All categories

3 years ago

Consider the figure below. Line FG is parallel to line JK.

Mathematics

1 answer:

Svet_ta [14]3 years ago

6 0

Answer:

A. a = 10

B. m<FNT = 120°

C. m<KTU = 60°

Step-by-step explanation:

A. (7a + 50)° and (14a - 20)° are corresponding angles. Therefore:

(7a + 50)° = (14a - 20)°

Use this equation to find the value of a

7a + 50 = 14a - 20

Combine like terms

7a - 14a = - 50 - 20

-7a = -70

Divide both sides by -7

-7a/-7 = -70/-7

a = 10

B. m<FNT = (14a - 20)° (alternate interior angles are congruent)

Plug in the value of a

m<FNT = 14(10) - 20

m<FNT = 140 - 20

m<FNT = 120°

C. m<KTU + (14a - 20)° = 180° (linear pair)

Plug in the value of a

m<KTU + 14(10) - 20 = 180

m<KTU + 120 = 180

m<KTU + 120 - 120 = 180 - 120

m<KTU = 60°

You might be interested in

What are the zeros of f(x) = x2 + 3x – 10?

Nady [450]

Step One
Factor f(x)
f(x) = (x + 5)(x - 2)

Step Two
The zeros occur when either (x + 5) = 0 or (x - 2) = 0

x + 5 = 0
x = -5

x - 2 =0
x = 2

Step 3
Record the zeros
(-5,0) or (2,0) <<<<<< answer.

3 0

3 years ago

Calculate 11010-1101 in base 2

stiv31 [10]

Answer:

1101 base 2

Step-by-step explanation:

you first take the larger number up the the lower number down

11010 -1101
when dividing if it's zero you borrow one from the number on the left hand side
since it's in base 2 the number you borrow would be 2 then u subtract
I hoped it helped u

6 0

3 years ago

Here is a triangular pyramid and its net.

bazaltina [42]

Answer:

a) Area of the base of the pyramid = $15.6\ mm^{2}$

b) Area of one lateral face = $24\ mm^{2}$

c) Lateral Surface Area = $72\ mm^{2}$

d) Total Surface Area = $87.6\ mm^{2}$

Step-by-step explanation:

We are given the following dimensions of the triangular pyramid:

Side of triangular base = 6mm

Height of triangular base = 5.2mm

Base of lateral face (triangular) = 6mm

Height of lateral face (triangular) = 8mm

a) To find Area of base of pyramid:

We know that it is a triangular pyramid and the base is a equilateral triangle. $\text{Area of triangle = } \dfrac{1}{2} \times \text{Base} \times \text{Height} ..... (1)\\$