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andrew-mc [135]
3 years ago
12

How many significant figures are there in 0.980 7

Mathematics
2 answers:
Varvara68 [4.7K]3 years ago
7 0
There are four significant figures
agasfer [191]3 years ago
5 0

Answer:

if you count the numbers after the decimal point there are 4 significant figures.

Step-by-step explanation:

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(3x−1)+x2−4 when x=3 how do i solve it
Reil [10]

(3×3-1)+3×2-5

=(9-1)+3×2-5

=8+6-5

=14-5

=9

8 0
3 years ago
If f(x)=(x-4 +6, then f(2)= ?<br> (1) 12<br> (3) -4<br> (2) 8<br> (4) 4
kykrilka [37]

Answer:

The answer to it is 4

Step-by-step explanation:

plug 2 to x and solve it

((2)-4)+6

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4

4 0
3 years ago
What is 24^2? Use mental math.
AnnZ [28]

Answer:

576

Step-by-step explanation:

calculator

5 0
3 years ago
Show that ( 2xy4 + 1/ (x + y2) ) dx + ( 4x2 y3 + 2y/ (x + y2) ) dy = 0 is exact, and find the solution. Find c if y(1) = 2.
fredd [130]

\dfrac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-\dfrac{2y}{(x+y^2)^2}

\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}

so the ODE is indeed exact and there is a solution of the form F(x,y)=C. We have

\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)

\dfrac{\partial F}{\partial y}=4x^2y^3+\dfrac{2y}{x+y^2}=4x^2y^3+\dfrac{2y}{x+y^2}+f'(y)

f'(y)=0\implies f(y)=C

\implies F(x,y)=x^2y^3+\ln(x+y^2)=C

With y(1)=2, we have

8+\ln9=C

so

\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}

8 0
3 years ago
A number cube is rolled. Find the Probability. (4 or 5)​
nikklg [1K]

Answer: 2/6 or 33%

Step-by-step explanation: Hope this helps <3

6 0
3 years ago
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