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2 years ago

Please help 20 points

Mathematics

1 answer:

S_A_V [24]2 years ago

6 0

Answer:

can you post a clearer picture ?

Step-by-step explanation:

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29847839298273827736+83839302839288383

Nikolay [14]

Answer:292820839382

Step-by-step explanation:

282739282992

7 0

2 years ago

Which piece of additional information can be used to prove △CEA ~ △CDB?

Slav-nsk [51]

Answer-

The correct answer is

∠BDC and ∠AED are right angles

Solution-

In the ΔCEA and ΔCDB,

$m\angle BCD=m\angle ACE$

As this common to both of the triangle.

If ∠BDC and ∠AED are right angles, then $m\angle E=90=m\angle D$

Now as

∠BCD = ∠ACE and ∠BDC = ∠AED,

∠DBC and ∠EAC will be same. (as sum of 3 angles in a triangle is 180°)

Then, ΔCEA ≈ ΔCDB

Therefore, additional information can be used to prove ΔCEA ≈ ΔCDB is ∠BDC and ∠AED are right angles.

5 0

2 years ago

Read 2 more answers

How can you use the backward problem-solving strategy to solve a two-step equation

neonofarm [45]

Let's use the equation 5x - 20 = 10 as an example for this.

To work backwards from this, you would add the same amount to each side.

You want to isolate the 5x so that you can solve for x.

To do that, you would add 20.

But again, you must add and subtract the same amount to both sides.

So it would look a bit like this:

5x - 20 + 20 = 10 + 20

That would simplify to 5x = 30, or in other words, x = 6.

Basically, you isolate the multiple of the variable then divide the answer to the equation from that multiple.

In the example, 5 is the multiple and after isolating it, the answer was 30.

30 ÷ 5 = 6

x = 6

Hope this helps :)

5 0

2 years ago

Help pls :)

nataly862011 [7]

a or d dont read this heiejebdyueneidyebjwow isi ehbr

6 0

2 years ago

Pls help me simplify thisss

Step2247 [10]

9514 1404 393

Answer:

2a(a^3 +a^3 b^2 +a b^2 +b^3)/(a^2 +b^2)

Step-by-step explanation:

Division is the same as multiplication by the inverse. Multiplication and division proceed left-to-right. So, we can rewrite the expression as ...

$\dfrac{a^3+a^3b^2+ab^2+b^3}{2a(a-b)^2}\times\dfrac{a^2-b^2}{a+2b}\times\dfrac{a^2+ab-2b^2}{a+b}\times\dfrac{4a^2}{a^2+b^2}\\\\=\dfrac{(a^3+a^3b^2+ab^2+b^3)(a-b)(a+b)(a+2b)(a-b)(4a^2)}{2a(a-b)^2(a+2b)(a+b)(a^2+b^2)}\\\\=\dfrac{4a^2}{2a}\cdot\dfrac{(a-b)^2}{(a-b)^2}\cdot\dfrac{a+2b}{a+2b}\cdot\dfrac{a^3+a^3b^2+ab^2+b^3}{a^2+b^2}\\\\=\boxed{\dfrac{2a(a^3+a^3b^2+ab^2+b^3)}{a^2+b^2}}$

5 0

2 years ago