle and their included angle is equal to the two sides of a triangle and their included angle of another triangle then the two triangles are congruent.

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3 years ago

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