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3 years ago

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LAIN!

1 answer:

natima [27]3 years ago

8 0

Answer:

Step-by-step explanation:

graph_1 and graph_2 are for this task

graph_3 is for previous task.

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Help please it’s for a test I’m so clueless!!!!!!

kumpel [21]

Answer:

h=12 A=144 sq. cm.

Step-by-step explanation:

pythagoream therom:

9^2+h^2=15^2

81+h^2=225

h^2=225-81

h^2=144

Sq. rt h^2=sq. rt 144

h=12cm

1/2(b1+b2)h

1/2(5+19)12

1/2(24)12

1/2(288)

a=144

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3 years ago

Help me please. Quick!

Elden [556K]

The length of one side is 5

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3 years ago

Read 2 more answers

1 a. A boat can travel 378 miles on 31.5

Yuki888 [10]

I’m here for the points srrr

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3 years ago

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If f(x)=x2+3x+5 what's is f(3+h)

Nikolay [14]

(3+h)2+3(3+h)+5

Distribute the 2 to 3+h and the 3 to 3+h

6+2h+9+3h+5

add like terms

20 + 5h

7 0

4 years ago

Given AG bisects CD, IJ bisects CE, and BH bisects ED. Prove KE = FD.

OverLord2011 [107]

When a line is bisected, the line is divided into equal halves.

See below for the proof of $\mathbf{KE \cong FD}$

The given parameters are:

<em>AC bisects CD</em>
<em>IJ bisects CE</em>
<em>BH bisects ED</em>

<em />

By definition of segment bisection, we have:

$\mathbf{CK \cong KE}$
$\mathbf{EF \cong FD}$
$\mathbf{CE \cong ED}$

By definition of congruent segments, the above congruence equations become:

$\mathbf{CK = KE}$
$\mathbf{EF = FD}$
$\mathbf{CE = ED}$

By segment addition postulate, we have:

$\mathbf{CE = CK + KE}$
$\mathbf{ED = EF + FD}$

Substitute $\mathbf{ED = EF + FD}$ in $\mathbf{CE = ED}$

$\mathbf{CK + KE = EF + FD}$

Substitute $\mathbf{CK = KE}$ and $\mathbf{EF = FD}$

$\mathbf{KE + KE = FD + FD}$

Simplify

$\mathbf{2KE = 2FD}$

Apply division property of equality

$\mathbf{KE = FD}$

By definition of congruent segments

$\mathbf{KE \cong FD}$

Read more about proofs of congruent segments at:

brainly.com/question/11494126

6 0

3 years ago