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3 years ago

What is the answer to this geometry problem?

Mathematics

1 answer:

viva [34]3 years ago

8 0

Answer:

$x=29$

Step-by-step explanation:

Here, MN= NP

→ $128=2x+70$

→ $2x=128-70$

Subtract :-

→ $2x=58$

Divide both sides by 2:-

→ $\frac{2x}{2}=\frac{58}{2}$

→ $x=29$

OAmalOHopeO

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22. Find the perimeter and area.

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Answer:

Part 22) The area is $A=15a^3b^6\ units^2($ and the perimeter is $P=10a^2b^4+6ab^2\ unit$

Part 24) The area is $A=16m^3n\ units^2$ and the perimeter is $P=24mn\ units$

Part 26) The area is equal to $A=9\pi x^6y^{2}\ units^2$

Step-by-step explanation:

Part 22) Find the perimeter and area

step 1

The area of a rectangle is equal to

$A=LW$

we have

$L=5(a^2)(b^4)\ units$

$W=3(a)(b^2)\ units$

Remember that

When multiply exponents with the same base, adds the exponents and maintain the base

substitute in the formula

$A=(5(a^2)(b^4))(3(a)(b^2))$

$A=15a^3b^6\ units^2$

step 2

The perimeter of a rectangle is equal to

$P=2(L+W)$

we have

$L=5(a^2)(b^4)\ units$

$W=3(a)(b^2)\ units$

substitute in the formula

$P=2(5(a^2)(b^4)+3(a)(b^2))$

$P=10a^2b^4+6ab^2\ unit$

Part 24) Find the perimeter and area

step 1

The area of triangle is equal to

$A=\frac{1}{2}bh$

where

$b=8mn\ units$

$h=4m^2\ units$

Remember that

When multiply exponents with the same base, adds the exponents and maintain the base

substitute the given values

$A=\frac{1}{2}(8mn)(4m^2)$

$A=16m^3n\ units^2$

step 2

Find the perimeter

I will assume that is an equilateral triangle (has three equal length sides)

The perimeter of an equilateral triangle is

$P=3b$

where

$b=8mn\ units$

substitute

$P=3(8mn)$

$P=24mn\ units$

Part 26) Find the area

The area of a circle is equal to

$A=\pi r^{2}$

where

$r=3x^3y\ units$

Remember the property

$(a^{m})^{n}=a^{m*n}$

substitute in the formula the given value

$A=\pi (3x^3y)^{2}$

$A=9\pi x^6y^{2}\ units^2$

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Triangle ABC has vertices A(-2, 3), B(0, 3), and C(-1,-1). Find the coordinates of the image after a reflection over the

motikmotik

Given:

Given that the triangle ABC has vertices A(-2,3), B(0,3) and C(-1,-1).

We need to determine the coordinates of the image after a reflection over the x - axis.

Let A'B'C' denote the coordinates of the triangle after a reflection over the x - axis.

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The general rule to reflect the coordinate across the x - axis is given by

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

Substituting the point A(-2,3), we get;

$(-2,3)\rightarrow (-2,-3)$

Thus, the coordinates of the point A' is (-2,-3)

Coordinates of the point B':

The general rule to reflect the coordinate across the x - axis is given by

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

Substituting the point B(0,3), we get;

$(0,3)\rightarrow (0,-3)$

Thus, the coordinates of the point B' is (0,-3)

Coordinates of the point C':

The general rule to reflect the coordinate across the x - axis is given by

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

Substituting the point C(-1,-1), we get;

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Thus, the coordinates of the point C' is (-1,1)

Hence, the coordinates of the image after a reflection over the x - axis is A'(-2,-3), B(0,-3) and C(-1,1)

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