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puteri [66]
3 years ago
8

If DCAE is a parallelogram, what is the perimeter of triangle DBE?

Mathematics
1 answer:
vlada-n [284]3 years ago
5 0

∆DBE ≅ ∆ABC by SAA

by puthagorean

BC² = AC² - AB²

BC = 4

hence, the perimeter is

3 + 4 + 5 = 12

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It would be appreciated if there could be an explanation along the answer
soldi70 [24.7K]

Answer:

i think probably 80°

Step-by-step explanation:

PQ = RS

2x+3 = 4x-3

x = 3

PTQ = RTS

9y-3x-1 = 5x+7y-5

2y-8x = -4

y-4x = -2

y-4(3) = -2

y-12 = -2

y = -2+12

y = 10

so PTQ = 9y-3x-1

9(10)-3(3)-1

90-9-1

80°

8 0
2 years ago
18. Curry powder is sold at $7.45 per 100 g. How much money will Thuy need in order to buy 3 kg of curry powder?
MissTica

Answer:

Step-by-step explanation:

$7.45 = 100g

x = 3kg( 3 × 1000) = 3000g

∵ $7.45 = 100g

   x        = 3000g

(100 × x) = (7.45 ×  3000)

100x = 22350

100x/100 = 22350/100

x  = $223.5

4 0
2 years ago
Read 2 more answers
If a polynomial function f(x) has roots -8, 1, and 6i, what must also be a root of f(x)?
Naddik [55]

Answer:

it must also have the root : - 6i

Step-by-step explanation:

If a polynomial is expressed with real coefficients (which must be the case if it is a function f(x) in the Real coordinate system), then if it has a complex root "a+bi", it must also have for root the conjugate of that complex root.

This is because in order to render a polynomial with Real coefficients, the binomial factor  (x - (a+bi)) originated using the complex root would be able to eliminate the imaginary unit, only when multiplied by the binomial factor generated by its conjugate: (x - (a-bi)). This is shown below:

(x-(a+bi))*(x-(a-bi))=\\(x-a-bi)*(x-a+bi)=\\([x-a]-bi)*([x-a]+bi)=\\(x-a)^2-(bi)^2=\\(x-a)^2-b^2(-1)=\\(x-a)^2+b^2

where the imaginary unit has disappeared, making the expression real.

So in our case, a+bi is -6i (real part a=0, and imaginary part b=-6)

Then, the conjugate of this root would be: +6i, giving us the other complex root that also may be present in the real polynomial we are dealing with.

5 0
3 years ago
When solving an equation like x3 = 64, why does our solution not include the negative cube root?
MAVERICK [17]
So I’m only answering so I can ask you a question
4 0
2 years ago
Read 2 more answers
What is the effect on the graph of the function f(x) = 2x when f(x) is replaced with f(−
olasank [31]
It would be replaced with f(-2) which is the reciperacle of 2x
4 0
3 years ago
Read 2 more answers
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