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2 years ago

Is 7.5 greater or less than 48.75?

Mathematics

2 answers:

fomenos2 years ago

7 0

Answer:

Less than

Step-by-step explanation:

7.5 < 48.75

48.75 is a bigger number than 7.5

Therefore, 7.5 is less than it.

telo118 [61]2 years ago

7 0

Answer:

Less

Step-by-step explanation:

The first thing to look at is the number to the LEFT of the decimal, if it is greater if it is greater the number will be greater no matter what's on the other side. If they are the same look at the first number to the RIGHT of the decimal. If they are also the same, keep moving one place to the right until you find a difference.

Hope this helped!!

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2 years ago

△ABC≅△DEF, BC=4x−12, and EF=−3x+16. Find x and EF.

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x = 4 and EF = 4 ⇒ 3rd answer

Step-by-step explanation:

If two triangles are congruent, then

1. Their corresponding sides are equal

2. Their corresponding angles are equal

3. Their areas and perimeters are equal

∵ △ ABC ≅ △ DEF

∴ AB ≅ DE

∴ BC ≅ EF

∴ AC ≅ DF

∵ BC = 4 x - 12

∵ EF = -3 x + 16

∵ BC = EF

∴ 4 x - 12 = -3 x + 16

Let us solve the equation to find x

∵ 4 x - 12 = -3 x + 16

- Add 3 x for both sides

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- Add 12 to both sides

∴ 7 x = 28

- Divide both sides by 7

∴ x = 4

Substitute the value of x in the expression of EF

∵ EF = -3 x + 16

∵ x = 4

∴ EF = -3(4) + 16 = -12 + 16

∴ EF = 4

x = 4 and EF = 4

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krek1111 [17]

Answer:

Part a) The radii are segments AC and AD and the tangents are the segments CE and DE

Part b) $DE=4\sqrt{10}\ cm$

Step-by-step explanation:

Part a)

we know that

A <u>radius</u> is a line from any point on the circumference to the center of the circle

A <u>tangent</u> to a circle is a straight line which touches the circle at only one point. The tangent to a circle is perpendicular to the radius at the point of tangency.

In this problem

The radii are the segments AC and AD

The tangents are the segments CE and DE

Part b)

we know that

radius AC is perpendicular to the tangent CE

radius AD is perpendicular to the tangent DE

CE=DE

Triangle ACE is congruent with triangle ADE

Applying the Pythagoras Theorem

$AE^{2}=AC^{2}+CE^{2}$