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3 years ago

Find WX. Assume that segments which appear to be tangent are tangent.

Mathematics

1 answer:

makkiz [27]3 years ago

5 0

Answer:

Step-by-step explanation:

Tangent from the point outside the circle are equal

WX = XY

7x - 29 = 2x + 16 {Add 29 to both sides}

7x = 2x + 16 +29

7x = 2x + 45 {Subtract 2x from both sides}

7x - 2x = 45

5x = 45 {Divide both sides by 5}

x = 45/5

x = 9

WX = 7x - 29

= 7*9 - 29

= 63 - 29

WX = 34

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larisa [96]

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3 years ago

Write the summation to estimate the area under the curve y = 1 + x^2 from x = –1 to x = 2 using 3 rectangles and right endpoints

Minchanka [31]

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3 years ago

Bo and Erica are yoga instructors. Between the two of them, they teach 45 yoga classes each week. If Erica teaches 15 fewer than

Solnce55 [7]

Answer:

20 and 25.

Step-by-step explanation:

Given:

They both teach total 45 yoga classes each week.

Erica teaches 15 fewer than twice as many as Bo.

To find:

How many classes does each instructor teach per week = ?

Solution:

Let number of classes are taken by Bo = $x$

Then number of classes are taken by Erica = $2x - 15$ (given)

Total number of classes are taken by both = 45 (given)

According to the question.

$x$ + $2x - 15$ = 45

$3x - 15 = 45$

By adding both side by 15

$3x = 60\\$

By dividing both side by 3

$x = 20$

number of classes are taken by Bo = $x$ = 20

number of classes are taken by Erica = $2x - 15$

= $2\times20 - 15 = 40- 15 = 25\\$

Therefore, number of classes are taken by Bo and Erica is 20 and 25.

6 0

4 years ago

Help me Please....................

MakcuM [25]

9514 1404 393

Answer:

1.3363

Step-by-step explanation:

The basic idea here is to find an expression for the direction vector between a point on L1 and a point on L2. Then, solve for the points on L1 and L2 that make that vector perpendicular to both lines L1 and L2. (The dot product of direction vectors is zero.) The distance between the points found is the shortest distance between the lines.

Let P be a point on L1. Then the parametric equation for P is ...

P = (6t, 0, -t) . . . . . . origin + t × direction vector

Let Q be a point on L2. The direction vector for L2 is given by the difference between the given points. It is (4-1, 1-(-1), 6-1) = (3, 2, 5). Then the parametric equation for Q is ...

Q = (3s+1, 2s-1, 5s+1) . . . . (1, -1, 1) + s × direction vector

The direction vector for PQ is ...

Q -P = (3s+1-6t, 2s-1, 5s+1+t)

The dot product of this and the two lines' direction vectors will be zero:

(3s+1-6t, 2s-1, 5s+1+t)·(6, 0, -1) = 0 = 13s -37t +5 . . . perpendicular to L1

(3s+1-6t, 2s-1, 5s+1+t)·(3, 2, 5) = 0 = 38s -13t +6 . . . perpendicular to L2

The solution to these equations is ...

s = -157/1237

t = 112/1237

Then (Q-P) becomes (94, -1551, 564)/1237, and its length is ...

|PQ| = √(94² +1551² +564²)/1237 ≈ 1.3363

The distance between the two lines is about 1.3363 units.

8 0

3 years ago

Please solve with explanation 9 points

Levart [38]

Step-by-step explanation:

try the option shown in the attachment, note, all the answers are marked with red colour.

8 0

2 years ago