Ask question

Ask question

All categories

3 years ago

14

Help plz How do you solve

1 answer:

tensa zangetsu [6.8K]3 years ago

7 0

Answer:

$y=4\sqrt{3}\ units$

Step-by-step explanation:

Triangle MRN is created by folding an equilateral triangle by half. Since triangle MRN is right triangle, the equilateral triangle must be folded along the height drawn to one side.

In equilateral triangle with side length of a units, the height is $\dfrac{a\sqrt{3}}{2}\ units$ and the height is greater than half the side because

$\dfrac{a\sqrt{3}}{2}>\dfrac{a}{2}.$

The leg with length x units has the smaller projection (2 units long) than the leg with length y units has (6 units long).

This means the leg with length y units is the height of the equilateral triangle. If the leg with length y units is the height of the equilateral triangle, then the hypotenuse MN = 8 units is the side of the equilateral triangle and lex with length x units is half the side, so x = 4 units.

By the Pythagorean theorem,

$y^2=8^2-4^2\\ \\y^2=64-16\\ \\y^2=48\\ \\y=4\sqrt{3}\ units$

You might be interested in

Transformations can be thought of as functions that take the points from one object (the pre-image) to the corresponding points

sweet-ann [11.9K]

1. Translation of a units to the right and b units up has rule:

(x,y)→(x+a,y+b).

Then the function is

$T(x,y)=(x+a,y+b).$

2. Reflection across the y-axis acts in following way:

(x,y)→(-x,y).

Then the function is

$R_y(x,y)=(-x,y).$

3. Reflection across the x-axis acts in following way:

(x,y)→(x,-y).

Then the function is

$R_x(x,y)=(x,-y).$

4. Rotation of 90 degrees counterclockwise about the origin, point O acts bu the rule:

(x,y)→(-y,x).

Then the function is

$R_{O,90^{\circ}}(x,y)=(-y,x).$

5. Rotation of 180 degrees counterclockwise about the origin, point O acts bu the rule:

(x,y)→(-x,-y).

Then the function is

$R_{O,180^{\circ}}(x,y)=(-x,-y).$

6. Rotation of 270 degrees counterclockwise about the origin, point O acts bu the rule:

(x,y)→(y,-x).

Then the function is

$R_{O,270^{\circ}}(x,y)=(y,-x).$

4 0

3 years ago

What is the answer to 14×67

Solnce55 [7]

14
× 67
--------
938 I think

3 0

2 years ago

VERY EASY QUESTION. Write out this equation: ten times the sum of half a number and 6 is 8

bija089 [108]

Answer:

x

=

58

5

Step-by-step explanation:

10 times:

→

10

×

The sum of:

→

10

×

(

?

+

?

)

half a number:

→

10

×

(

x

2

+

?

)

and 6:

→

10

×

(

x

2

+

6

)

is 8 :

→

10

×

(

x

2

+

6

)

=

8

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

10

(

x

2

+

6

)

=

8

Multiply out the bracket

5

x

+

60

=

8

Subtract 60 from both sides

5

x

=

58

Divide both sides by 5

x

=

58

5

6 0

2 years ago

Read 2 more answers

One leg of a right triangle measures 6 inches. the remaining leg measures 6sqrt3 inches. what is the measure of the angle opposi

Amanda [17]

Hope this helps you.

4 0

2 years ago

Read 2 more answers

Find the following limit or state that it does not exist. ModifyingBelow lim With x right arrow minus 2 StartFraction 3 (2 x min

leva [86]

Answer:

-60

Step-by-step explanation:

The objective is to state whether or not the following limit exists

$\lim_{x \to -2} \frac{3(2x-1)^2 - 75}{x+2}$ .

First, we simplify the expression in the numerator of the fraction.

$3(2x-1)^2 -75 = 3(4x^2 - 4x +1) -75 = 12x^2 - 12x + 3 - 75 = 12x^2 - 12x -72$

Now, we obtain

$12(x^2-x-6) = 12(x+2)(x-3)$

and the fraction is transformed into

$\frac{3(2x-1)^2 - 75}{x+2} = \frac{12(x+2)(x-3)}{x+2} = 12 (x-3)$

Therefore, the following limit is

$\lim_{x \to -2} \frac{3(2x-1)^2 - 75}{x+2} = \lim_{x \to -2} 12(x-3) = 12 \lim_{x \to -2} (x-3)$

You can plug in $-2$ in the equation, hence

$12 \lim_{x \to -2} (x-3) = 12 (-2-3) = -60$

6 0

3 years ago