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3 years ago

12

Round 8.95 to the nearest tenth.did the ones place change? explain

2 answers:

Gekata [30.6K]3 years ago

6 0

8.95 to the nearest tenth.

The tenth digit is the first after the decimal point to the right. And that is 9.
Looking at the next digit after it is 5, and it is up to 5 so we increase the 9 to 10.

And of course the 10 would cause an increase to the 8 for it to become 9.

≈ 9.0 to the nearest tenth.

Papessa [141]3 years ago

6 0

Yes the ones place changed to 9 because when u round numbers you look to the digit to the right of it. If it's 5 or higher you round up. So the digit to the right of 9 being 5 u round up. But 9 rounds up to 10 and then u put the zero in the tenths place and you add one to 8

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ozzi

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2 years ago

If H is in the interior of ∠EFG, m∠EFH = 75°, and m∠HFG = (10x)°, and m∠EFG = (20x − 5)°, then x = ? and m∠HFG = ?

Diano4ka-milaya [45]

Answer:

$x=8$

$\m\angle HFG= 80^\circ$

Step-by-step explanation:

Given that:

Point $H$ is interior of $\angle EFG$ .

$m\angle EFH = 75^\circ\\m\angle HFG = (10x)^\circ\\m\angle EFG = (20x-5)^\circ$

To find:

$x = ?\\m\angle HFG = ?$

Solution:

First of all, let us represent the given values in the form of a diagram.

Kindly refer to the attached image for the given points and values of angles.

We can clearly see that:

$m\angle EFG = m\angle EFH + m\angle HFG$

Putting all the values given in above equation, we get:

$(20x-5)^\circ = 75^\circ + 10x^\circ\\\Rightarrow 20x-10x=75+5\\\Rightarrow 10x =80\\\Rightarrow \bold{x =8}$

$m\angle HFG =10x^\circ\\\Rightarrow m\angle HFG =10\times 8^\circ\\\Rightarrow m\angle HFG = \bold{80^\circ}$

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