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2 years ago

The measures of the angles of LMN are in ratio 2:4:6 what are the measures of the angle

1 answer:

5 0

The three interior angles of a triangle make a sum of 180
2+4+6=12 there are 12 shares, each share is 180/12=15
so the angles are: 30, 60, 90

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Solve for the dimensions of the area model below and then write an equation showing how the area as a sum is equivalent to the a

dusya [7]

Full question:

For each multiplication expression, sketch an area model. Label the dimensions and the area of each part. Then write an equation showing that the area as a product equals the area as a sum. a. (x+1)(x+2), b. 3(2x+5), c. (2x-3)(x+2), d. (x-1)(y-1), e. -2y(y+3), f. (-x+1)(3x+y-4)

Answer and explanation:

a. (x+1)(x+2)= x×x+x×2+1×x+1×2

The dimensions (length and width) is x+1 and x+2

b. 3(2x+5) = 3×2x+3×5

The dimensions is 3 and 2x+5

c. (2x-3)(x+2)= 2x×x+2x×2-3×x-3×+2

The dimensions are 2x-3 and x+2

d. (x-1)(y-1)= x×y+x×-1-1×y-1×-1

Dimensions are x-1 and y-1

e. -2y(y+3)= -2y×y-2y×3

Dimensions are -2y and y+3

f. (-x+1)(3x+y-4)= -x×3x-x×y-x×-4+1×3x+1×y+1×-4

Dimensions are -x+1 and 3x+y-4

7 0

2 years ago

What is the ratio of a nickel to one dollar? reduce to simplest form

makkiz [27]

1:0.05

I think this is the answer

3 0

3 years ago

For her presentation on the Wonders of the World, Mary baked a square pyramid-shaped cake as pictured below. The slant height of

mihalych1998 [28]

Answer:

$V=196in^3$

Step-by-step explanation:

The volume of a pyramid is:

$V=\frac{A_{b}h}{3}$

where $A_{b}$ is the area of the base and $h$ is the height (the perpendicular measurement between base and highest point, not the slant height)

Since the base is a square, the area is given by:

$A_{b}=l^2$

where $l$ is the length of the side: $l=8in$ , thus:

$A_{b}=(8in)^2\\A_{b}=64in^2$

Now we need to find the height, for this we use the right triangle that forms with half of a square side (8in/2 = 4in), the slant height (10in), and the height.

In this right triangle, the slant height is the hypotenuse, the leg 1 is the unknown height, and leg 2 is half of the square side.

Using pythagoras:

$hypotenuse^2=leg1^2+leg2^2$

substituting our values, and indicating that leg 1 is height h:

$(10in)^2=h^2+(4in)^2$

$100in^2=h^2+16in^2$

and solving for the height:

$h^2=100in^2-16in^2\\h^2=84in^2\\h=\sqrt{84in^2}\\ h=9.165in$

and finally we calculate the volume using this height and the area of the base:

$V=\frac{A_{b}h}{3}$

$V=\frac{(64in^2)(9.165in)}{3} \\V=195.5in^3$

rounding to the nearest cubic inch: $V=196in^3$

4 0

3 years ago

Help!!! Please Answer

adell [148]

Answer:

d sorry if I'm wrong...

6 0

3 years ago

X = ? y = ? 16 45 degrees

mezya [45]

Let's put more details in the figure to better understand the problem:

Let's first recall the three main trigonometric functions:

$\text{ Sine }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Hypotenuse}}$ $\text{ Cosine }\theta\text{ = }\frac{\text{ Adjacent Side}}{\text{ Hypotenuse}}$ $\text{ Tangent }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Adjacent Side}}$

For x, we will be using the Cosine Function:

$\text{ Cosine }\theta\text{ = }\frac{\text{ Adjacent Side}}{\text{ Hypotenuse}}$ $Cosine(45^{\circ})\text{ = }\frac{\text{ x}}{\text{ 1}6}$ $(16)Cosine(45^{\circ})\text{ = x}$ $(16)(\frac{1}{\sqrt[]{2}})\text{ = x}$ $\text{ }\frac{16}{\sqrt[]{2}}\text{ x }\frac{\sqrt[]{2}}{\sqrt[]{2}}\text{ = }\frac{16\sqrt[]{2}}{2}$ $\text{ 8}\sqrt[]{2}\text{ = x}$

Therefore, x = 8√2.

For y, we will be using the Sine Function.

$\text{ Sine }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Hypotenuse}}$ $\text{ Sine }(45^{\circ})\text{ = }\frac{\text{ y}}{\text{ 1}6}$ $\text{ (16)Sine }(45^{\circ})\text{ = y}$ $\text{ (16)(}\frac{1}{\sqrt[]{2}})\text{ = y}$ $\text{ }\frac{16}{\sqrt[]{2}}\text{ x }\frac{\sqrt[]{2}}{\sqrt[]{2}}\text{ = }\frac{16\sqrt[]{2}}{2}$ $\text{ 8}\sqrt[]{2}\text{ = y}$

Therefore, y = 8√2.

5 0

1 year ago