Mia wants to construct a triangle with angles measuring 51*, 102*and 37*. how many different triangles can she construct?

What is the value of 40/z, when z = 10? <br><br>I WILL GIBE BRAINLY POINTS

Mamont248 [21]

Replace x with 10 and simplify the fraction.

40/10 = 4/1 = 4

The answer is 4

5 0

2 years ago

This trapezoid is made up of two right triangles and one rectangle.

artcher [175]

9514 1404 393

Answer:

each triangle: 29.75 in²
rectangle: 374 in²
total area: 433.5 in²

Step-by-step explanation:

The base of each triangle is half the difference between the bases of the trapezoid:

(29 in -22 in)/2 = 7 in/2 = 3.5 in

The area of the triangle is given by the formula ...

A = 1/2bh

A = 1/2(3.5 in)(17 in) = 29.75 in² . . . triangle area

__

The area of the central rectangle is the product of its base and height.

A = bh = (22 in)(17 in) = 374 in² . . . rectangle area

__

The total area of the trapezoid is the sum of the areas of the two triangles and the area of the rectangle.

total area = 2(29.75 in²) + 374 in² = 433.5 in² . . . total trapezoid area

5 0

3 years ago

Which graph represents a function?

sweet [91]

Answer:

This graph does not represent a function.

Step-by-step explanation:

This graph does not pass the vertical line test. The vertical line test states that if a vertical line drawn intersects the graph of the relation more than once, then the relation is a NOT a function.

5 0

3 years ago

Read 2 more answers

The midpoint of the line segment from P4 to P2 is (-5,1). If P1 = (-7,9), what is P2?

balandron [24]

Let A(2,0),B(0,92)

Let A(2,0),B(0,92)and let mid-point be C(1,3p)

Let A(2,0),B(0,92)and let mid-point be C(1,3p)Finding coordinates of C

Let A(2,0),B(0,92)and let mid-point be C(1,3p)Finding coordinates of C(1,3p)=⎝⎜⎜⎛22+0,20+92⎠⎟⎟⎞

Let A(2,0),B(0,92)and let mid-point be C(1,3p)Finding coordinates of C(1,3p)=⎝⎜⎜⎛22+0,20+92⎠⎟⎟⎞⇒(1,3p)=(1,91)

Let A(2,0),B(0,92)and let mid-point be C(1,3p)Finding coordinates of C(1,3p)=⎝⎜⎜⎛22+0,20+92⎠⎟⎟⎞⇒(1,3p)=(1,91)Comparing y-coordinate

Let A(2,0),B(0,92)and let mid-point be C(1,3p)Finding coordinates of C(1,3p)=⎝⎜⎜⎛22+0,20+92⎠⎟⎟⎞⇒(1,3p)=(1,91)Comparing y-coordinate3p=91

Let A(2,0),B(0,92)and let mid-point be C(1,3p)Finding coordinates of C(1,3p)=⎝⎜⎜⎛22+0,20+92⎠⎟⎟⎞⇒(1,3p)=(1,91)Comparing y-coordinate3p=91∴p=31

Let A(2,0),B(0,92)and let mid-point be C(1,3p)Finding coordinates of C(1,3p)=⎝⎜⎜⎛22+0,20+92⎠⎟⎟⎞⇒(1,3p)=(1,91)Comparing y-coordinate3p=91∴p=31We need to show that line 5x+3y+2=0 passes through point (−1,3p)

Let A(2,0),B(0,92)and let mid-point be C(1,3p)Finding coordinates of C(1,3p)=⎝⎜⎜⎛22+0,20+92⎠⎟⎟⎞⇒(1,3p)=(1,91)Comparing y-coordinate3p=91∴p=31We need to show that line 5x+3y+2=0 passes through point (−1,3p)Point =(−1,3p)=(−1,3×31)=(−1,1)

Let A(2,0),B(0,92)and let mid-point be C(1,3p)Finding coordinates of C(1,3p)=⎝⎜⎜⎛22+0,20+92⎠⎟⎟⎞⇒(1,3p)=(1,91)Comparing y-coordinate3p=91∴p=31We need to show that line 5x+3y+2=0 passes through point (−1,3p)Point =(−1,3p)=(−1,3×31)=(−1,1)Putting (−1,1) in line

Let A(2,0),B(0,92)and let mid-point be C(1,3p)Finding coordinates of C(1,3p)=⎝⎜⎜⎛22+0,20+92⎠⎟⎟⎞⇒(1,3p)=(1,91)Comparing y-coordinate3p=91∴p=31We need to show that line 5x+3y+2=0 passes through point (−1,3p)Point =(−1,3p)=(−1,3×31)=(−1,1)Putting (−1,1) in line5x+3y+2=0

Let A(2,0),B(0,92)and let mid-point be C(1,3p)Finding coordinates of C(1,3p)=⎝⎜⎜⎛22+0,20+92⎠⎟⎟⎞⇒(1,3p)=(1,91)Comparing y-coordinate3p=91∴p=31We need to show that line 5x+3y+2=0 passes through point (−1,3p)Point =(−1,3p)=(−1,3×31)=(−1,1)Putting (−1,1) in line5x+3y+2=0⇒5(−1)+3(1)+2=0

4 0

3 years ago

Mcd de 120 , 160 , 180 , 200 paso por paso (el que responda le doy corona)

olga55 [171]

Answer:

The Highest Common Factor is 20

Step-by-step explanation:

Aquí le mostraremos paso a paso como hallar el Máximo Común Divisor de 120 y 180. (mcd de 120 y 180) Paso 1: Creamos una lista de todos los divisores de 120, que incluye 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, y 120. Paso 2: Siguiente, creamos una lista de todos los divisores de 180, que incluye 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, y 180. Paso 3: Finalmente, comparamos las listas de Paso 1 y Paso 2 y hallamos que el mayor número que tienen en común es 60. Esto significa que el Máximo Común Divisor de 120 y 180 es 60. MCD de 120 y 180 = 60

4 0

3 years ago