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olga2289 [7]
3 years ago
8

Solve for x: 0.2(2x-1)=0.2+0.08 A.1.4 B.5 C.2 D.0.3

Mathematics
1 answer:
Umnica [9.8K]3 years ago
8 0

Answer:

A

Step-by-step explanation:

0.4x-0.2=0.28

0.4x=.28+0.2

0.4x=0.48

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Need help on A and B thank you
lesya [120]
A 78 lollll I need points so I’m guessing
7 0
3 years ago
PLEASE HURRY
Delvig [45]

Answer:

Answer 2) When dividing by -5, he did not change the \leq to \geq.

Step-by-step explanation:

If you divide or multiply a negative, you ALWAYS change the sign to its opposite.

Hope this helps! :)

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3 years ago
Find the area of the shaded region
Alex17521 [72]
I think it could be b
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3 years ago
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What's the answer to this problem
zhenek [66]
The answer is 0.08 or you could say 8 * 10^-2.

Reasoning..
6.4 * 10^3 = 6400
8 * 10^4 = 80000

Divide those two numbers like shown in the problem and boom!
6 0
2 years ago
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(Prove) The angle subtended by an arc at the center is double the angle subtended by it at any
Pavlova-9 [17]

Answer:

The angle subtended by an arc at the center is double the angle subtended by it at any  point on the remaining part of the circle.

Step-by-step explanation:

Let us consider the image attached.

Center of circle be O.

Arc AB subtends the angle \angle APB on the circle and \angle AOB on the center of the circle.

To prove:

\angle AOB = 2 \times \angle APB

Proof:

In \triangle PAO: AO and PO are radius of the circles so AO = PO

And angles opposite to equal sides of a triangle are also equal in a triangle.

So, \angle PAO = \angle OPA

Using external angle property, that external angle is equal to sum of two opposite internal angles of a triangle.

\angle AOQ = \angle PAO + \angle OPA=2 \times \angle APO .... (1)

Similarly,

In \triangle PBO: BO and PO are radius of the circles so BO = PO

And angles opposite to equal sides of a triangle are also equal in a triangle.

So, \angle PBO = \angle OPB

Using external angle property, that external angle is equal to sum of two opposite internal angles of a triangle.

\angle BOQ = \angle PBO + \angle OPB=2 \times \angle BPO .... (2)

Now, we can see that:

\angle AOB = \angle AOQ+\angle BOQ

Using equations (1) and (2):

\angle AOB = 2\angle APO+2\angle BPO\\\angle AOB = 2(\angle APO+\angle BPO)\\\bold{\angle AOB = 2(\angle APB)}

Hence, proved.

4 0
3 years ago
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