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4 years ago

Find the lengths of DE and EF. Note: Segments BC and DF are parallel

Mathematics

2 answers:

rjkz [21]4 years ago

7 0

Answer:

Step-by-step explanation:

Given that Segments BC and DF are parallel, then the following proportion must be satisfied:

BE/EC = DE/EF

Replacing with data from the picture:

3/5 = (5x - 4 + 3)/(5 + 4x + 2)

3*(7 + 4x) = (5x - 1)*5

21 + 12x = 25x - 5

21 + 5 = 25x - 12x

26 = 13x

x = 2

Then,

DE = 5x - 4 + 3 = 5(2) - 4 + 3 = 9 cm

EF = 5 + 4x + 2 = 5 + 4(2) + 2 = 15 cm

wlad13 [49]4 years ago

5 0

Answer:

the answer is c

Step-by-step explanation:

you have to create a proportion ( it being 5x-4/3=4x+2/5) then after you cross multiply from that proportion you plug in x. and don't forget to add the other part of the triangle as well.

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Let's solve the equation for x! 3x-3=180 *

Trava [24]

I think this is the answer

3 0

3 years ago

Read 2 more answers

Multiply the following<br> (-7pq^2r^3) * (-13p^3q^2r)

artcher [175]

Answer:

$91p^{1+3q^{2r}}q^{2r^3}$

Step-by-step explanation:

$(-7pq^2r^3) * (-13p^3q^2r)$

= 91 $p^{1+3q^{2r} }$ * $q^{2r^{3} }$

= $91p^{1+3q^{2r}}q^{2r^3}$

7 0

3 years ago

Using the following image, if you EG = 59, what are x, EF, and FG.

Harman [31]

Answer:

$x=6\,\,units\,,\,EF=34 \,\,units\,,\,FG=25\,\,units$

Step-by-step explanation:

Given: $EG =59 \,\,units$

To find: $x,EF,FG$

Solution:

From the image, $EG=EF+FG$

Put $EF=8x-14\,,\,FG=4x+1\,,\,EG=59$

$8x-14+4x+1=59\\12x-13=59\\12x=59+13\\\\12x=72\\x=\frac{72}{12}=6$

$EF=8x-14=8(6)-14=48-14=34\\FG=4x+1=4(6)+1=24+1=25$

Therefore,

$x=6\,\,units\,,\,EF=34 \,\,units\,,\,FG=25\,\,units$

8 0

3 years ago

How to solve??????????HELP

nadezda [96]

I presume you're talking about number 3?

Solution

I = prt (I being the interest)

I =

$5000 \times 5 \times.05$

I = 1,250$

3 0

3 years ago

Ronin kicks a ball over the gate of the first floor balcony. The function h(t)=-36t^2+288t+9 can be used to model the height of

ale4655 [162]

Answer: 4 seconds

Step-by-step explanation:

Given: The function $h(t)=-36t^2+288t+9$ can be used to model the height of the ball where t is the time in seconds after the ball kicked and h(t) is the height in feet.

Differentiate the given function with respect to t , we get

$h'(t)=-36(2)t+288=-72t+288$

Put h'(t)=0

$\Rightarrow\ -72t+288=0\Rightarrow\ t=\dfrac{288}{72}\\\\\Rightarrow\ t=4$

Also, $h''(t)=-72$

By second derivative test, h(t) is maximum at t=4.

Hence, the time in Devon which the ball reaches it maximum height = 4 seconds

3 0

3 years ago