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djverab [1.8K]
3 years ago
10

Help with this question please also explain how you got your answer. Will mark brainliest.

Mathematics
1 answer:
sergij07 [2.7K]3 years ago
3 0
Green.


You first must find the Least Common Denominator(LCD) of the fractions. You find that it is 20. 4/5 will turn into 16/20 because you must make it have a denominator with 20, and by doing so, you must multiply to get 20. 4×4=16, 5×4=20: 16/20. For 1/2, simply multiply by 10 to get 20 as the denominator. 1×10=10, 2×10=20: 10/20. Compare the two. You may convert the fractions to decimals to make your answer reasonable. 16/20 is clearly more than 10/20, so there is your answer.

P.S

16/20 is 0.8 as a decimal.
10/20 is 0.5 as a decimal.

Oh, and You're welcome!
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Need help can anyone help me???​
Paladinen [302]

Answer:

  it is greater than 45°

Step-by-step explanation:

From the relationship of angles and secants/tangents, we have ...

  m∠GSO = (long arc GO -arc GT)/2

Solving for (long arc GO), we have ...

  2(m∠GSO) +arc GT = (long arc GO)

We know that (long arc GO) > 180°, so we can write ...

  2(m∠GSO) +90° > 180° . . . . arc GT = 90°

  2(m∠GSO) > 90° . . . . . . subtract 90°

  m∠GSO > 45° . . . . . . . . divide by 2

_____

<em>Alternate solution</em>

Inscribed ∠GOT has half the measure of arc GT, so is 45°.

You know that if angle G were 90°, then the right triangle would be isosceles, and angle S would also be 45°. In this triangle, arc GTO is less than 180°, so angle G is less than 90°.

When angle G gets smaller, the sum of angles remains the same, so angle S must be larger than 45°.

This reasoning is written more formally in the math above.

8 0
3 years ago
In the figure shown, BC is parallel to
arlik [135]

Answer:

52 units.

Step-by-step explanation:

If we drop a perpendicular line from point C to AD and call the point ( on AD) E we have a right triangle CED.

Now CE = 6 and as the whole figure is symmetrical about the dashed line,

ED = (26 - 10)/2

= 8.

So by Pythagoras:

CD^2 = 6^2 + 8^2 = 100

CD = 10.

So, as AB = CD,

the perimeter = 10 + 26 + 2(8)

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5 0
3 years ago
Triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C'.
frozen [14]

Note: Consider we need to find the vertices of the triangle A'B'C'

Given:

Triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C'.

Triangle A,B,C with vertices at A(-3, 6), B(2, 9), and C(1, 1).

To find:

The vertices of the triangle A'B'C'.

Solution:

If triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C', then

(x,y)\to (y,-x)

Using this rule, we get

A(-3,6)\to A'(6,3)

B(2,9)\to B'(9,-2)

C(1,1)\to C'(1,-1)

Therefore, the vertices of A'B'C' are A'(6,3), B'(9,-2) and C'(1,-1).

7 0
4 years ago
If a+11=20 what does a equal
luda_lava [24]

Answer:

a = 9

Step-by-step explanation:

a+11=20

Subtract 11 from each side

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a = 9

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4 years ago
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