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2 years ago

A scarf is a triangle with two perpendicular sides that are the same length which two terms describe a triangular scarf

Mathematics

1 answer:

Fittoniya [83]2 years ago

6 0

Answer:

A triangular scarf is an isosceles right triangle.

Step-by-step explanation:

Given: A scarf is a triangle with two perpendicular sides that are the same length.

We have to name which two terms describe a triangular scarf.

Isosceles triangle is a triangle with length of two of its sides equal.

Right angled triangle is a triangle in which one angle is a right angle.

Given :

The two sides of scarf are perpendicular to each other.

Thus, the angle formed between them is 90°. Hence, right angled triangle.

Also, length of two sides are same, Hence isosceles triangle.

Thus,a triangular scarf is an isosceles right triangle.

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Evaluate (c − b)2 + a2 for a = –5, b = –2, and c = –4.

frosja888 [35]

Answer:

Step-by-step explanation:

(c − b)^2 + a^2

Let a = –5, b = –2, and c = –4.

(-4--2)^2 + (-5)^2

(-4+2)^2 + (-5)^2

(-2)^2+ (-5)^2

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2 years ago

Help please urgent 25 point

uranmaximum [27]

Answer:

Step-by-step explanation:

b = the hypotenuse of a right angle triangle

a = the adjacent side.

R is being defined by the cosine

cos(R) = adjacent / hypotenuse

adjacent = 16* sqrt(2)

hypotenuse = 32*sqrt(2)

Cos(R) = 16*sqrt(2) / 32*sqrt(2) sqrt(2) cancels.

cos(R) = 1/2

R = cos-1(1/2)

R = 60 degrees.

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2 years ago

Please help me Show your work

GrogVix [38]

ANSWER TO PART A

The given triangle has vertices $J(-4,1), K(-4,-2),L(-3,-1)$

The mapping for rotation through $90\degree$ counterclockwise has the mapping

$(x,y)\rightarrow (-y,x)$

Therefore

$J(-4,1)\rightarrow J'(-1,-4)$

$K(-4,-2)\rightarrow K'(2,-4)$

$L(-3,-1)\rightarrow L'(1,-3)$

We plot all this point and connect them with straight lines.

ANSWER TO PART B

For a reflection across the y-axis we negate the x coordinates.

The mapping is

$(x,y)\rightarrow (-x,y)$

Therefore

$J(-4,1)\rightarrow J''(4,1)$

$K(-4,-2)\rightarrow K''(4,-2)$

$L(-3,-1)\rightarrow L''(3,-1)$

We plot all this point and connect them with straight lines.

See graph in attachment

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3 years ago

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Answer:

Step-by-step explanation:

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2 years ago

A circle has the equation 2x²+12x+2y²−16y−150=0.

KonstantinChe [14]

Answer: B. The coordinates of the center are (-3,4), and the length of the radius is 10 units.

Step-by-step explanation:

The equation of a circle in the center-radius form is:

$(x-h)^{2} +(y-k)^{2}=r^{2}$ (1)

Where $(h,k)$ are the coordinates of the center and $r$ is the radius.

Now, we are given the equation of this circle as follows:

$2x^{2}+12x+2y^{2}-16y-150=0$ (2)

And we have to write it in the format of equation (1). So, let's begin by applying common factor 2 in the left side of the equation:

$2(x^{2}+6x+y^{2}-8y-75)=0$ (3)

Rearranging the equation:

$x^{2}+6x+y^{2}-8y=75$ (4)

$(x^{2}+6x)+(y^{2}-8y)=75$ (5)

Now we have to complete the square in both parenthesis, in order to have a perfect square trinomial in the form of $(a\pm b)^{2}=a^{2}\pm+2ab+b^{2}$ :

For the first parenthesis:

$x^{2}+6x+b^{2}$

We can rewrite this as:

$x^{2}+2(3)x+b^{2}$

Hence in this case $b=3$ and $b^{2}=9$ :

$x^{2}+2(3)x+3^{2}=x^{2}+6x+9=(x+3)^{2}$

For the second parenthesis:

$y^{2}-8y+b^{2}$

We can rewrite this as:

$y^{2}-2(4)y+b^{2}$

Hence in this case $b=-3$ and $b^{2}=9$ :

$y^{2}-2(4)y+4^{2}=y^{2}-8y+16=(y-4)^{2}$

Then, equation (5) is rewritten as follows:

$(x^{2}+6x+9)+(y^{2}-8y+16)=75+9+16$ (6)

Note we are adding 9 and 16 in both sides of the equation in order to keep the equality.

Rearranging:

$(x-3)^{2}+(y-4)^{2}=100$ (7)

At this point we have the circle equation in the center radius form $(x-h)^{2} +(y-k)^{2}=r^{2}$

Hence:

$h=-3$

$k=4$

$r=\sqrt{100}=10$

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3 years ago