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2 years ago

Solve the equation

Mathematics

1 answer:

motikmotik2 years ago

3 0

Answer:

Step-by-step explanation:

$\frac{1}{2}$ (2x + 2) - $\frac{1}{4}$ (3x - 1) = 15

multiply through by 4 ( the LCM of 2 and 4 ) to clear the fractions

2(2x + 2) - (3x - 1) = 60 ← distribute parenthesis and simplify left side

4x + 4 - 3x + 1 = 60

x + 5 = 60 ( subtract 5 from both sides )

x = 55

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What is the scale factor of triangle abc and triangle def

Thepotemich [5.8K]

Answer:

The answer is 6

Step-by-step explanation:

ABC is 72, 66, 42 and DEF is 12, 11, 7

simply divide the largest shape measurements by the smaller one.

72÷12

66÷11

42÷7

you get 6 in all of them, this technique may not work all the time.

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3 years ago

On the first play, the football team lost 6 yards. On the second play, the team lost 5yards. What was their total change in yard

WINSTONCH [101]

The total change in yards the team lost was 11

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3 years ago

James has 6 stamps in his stamp collection. Roy has 12 stamps in his stamp collection. James adds 2 stamps to his stamp collecti

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3 years ago

ASAP Can someone help me with this pleaseeee

Rashid [163]

It’s option D.) I did a quiz on this !

4 0

2 years ago

PLEASE HELP.

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 radius is a line from any point on the circumference to the center of the circle

A tangent 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

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

substitute the values and solve for CE

$14^{2}=6^{2}+CE^{2}$

$CE^{2}=14^{2}-6^{2}$

$CE^{2}=160$

$CE=\sqrt{160}\ cm$

$CE=4\sqrt{10}\ cm$

remember that

CE=DE

$DE=4\sqrt{10}\ cm$

7 0

3 years ago