MATHEMATICS
COLLEGE

Answer:

Correct to the nearest degree, the three angles of the triangle are , , and and this can be determined by using the properties of the triangle and dot product formula.

Given :

The vertices of triangle: a(1, 0, −1), b(5, −4, 0), c(1, 4, 5)

Let the angles are , , and . Now, the magnitude of AB will be:

Now, the magnitude of AC will be:

Using dot product for finding the angles:

Now, the magnitude of BA will be:

Now, the magnitude of BC will be:

Using dot product for finding the angles:

Now, the sum of interior angles of the triangle is , then:

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Answer:

Answer:

The three angles are 43.11°, 103.95° and 32.94°.

Step-by-step explanation:

Let us name the corresponding angles to A, B and C to be ∠a, ∠b and ∠c respectively as shown in the figure below.

Using the vector representation of lines, we get,

Vector AB = < 1-5, 0+4, -1-0 > = < -4, 4, -1 >

Vector AC = < 1-5, 4+4, 5-0 > = < -4, 8, 5 >

Also, we have the modulus value given by,

|AB| = = = 5.75

|AC| = = = 10.25

Now, using the dot product, we have,

AB · AC = |AB| |AC| cos(a)

< -4,8,5> }{5.75 \times 10.25}" alt="\cos a=\frac{< -4,4,-1> < -4,8,5> }{5.75 \times 10.25}" align="absmiddle" class="latex-formula">

i.e.

i.e.

i.e.

i.e.

i.e.

Hence, ∠a = 43.11°

Again, we see that,

Vector BA = < 5-1, -4-0, 0+1 > = < 4, -4, 1 >

Vector BC = < 1-1,4-0, 5+1 > = < 0, 4, 6 >

Also, we have the modulus value given by,

|Ba| = = = 5.75

|BC| = = = 7.21

Now, using the dot product, we have,

BA · BC = |BA| |BC| cos(b)

< 0,4,6> }{5.75 \times 7.21}" alt="\cos b=\frac{< 4,-4,1> < 0,4,6> }{5.75 \times 7.21}" align="absmiddle" class="latex-formula">

i.e.

i.e.

i.e.

i.e.

i.e.

Hence, ∠b = 103.95°

Since, the sum of all the angles in a triangle is 180°.

Thus, ∠a + ∠b + ∠c = 180°

i.e. 43.11° + 103.95° + ∠c = 180°

i.e. ∠c = 180° - 147.06°

i.e. ∠c = 32.94°

Hence, the three angles are 43.11°, 103.95° and 32.94°.

COLLEGE

X3 + 11x2 – 3x – 33 by grouping?

Simplify 11x2 to 22

X^3+22-3x-33

Collect like terms

X^3(22-33)-3x

Simplify

X^3-11-3x

X^3+22-3x-33

Collect like terms

X^3(22-33)-3x

Simplify

X^3-11-3x

HIGH SCHOOL

What is the following profuct?

( sqrt 14 - sqrt 3) ( sqrt 12 + sqrt 7)

I believe the answer is 8.10

Answer: 2 √42 + 7 √2 - 6 - √21

explanation: edge 2021

MIDDLE SCHOOL

What is 38,256.5 to the nearest whole number

38,257 this is because with rounding if the number is 5 or above you give it a shove up, if it's 4 or below let it go

MIDDLE SCHOOL

150 + 1000 -20 + 4488989 x 8

The answer is 35,920,952. Hope I helped (:

35913042 hope this helps

HIGH SCHOOL

Given the function f(x) = 3x, find the value of f−1(27). f−1(27) = 1 f−1(27) = 3 f−1(27) = 9 f−1(27) = 81

Fun fact

if f(a)=b then f⁻¹(b)=a

so

if f⁻¹(b)=a then f(a)=b

so we find f(x)=27 and the solve for x

f(x)=3x=27

3x=27

divide both sides by 3

x=9

so if f(9)=27 then f⁻¹(27)=9

problem solved

answer is 9

if f(a)=b then f⁻¹(b)=a

so

if f⁻¹(b)=a then f(a)=b

so we find f(x)=27 and the solve for x

f(x)=3x=27

3x=27

divide both sides by 3

x=9

so if f(9)=27 then f⁻¹(27)=9

problem solved

answer is 9

MIDDLE SCHOOL

Which is a solution to the equation y equals 3x + 1 ? A.

(3, 10)

B.

(2, 6)

C.

(1, 5)

D.

(0, 3)

It is a, if you plug it in, it works. 10=3*3+1 is true.

HIGH SCHOOL

F two angles form a linear pair, then they are adjacent and

A linear pair is two angles that are adjacent and whose non-common sides form a straight line. If two angles are a linear pair, then they are supplementary.

MIDDLE SCHOOL

Solve:

3x+5-13x = 25

3x+5-13x = 25

Subtract 13x from 3x

5 - 10x = 25

Subtract 5 from both sides

-10x = 20

Divide -10 on both sides so that the only thing remaining on the left side is the variable x.

Final Answer: -2

Subtract 13x from 3x

5 - 10x = 25

Subtract 5 from both sides

-10x = 20

Divide -10 on both sides so that the only thing remaining on the left side is the variable x.

Final Answer: -2

MIDDLE SCHOOL

(what is prime number )?

A number that can only be divided by itself and 1

Step-by-step explanation:

a number that cant have more than one factor pair