Ask question

Ask question

All categories

3 years ago

9

If segment AB is 4 units, segment AD is 23 units, and segment CD is 12 units, what is the length of segment BC?

1 answer:

Thepotemich [5.8K]3 years ago

3 0

Answer:

<h2>C . 7 units</h2>

Step-by-step explanation:

The question is lacking appropriate diagram. Find the diagram to the question attached.

From the diagram, it can be seen that AD = AB+BC+CD

Given the following

AD = 23units

AB = 4units

CD = 12units

To get C=BC, we will substitute the given segments into the expression above as shown;

23 = 4+BC+12

23 = BC + 4 + 12

23 = BC+16

Subtract 16 from both sides

23-16 = BC+16-16

7 = BC

<em></em>

<em>Hence the length of segment BC is 7units.</em>

You might be interested in

Find x and decide if that side (with the x) is an Altitude

irga5000 [103]

Answer:

b = 15
c = 51
c = 100
b = 25.75

Step-by-step explanation:

Using the Pythagoras theorem equation:

$c^{2}=a^{2}+b^{2}$

Where 'c' is the hypotenuse and 'b' and 'a' the sides of the right triangle.

1) Here c = 17, a = 8 and b = ?

Then, using the above equation we can find the side b.

$17^{2}=8^{2}+b^{2}$

$b=\sqrt{225}$

$b=15$

2) Here a = 24, b = 45 and c = ?

$c=\sqrt{24^{2}+45^{2}}$

$c=51$

3) Here a = 28, b = 96 and c = ?

$c=\sqrt{28^{2}+96^{2}}$

$c=100$

4) Here a = 17, b = ? and c = 28

$b=\sqrt{28^{2}-11^{2}}$

$b=25.75$

Just in the 4 case x is an altitude.

I hope it helps you!

3 0

3 years ago

Ryan charges his neighbors $17.50 to wash their car. How many cars must he wash next summer if his goal is to earn at least $1,5

stich3 [128]

Answer:

Approximately 86 cars

Step-by-step explanation:

$1,500/$17.50

= 86 cars approximately

4 0

4 years ago

Pls help I don’t understand this question very well

kiruha [24]

Answer:

$CD = 12.6866616739cm$

$CD \approx12.7cm$

Step-by-step explanation:

$CB = a$

$AB = b$

$AC = c$

${c}^{2} = {a}^{2} + {b}^{2}$

${a}^{2} = {c}^{2} - {b}^{2}$

$a = \sqrt{ {c}^{2} - {b}^{2} }$

$a = \sqrt{ {12}^{2} - {6}^{2} }$

$a = \sqrt{144 - 36}$

$a = \sqrt{108}$

$CB = \sqrt{108}$

$\sin(D) = \frac{opposite}{hypotenuse}$

$\sin(D) = \frac{CB}{CD}$

$\sin(55) = \frac{ \sqrt{108} }{CD}$

$(CD) \sin(55) = \frac{ \sqrt{108} }{CD} (CD)$

$(CD) \sin(55) = \sqrt{108}$

$\frac{(CD) \sin(55) }{ \sin(55) } = \frac{ \sqrt{108} }{ \sin(55) }$

$CD = \frac{ \sqrt{108} }{ \sin(55) }$

$CD = 12.6866616739cm$

$CD \approx12.7cm$

6 0

3 years ago

Consider functions f and g below.

Katyanochek1 [597]

Answer:

g(x) approaches negative infinity

Step-by-step explanation:

g(x) = -x² + 2x + 4

limit at g(x) approaches infinity = -(∞)² + 2(∞) + 4 = -∞

I can't solve for f(x) because you didnt give me f(x)

4 0

3 years ago

Read 2 more answers

The equation of line WX is 2x + y = −5. What is the equation of a line perpendicular to line WX in slope-intercept form that con

borishaifa [10]

Answer: $y=\dfrac12x-\dfrac{3}{4}$

Step-by-step explanation:

Given, The equation of line WX is 2x + y = −5.

It can be written as $y=-2x-5$ comparing it with slope-intercept form y=mx+c, where m is slope and c is y-intercept, we have

slope of WX = -2

Product of slopes of two perpendicular lines is -1.

So, (slope of WX) × (slope of perpendicular to WX)=-1

$-2\times\text{slope of WX}=-1\\\\\Rightarrow\ \text{slope of WX}=\dfrac{1}{2}$

Equation of a line passes through (a,b) and has slope m:

$y-b=m(x-a)$

Equation of a line perpendicular to WX contains point (−1, −2) and has slope $=\dfrac12$

$y-(-2)=\dfrac{1}{2}(x-(-1))\\\\\Rightarrow\ y+2=\dfrac12(x+1)\\\\\Rightarrow\ y+2=\dfrac12x+\dfrac12\\\\\Rightarrow\ y=\dfrac12x+\dfrac12-2\\\\\Rightarrow\ y=\dfrac12x-\dfrac{3}{4}$

Equation of a line perpendicular to line WX in slope-intercept form that contains point (−1, −2) $:y=\dfrac12x-\dfrac{3}{4}$

4 0

3 years ago