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MatroZZZ [7]
2 years ago
9

ABC IS A TRINAGLE WORK OUT ANGLE X GIVE YOUR ANSWER CORRECT TO 3 S.F​

Mathematics
1 answer:
kati45 [8]2 years ago
5 0

Answer:

2 {2 =  | \leqslant 3 > 3 |2 >  \times 2 \frac{?}{ \tan( \sec(\pi\% log_{?}( \beta ) ) ) } | | }^{?}

7=9/8ths

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Please Help! In which of the following intervals does the trigonometric inequality sec(x)
timofeeve [1]

Step-by-step explanation:

sec(x) = 1/cos(x)

cot(x) = cos(x)/sin(x)

sec is smaller than cot, if cos is negative (making sec negative) and sin is negative (making cot positive).

and that is true for the third quadrant.

which is pi < x < 3×pi/2

so, the third answer is right.

3 0
2 years ago
Please help me with this
Zolol [24]

Answer:

294.5 m²

Step-by-step explanation:

shaded region = area of major sector + area of triangle

central angle of major sector = 360° - 130° = 230°

area of sector = area of circle × fraction of circle

A = π × 11.1² × \frac{230}{360}

   = \frac{x11.1^2(230)\pi }{360} ≈ 247.3 m²

area of triangle = 0.5 × 11.1 × 11.1 × sin130° ≈ 47.2 m²

shaded area = 247.3 + 47.2 ≈ 294.5 m²

7 0
2 years ago
Read 2 more answers
Write the slope-intercept form of the equation of the line that passes through the given points.
zmey [24]

Answer:

Slope=1

Step-by-step explanation:

Slope=y1-y2/x1-x2

Where x1,y1= (1, 1) and X2,y2=(-3, -3)

Slop=1+3/1+4

=4/4

=1

So the slope is 1.

8 0
2 years ago
Read 2 more answers
Determine the values of xfor which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001.(Enter
MrMuchimi

Answer:

The values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001 is 0 < x < 0.3936.

Step-by-step explanation:

Note: This question is not complete. The complete question is therefore provided before answering the question as follows:

Determine the values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001. f(x) = e^x ≈ 1 + x + x²/2! + x³/3!, x < 0

The explanation of the answer is now provided as follows:

Given:

f(x) = e^x ≈ 1 + x + x²/2! + x³/3!, x < 0 …………….. (1)

R_{3} = (x) = (e^z /4!)x^4

Since the aim is R_{3}(x) < 0.001, this implies that:

(e^z /4!)x^4 < 0.0001 ………………………………….. (2)

Multiply both sided of equation (2) by (1), we have:

e^4x^4 < 0.024 ……………………….......……………. (4)

Taking 4th root of both sided of equation (4), we have:

|xe^(z/4) < 0.3936 ……………………..........…………(5)

Dividing both sides of equation (5) by e^(z/4) gives us:

|x| < 0.3936 / e^(z/4) ……………….................…… (6)

In equation (6), when z > 0, e^(z/4) > 1. Therefore, we have:

|x| < 0.3936 -----> 0 < x < 0.3936

Therefore, the values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001 is 0 < x < 0.3936.

3 0
3 years ago
Find the volume 7 height 8 radius <br> A.1600km<br> B.1907km<br> C.1407km<br> D.1100km
Oksi-84 [34.3K]

Answer:

 1407

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
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