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solniwko [45]
3 years ago
6

Please help don’t understand

Mathematics
1 answer:
yarga [219]3 years ago
7 0

Answer:

D

Step-by-step explanation:

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Please Answer Quickly
Natali5045456 [20]
ANSWER
i think it is six.
EXPLANATION
6 0
3 years ago
If AACB= ADCE, ZBAC = 3x-10,<br> ZECD = 45°, and ZEDC = 2x+10<br> D<br> C<br> (33<br> E<br> x= [?]
inna [77]

Answer: 20

Step-by-step explanation:

By CPCTC,

3x-10=2x+10\\\\x=20

6 0
2 years ago
How do I work out A &amp; B ?
Sonja [21]

Answer:

see explanation

Step-by-step explanation:

(a)

Given

x² + 5x + 6

Consider the factors of the constant term ( + 6) which sum to give the coefficient of the x- term ( + 5)

The factors are 3 and 2, since

3 × 2 = 6 and 3 + 2 = 5, hence

x² + 5x + 6 = (x + 3)(x + 2) ← in factored form

(b)

To solve

x² + 5x + 6 = 0 ← use the factored form, that is

(x + 3)(x + 2) = 0

Equate each factor to zero and solve for x

x + 3 = 0 ⇒ x = - 3

x + 2 = 0 ⇒ x = - 2

4 0
3 years ago
Read 2 more answers
A number x rounded to 1 significant figure is 40. What is the error interval for x.
Viktor [21]

Answer:sorry i need points for my test

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
I NEED HELP WITH THESE TWO QUESTIONS PLZ DO NOT ANSWER IF YOU CANNOT EXPLAIN THE ANSWER
Murljashka [212]

9514 1404 393

Answer:

  1. asymptotes: x=-6, x=6, y=0; zero: x=0

  2. 20 blue fish

Step-by-step explanation:

1. The vertical asymptotes are where the denominator is zero. It factors as (x-6)(x+6), so will have zeros at x=-6 and x=6.

The horizontal asymptote is at the limit when x goes to infinity. Here, the ratio of highest-degree terms is 6/x, so the value goes to zero as x gets large.

The zero is where the numerator is zero, at x=0.

In summary:

  • vertical asymptotes: x = ±6
  • horizontal asymptote: y = 0
  • zero: (0, 0)

__

2. 60% are red, so 40% are blue. The difference in percentages is 60%-40% = 20%. The percentage of blue fish (40%) is twice the difference (=2×20%), so the number of blue fish is twice the difference in numbers.

  blue fish = 2×10 = 20 fish

<em><u>Check</u></em>

There are then 20+10=30 red fish, so 20+30=50 fish total. Reds are 30/50 = 60% of the total.

<u><em>Additional comment</em></u>

I find it often works well to work with ratios, which I can often do in my head. Other folks like to see equations. Here, we could write equations like ...

  r = 60%(r + b) . . . . reds are 60% of the total

  r - b = 10 . . . . . . . . 10 more reds than blues

We want the value of b, so we can eliminate r by substituting for it:

  r = 10 + b . . . . from the second equation above

  (10 +b) = 0.60((10 +b) + b) . . . . substitute for r

  10 +b = 6 + 1.2b . . . simplify

  4 = 0.2b . . . . subtract b+6

  20 = b . . . . . . multiply by 5 (there are 20 blue fish)

6 0
3 years ago
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