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igor_vitrenko [27]
3 years ago
11

You look at an ant under a magnifying glass. Find the actual length of the ant. The image of the ant is 8 times the ants actual

size and has a length of 5.6cm.
The ants actual length is_____cm
Mathematics
1 answer:
ella [17]3 years ago
5 0
Divide 5.6 by 8 which is 0.7. Since the image of the ant is 8 times bigger we need to divide by 8 to make it actual size. 

The ant's actual length is 0.7 cm. 
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What is equivalent to log 25<br> 125
nexus9112 [7]
Log25 (5•25) = 3 over 2 aka 3/5
5 0
3 years ago
If 5-y^2=x^2 then find d^2y/dx^2 at the point (2, 1) in simplest form. ​
ratelena [41]

Answer:

y"(2, 1) = -5

Step-by-step explanation:

Step 1: Define implicit differentiation

5 - y² = x²

Step 2: Find dy/dx

  1. Take implicit differentiation: -2yy' = 2x
  2. Isolate y': y' = 2x/-2y
  3. Isolate y': y' = -x/y

Step 3: Find d²y/dx²

  1. Quotient Rule: y'' = [y(-1) - y'(-x)] / y²
  2. Substitute y': y" = [-y - (-x/y)(-x)] / y²
  3. Simplify: y" = [-y - x²/y] / y²
  4. Multiply top/bottom by y: y" = (-y² - x²) / y³
  5. Factor negative: y" = -(y² + x²) / y³

Step 4: Substitute and Evaluate

y"(2, 1) = -(1² + 2²) / 1³

y"(2, 1) = -(1 + 4) / 1

y"(2, 1) = -5/1

y"(2, 1) = -5

3 0
2 years ago
1. Consider the following hypotheses:
Andrej [43]

Answer:

See deductions below

Step-by-step explanation:

1)

a) p(y)∧q(y) for some y (Existencial instantiation to H1)

b) q(y) for some y (Simplification of a))

c) q(y) → r(y) for all y (Universal instatiation to H2)

d) r(y) for some y (Modus Ponens using b and c)

e) p(y) for some y (Simplification of a)

f) p(y)∧r(y) for some y (Conjunction of d) and e))

g) ∃x (p(x) ∧ r(x)) (Existencial generalization of f)

2)

a) ¬C(x) → ¬A(x) for all x (Universal instatiation of H1)

b) A(x) for some x (Existencial instatiation of H3)

c) ¬(¬C(x)) for some x (Modus Tollens using a and b)

d) C(x) for some x (Double negation of c)

e) A(x) → ∀y B(y) for all x (Universal instantiation of H2)

f)  ∀y B(y) (Modus ponens using b and e)

g) B(y) for all y (Universal instantiation of f)

h) B(x)∧C(x) for some x (Conjunction of g and d, selecting y=x on g)

i) ∃x (B(x) ∧ C(x)) (Existencial generalization of h)

3) We will prove that this formula leads to a contradiction.

a) ∀y (P (x, y) ↔ ¬P (y, y)) for some x (Existencial instatiation of hypothesis)

b) P (x, y) ↔ ¬P (y, y) for some x, and for all y (Universal instantiation of a)

c) P (x, x) ↔ ¬P (x, x) (Take y=x in b)

But c) is a contradiction (for example, using truth tables). Hence the formula is not satisfiable.

7 0
3 years ago
What is the solution of 3x+9&lt;15
Zolol [24]

Answer:

Third one

Step-by-step explanation:

<u>Two </u>things to solve      3x+9 <= 15

                                        3x <= 6

                                         <u> x <= 2</u>

      <u>   and</u>   - (3x+ 9) <= 15

                  - 3x - 9 <= 15

                   - 3x <= 24

                      3x >= -24

                     <u>   x > = - 8 </u>

<u />

<u />

<u>s o   -8 <= x <= 2 </u>

3 0
2 years ago
The first term of an arithmetic sequence is a1 = 2, and the third term is a3 = 6. Which of the following could be formula(s) for
Ulleksa [173]

Answer:

A

Step-by-step explanation:

an=2(1)=2

an=2(3)=6

8 0
2 years ago
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