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noname [10]
3 years ago
12

Hellen is using 3.508 grams of copper for her science experiment. How many kilograms of copper did Hellen use for her science ex

periment?
Mathematics
1 answer:
mash [69]3 years ago
7 0

Answer:

.003508 kg of copper

Step-by-step explanation:

1000 grams = 1 kg

3.508 g/1000 g = .003508 kg

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Log x + log (2x-1) = log 6
Brilliant_brown [7]

Answer:

x = 2

Step-by-step explanation:

logx + log(2x - 1) = log 6

Apply log rules,

x(2x - 1) = 6

2x^2 - 1x = 6

x = 2 (true) or x = -3/2 (False)

8 0
2 years ago
Solving for Unknown Angle Measures
Tresset [83]

Answer:

angle 4 is a full angle because it is represented full 110

4 0
3 years ago
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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}

8 0
3 years ago
5. Original price of concert tickets:$100 . Discount: 21%
lutik1710 [3]

Answer:

Concert ticket now costs $79.

Step-by-step explanation:

21% is being removed from the original cost, in this case, that's $21.  100-21=79. Hope this helps ;)

4 0
3 years ago
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Segments AB , CD , and EF intersect at point O, points A, E, C and points B, F, D are collinear so that AO ≅ OB , CO ≅ OD . Prov
katovenus [111]

If points A, E and C are colinear, then they lie on the same line. The same statement you can say about points B, F and D.

1. Consider triangles AOC and BOD. In these triangles:

  • AO≅OB (given);
  • CO≅OD (given);
  • ∠AOC≅∠BOD (as vertical angles).

Thus, ΔAOC≅ΔBOD by SAS Postulate (If any two corresponding sides and their included angle are the same in both triangles, then the triangles are congruent). Corresponding parts of congruent triangles are congruent, then

  • AC≅BD;
  • ∠ACO≅∠BDO;
  • ∠CAO≅∠DBO.

Since angles ACO and BDO are alternate interior angles between lines AE and BF with transversal CD and these angles are congruent, then lines AE and BF are parallel.

This gives you that

  • ∠CEO≅∠OFD;
  • ∠ECO≅∠ODF.

2. Consider triangles ECO and FDO. In these triangles

  • ∠CEO≅∠OFD (previous proof);
  • CO≅OD (given);
  • ∠ECO≅∠ODF (previous proof).

Therefore, ΔECO≅ΔFDO by AAS Postulate (if two angles and the non-included side one triangle are congruent to two angles and the non-included side of another triangle, then these two triangles are congruent). Then CE≅FD.

3. Note that

  • AE=AC+CE;
  • BF=BD+DF.

Since AC≅BD and CE≅DF, then AE=AC+CE=BD+DF=BF.

7 0
3 years ago
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