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olganol [36]
3 years ago
14

Find the pattern43) (76 (115)☺ 1 40 WEZZO​

Mathematics
1 answer:
Masja [62]3 years ago
3 0

Answer:

1.first add 5, then 8 then 11, then 14 so it goes 3 by 3

3. you add 25 each time

4. you add 9 each time

I couldn't solve second and fifth one

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If the test for a disease is accurate 45% of the time, How often will it come back negative if a patient has the disease?
Dafna1 [17]
D. 55% of the time

45% of the time it will say positive

The other 55% of the time will come back negative.
7 0
3 years ago
16. Which of the following are equivalent to(x) 16xA) g(r) 8.21B) g(c) 4096.16-3g (x) -4.4xD) g(x) 0.0625-16'+1E) g(x) 32.16E) g
swat32
F(x) = 16ˣ

A. g(x) = 8(2ˣ)
    g(x) = (2³)(2ˣ)
    g(x) = 2ˣ⁺³
The answer is not A.

B. g(x) = 4096(16ˣ⁻³)
    g(x) = (16³)(16ˣ⁻³)
    g(x) = 16ˣ
The answer is B.

C. g(x) = 4(4ˣ)
     g(x) = 4ˣ⁺¹
The answer is not C.

D. g(x) = 0.0625(16ˣ⁺¹)
     g(x) = (16⁻¹)(16ˣ⁺¹)
     g(x) = 16ˣ
The answer is D.

E. g(x) = 32(16ˣ⁻²)
    g(x) = (2⁵)(2⁴ˣ⁻⁸)
    g(x) = 2(⁴ˣ⁻³)
The answer is not E.

F. g(x) = 2(8ˣ)
    g(x) = 2(2³ˣ)
    g(x) = 2³ˣ⁺¹
The answer is not F.

The answer is B and D.
6 0
3 years ago
Does anyone know how to solve a system of linear equations using elimination? I am having a hard time on how to do it for differ
Mariana [72]
1. Multiply each equation so they end up with the same coefficient

2. Subtract your second equation from the first

3. Solve for one of the variables (I tend to solve an equation that only contains 2 variables if possible. So it would be if you have one question with x and y, solve for the easier one)

4. Substitute the variable you found in the least step into one of the other equations and find a second variable.

5. Substitute both variables you found into the last equation and there you should be left with x, y, and z :))

I hope this helped sksjsk if it didn’t I could write it out to hopefully help more :)
3 0
3 years ago
Find y' by implicit differentiation: xy + 2x + 3x^2 = 4
NeX [460]
(xy)' + (2x)' + (3x^2)' = (4)'

y + xy' + 2 + 6x = 0

xy' = -y  -2 -6x

y' = [-y -2 -6x] / x

Now solve y from the original equation and substitue

xy + 2x + 3x^2 = 4 => y = [-2x - 3x^2 + 4] / x

y' =  [(-2x - 3x^2 +4) / x - 2 - 6x ] / x

y' = [-2x - 3x^2 + 4 -2x -6x^2 ] x^2 = [ -4x - 9x^2 + 4] / x^2 =

= [-9x^2 - 4x + 4] / x^2
3 0
3 years ago
Caroline measured the weight of a dog that came into the veterinary clinic. She determined that her measurement had a margin of
Andrews [41]

Weight of the dog in kgs-0.05 kgs

3 0
3 years ago
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