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3 years ago

How many centimeters are equal to 0.4 inches? Remember, 2.54 cm = 1 in.

Mathematics

2 answers:

Mkey [24]3 years ago

7 0

1.02 centimeters is your answer

Andrew [12]3 years ago

6 0

Answer:

1.016 cm

Step-by-step explanation:

1 inch = 2.54 cm

0.4 inch = 2.54 * 0.4 cm = 1.016 cm

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A ladder leans against a wall 6 ft above the ground. The base of the ladder is 3 ft from the wall. How long is the ladder?

Igoryamba

Answer:

6.71

Step-by-step explanation:

Hopefully this helps! :)

7 0

3 years ago

What percent of 98is14

Ivanshal [37]

Answer:

14.29

Step-by-step explanation:

14:98*100 =

(14*100):98 =

1400:98 = 14.29

4 0

3 years ago

If 3x - 2 (4x - 8) = 16 , then x = 0. True False

Andrew [12]

Answer:

try your best

Step-by-step explanation:

6 0

4 years ago

Read 2 more answers

.?????????????????????????

weqwewe [10]

Step-by-step explanation:

To divide fractions, we have to focus on 4 steps, which are:

1: Change to improper fractions

2: Keep the same

3: Change the sign to multiplication

4: Flip the 2nd fraction in the equation

So, step one:

1/3 / 4/5

Step two:

1/3 / 4/5

Step three:

1/3 x 4/5

Step four:

1/3 x 4/5 = (4 / 15)

4/15 is in simplist for.

Therefore, your answer is 4/15.

Best of Luck to you!!

4 0

3 years ago

1.) Find the length of the arc of the graph x^4 = y^6 from x = 1 to x = 8.

xxTIMURxx [149]

First, rewrite the equation so that y is a function of x :

$x^4 = y^6 \implies \left(x^4\right)^{1/6} = \left(y^6\right)^{1/6} \implies x^{4/6} = y^{6/6} \implies y = x^{2/3}$

(If you were to plot the actual curve, you would have both $y=x^{2/3}$ and $y=-x^{2/3}$ , but one curve is a reflection of the other, so the arc length for 1 ≤ x ≤ 8 would be the same on both curves. It doesn't matter which "half-curve" you choose to work with.)

The arc length is then given by the definite integral,

$\displaystyle \int_1^8 \sqrt{1 + \left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx$

We have

$y = x^{2/3} \implies \dfrac{\mathrm dy}{\mathrm dx} = \dfrac23x^{-1/3} \implies \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 = \dfrac49x^{-2/3}$

Then in the integral,

$\displaystyle \int_1^8 \sqrt{1 + \frac49x^{-2/3}}\,\mathrm dx = \int_1^8 \sqrt{\frac49x^{-2/3}}\sqrt{\frac94x^{2/3}+1}\,\mathrm dx = \int_1^8 \frac23x^{-1/3} \sqrt{\frac94x^{2/3}+1}\,\mathrm dx$

Substitute

$u = \dfrac94x^{2/3}+1 \text{ and } \mathrm du = \dfrac{18}{12}x^{-1/3}\,\mathrm dx = \dfrac32x^{-1/3}\,\mathrm dx$

This transforms the integral to

$\displaystyle \frac49 \int_{13/4}^{10} \sqrt{u}\,\mathrm du$

and computing it is trivial:

$\displaystyle \frac49 \int_{13/4}^{10} u^{1/2} \,\mathrm du = \frac49\cdot\frac23 u^{3/2}\bigg|_{13/4}^{10} = \frac8{27} \left(10^{3/2} - \left(\frac{13}4\right)^{3/2}\right)$

We can simplify this further to

$\displaystyle \frac8{27} \left(10\sqrt{10} - \frac{13\sqrt{13}}8\right) = \boxed{\frac{80\sqrt{10}-13\sqrt{13}}{27}}$

7 0

3 years ago