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Leya [2.2K]
3 years ago
7

Work out the size of angle x.x35​

Mathematics
1 answer:
Naddik [55]3 years ago
5 0

Answer:

145.

Step-by-step explanation:

x+35=180

x=180-35

x=145

please mark as brainliest.

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Find k so that the following function is continuous on any interval: . . f(x)=kx if 0<=x<3 and f(x)=8x^2 if 3<=x. . K=?
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4 years ago
The following is a special trapezoid. Given AB || DC and m∠A = 113° Find the measure of ∠D.
Alexandra [31]
The answer is 67°.
Since the parallel sides of a trapezoid are called bases, the bases in our given trapezoid are AB and DC. We know that the corresponding pairs of base angles, such as ∠A and ∠D, or ∠B and ∠C, are supplementary, therefore, their angles add up to 180 degrees:
     ∠A + ∠D = 180°
     ∠B + ∠C = 180°
Given that m∠A = 113°, we can calculate for the measure of ∠D:
     113° + ∠D = 180°
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6 0
3 years ago
Use a 30degrees and -60degrees right triangle to find the exact value of the following trigonometric expression. Cot 30 degrees=
LuckyWell [14K]

The dirty little secret about trig is students mostly just need to understand the 45/45/90 right isosceles triangle and the 30/60/90 triangle, which is half an isosceles triangle.  

I'm sure you have seen it many times already, so I'll remind you in the 30/60/90 right triangle the sides opposite those angles are respectively in the ratio 1: \sqrt{3}:2.  Remember the smallest side is always opposite the smallest angle, etc.   These make a right triangle because

1^2 + \sqrt{3}^2 = 2^2

So in this hypotenuse 2 triangle, the length 1 side is opposite the 30 degree angle and the \sqrt{3} side is adjacent to it.

We could jump right to the answer but let's enumerate all the trig functions.  Really you should memorize the sine and cosine so you can get these quickly.

\cos 30^\circ = \dfrac{ \textrm{adjacent} }{ \textrm{hypotenuse} } = \dfrac{\sqrt{3}}{2}

\sin 30^\circ = \dfrac{ \textrm{opposite} }{ \textrm{hypotenuse} } = \dfrac{1}{2}

\tan 30^\circ = \dfrac{ \textrm{opposite} }{ \textrm{adjacent} }= \dfrac{1}{\sqrt{3}} =\dfrac{\sqrt{3}}{3}

\cot 30^\circ = \dfrac{ \textrm{adjacent} }{ \textrm{opposite}} = \dfrac{\sqrt{3}}{1} =\sqrt{3}

\sec 30^\circ = \dfrac{ \textrm{hypotenuse} }{ \textrm{adjacent}}= \dfrac{2}{\sqrt{3}}

\csc 30^\circ = \dfrac{ \textrm{hypotenuse} }{ \textrm{opposite}}= \dfrac{2}{1} = 2

We want the cotangent, another way to get it is

\cot 30^\circ = \dfrac{\cos 30^\circ}{\sin 30^\circ} = \dfrac{\sqrt{3}/2}{1/2} = \sqrt{3}

8 0
3 years ago
Read 2 more answers
PLEASE HELP ME OUT!!!
Shkiper50 [21]

Answer:

154

Step-by-step explanation:

Area = Perimeter times Apothem divided by 2

A = \frac{p x a}{2} = 28 x 11 / 2 = 154

6 0
2 years ago
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