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PilotLPTM [1.2K]
2 years ago
11

Round 0.000045389 to the three significant figures

Mathematics
1 answer:
erica [24]2 years ago
3 0

Answer:

0.0000454

Step-by-step explanation:

The zeros are not significant since they come before the numbers greater than 0. And I rounded 3 to 4 since the 8 was above 5 so it gets rounded up

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A preimage includes a line segment of length x and slope m. If the preimage is dilated by a scale factor of n, what are the leng
harkovskaia [24]
The corresponding line segment is B. because the slope does not change. A dilation will always have the effect of shrinking or even stretching the line segment by the factor of n. That is why the length will nx and the slope is m.

Hope this helped.
6 0
3 years ago
Find the solution to the system of equations.
Vikentia [17]

Answer:

C

Step-by-step explanation:

4 0
3 years ago
Which are the solutions to the quadratic equation x^2=9x+6
AlekseyPX
I will use a quadratic formula:

x^2=9x+6\ \ \ \ |-9x-6\\\\x^2-9x-6=0\\\\a=1;\ b=-9;\ c=-6\\\\b^2-4ac=(-9)^2-4\cdot1\cdot(-6)=81+24=105\\\\\sqrt{b^2-4ac}=\sqrt{105}\\\\x_1=\dfrac{-b-\sqrt{b^2-4ac}}{2a};\ x_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a}\\\\x_1=\dfrac{-(-9)-\sqrt{105}}{2\cdot1}=\dfrac{9-\sqrt{105}}{2}\\\\x_2=\dfrac{-(-9)+\sqrt{105}}{2\cdot1}=\dfrac{9+\sqrt{105}}{2}
5 0
3 years ago
Jane bought 12 mangoes for $600 how many mangoes can she buy with $1000​
Ghella [55]

Answer:

20 mangos

Step-by-step explanation:

1. Divide $600 by 12. Your answer should be $50. This means that every mango costs $50.

2. You now need to solve the expression: 1000 ÷ 50 which equals 20. This means that with $1000 Jane can buy 20 mangos.

3 0
2 years ago
(b) dy/dx = (x - y+ 1)^2
Elanso [62]

Substitute v(x)=x-y(x)+1, so that

\dfrac{\mathrm dv}{\mathrm dx}=1-\dfrac{\mathrm dy}{\mathrm dx}

Then the resulting ODE in v(x) is separable, with

1-\dfrac{\mathrm dv}{\mathrm dx}=v^2\implies\dfrac{\mathrm dv}{1-v^2}=\mathrm dx

On the left, we can split into partial fractions:

\dfrac12\left(\dfrac1{1-v}+\dfrac1{1+v}\right)\mathrm dv=\mathrm dx

Integrating both sides gives

\dfrac{\ln|1-v|+\ln|1+v|}2=x+C

\dfrac12\ln|1-v^2|=x+C

1-v^2=e^{2x+C}

v=\pm\sqrt{1-Ce^{2x}}

Now solve for y(x):

x-y+1=\pm\sqrt{1-Ce^{2x}}

\boxed{y=x+1\pm\sqrt{1-Ce^{2x}}}

3 0
3 years ago
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