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3 years ago

Find the value of x. If necessary, round your answer to the nearest tenth. The figures are not drawn to scale.

Mathematics

1 answer:

REY [17]3 years ago

4 0

Answer:

9.93

Step-by-step explanation:

Secant-Tangent theorem tells us that the product of the secant segment with its external segment is equal to the square of the tangent segment.

From the diagram, we can say (let the unknown part of secant line, the part left of the segment length 5, be y):

(15+y)(10) = 17^2

Solving for y we get:

$(15+y)(10) = 17^2\\150+10y=289\\10y=289-150\\10y=139\\y=13.9$

Now we can use the chord theorem to solve for x. Chord theorem tells us that when 2 intersecting chords create 4 segments, the product of the individual chord segments are equal. Thus we can say:

5 * 13.9 = 7 * x

Now solving, we get:

$5 * 13.9 = 7 * x\\69.5=7x\\x=\frac{69.5}{7}\\x=9.93$

Thus x = 9.93

last answer choice is right.

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A six sided number cube is laveled with number from 1 to 6 . If a student tolls the cube 150 times howany times should they expe

andrey2020 [161]

Answer:

75 times

Step-by-step explanation:

6 sided die - 1,2,3,4,5,6

3 of these numbers are even

P(even) = 3/6 = 1/2

1/2 * 150 = 75

7 0

3 years ago

Which is the inverse of the function a(d)=5d-3? And use the definition of inverse functions to prove a(d) and a-1(d) are inverse

Drupady [299]

Answer:

$a'(d) = \frac{d}{5} + \frac{3}{5}$

$a(a'(d)) = a'(a(d)) = d$

Step-by-step explanation:

Given

$a(d) = 5d - 3$

Solving (a): Write as inverse function

$a(d) = 5d - 3$

Represent a(d) as y

$y = 5d - 3$

Swap positions of d and y

$d = 5y - 3$

Make y the subject

$5y = d + 3$

$y = \frac{d}{5} + \frac{3}{5}$

Replace y with a'(d)

$a'(d) = \frac{d}{5} + \frac{3}{5}$

Prove that a(d) and a'(d) are inverse functions

$a'(d) = \frac{d}{5} + \frac{3}{5}$ and $a(d) = 5d - 3$

To do this, we prove that:

$a(a'(d)) = a'(a(d)) = d$

Solving for $a(a'(d))$

$a(a'(d)) = a(\frac{d}{5} + \frac{3}{5})$

Substitute $\frac{d}{5} + \frac{3}{5}$ for d in $a(d) = 5d - 3$

$a(a'(d)) = 5(\frac{d}{5} + \frac{3}{5}) - 3$

$a(a'(d)) = \frac{5d}{5} + \frac{15}{5} - 3$

$a(a'(d)) = d + 3 - 3$

$a(a'(d)) = d$

Solving for: $a'(a(d))$

$a'(a(d)) = a'(5d - 3)$

Substitute 5d - 3 for d in $a'(d) = \frac{d}{5} + \frac{3}{5}$

$a'(a(d)) = \frac{5d - 3}{5} + \frac{3}{5}$

Add fractions

$a'(a(d)) = \frac{5d - 3+3}{5}$

$a'(a(d)) = \frac{5d}{5}$

$a'(a(d)) = d$

Hence:

$a(a'(d)) = a'(a(d)) = d$

7 0

2 years ago

<img src="https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B2%7D%20%20%2B%2010%20-%2021%20%3D%20y" id="TexFormula1" title=" {x}^{2} + 10

myrzilka [38]

Answer:

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

(-4, -2); y = - 2x + 4

padilas [110]

Answer:

y = -2x - 10

Step-by-step explanation:

Slope intercept form of equation is of form

y = mx+c

where m is the slope of line and c is the y intercept of the line.

Y intercept is point on y axis where the line intersects the y axis.

_____________________________

Given equation

y = -2x +4

comparing it with y = mx+ c

m = -2 , c = 4

_____________________________

when two lines are parallel, their slopes are equal.

Let the equation of new line in slope intercept form be y = mx + c

Thus slope of of new required line is -2

Thus m for new line is -2.

now, equation of required line : y = -2x+c

Given that this line passes through (-4, -2). This point shall should satisfy equation y = -2x+c.

Substituting the value of (-4, -2) we have

-2 = -2(-4)+c

=> -2 = 8 +c

=> -2 -8 = c

=> c = -10.

Thus , equation of required line is y = -2x - 10.

8 0

3 years ago

I will give u brainlyist!! pls help

Karolina [17]

Answer:

i believe C

Step-by-step explanation:

5 0

3 years ago