All categories

3 years ago

2. The lengths of the sides of a triangle are 4cm, 5cm and 6cm. Find, in degrees to one decimal

Mathematics

1 answer:

aivan3 [116]3 years ago

4 0

Answer:

82.8 degrees.

Step-by-step explanation:

The largest angle is opposite the longest side.

So we find the angle opposite the 6cm line using The Cosine Rule:

6^2 = 4^2 + 5^2 - 2.4.5 cos A

cos A = (6^2 - 4^2 - 5^2) / (-2*4*5)

= -5/ -40

= 0.125.

So m angle A = 82.819 degrees.

You might be interested in

Solve problems <br>5 1/2×3 1/3

makvit [3.9K]

5 1/2 X 3 1/3=18 1/3

6 0

2 years ago

Use exponents to write each expression. Then evaluate the expression. 3x3x3x3x3

nalin [4]

Step-by-step explanation:

3 to the power of 5

3^5

= 3x3x3x3x3

= 243

5 0

2 years ago

Read 2 more answers

A woman is 59 and one twelfth inches tall and her son is 57 and five ninths inches tall. How much taller is the woman?

mihalych1998 [28]

Answer:

1 and 19/36 inches

Step-by-step explanation:

All you have to do is subtract the woman's height by her son's height.

59 and 1/12 - 57 and 5/9 is 1 19/36

7 0

2 years ago

Solve -5(3n + 4) = 40.

Sveta_85 [38]

N=-4
Distribute the -5
-15n-20=40
-15n=60
n=-4

3 0

2 years ago

Factor f(x) = 15x^3 - 15x^2 - 90x completely and determine the exact value(s) of the zero(s) and enter them as a comma separated

Illusion [34]

Answer:

$x=-2,0,3$

Step-by-step explanation:

We have been given a function $f(x)=15x^3-15x^2-90x$ . We are asked to find the zeros of our given function.

To find the zeros of our given function, we will equate our given function by 0 as shown below:

$15x^3-15x^2-90x=0$

Now, we will factor our equation. We can see that all terms of our equation a common factor that is $15x$ .

Upon factoring out $15x$ , we will get:

$15x(x^2-x-6)=0$

Now, we will split the middle term of our equation into parts, whose sum is $-1$ and whose product is $-6$ . We know such two numbers are $-3\text{ and }2$ .

$15x(x^2-3x+2x-6)=0$

$15x((x^2-3x)+(2x-6))=0$

$15x(x(x-3)+2(x-3))=0$

$15x(x-3)(x+2)=0$

Now, we will use zero product property to find the zeros of our given function.

$15x=0\text{ (or) }(x-3)=0\text{ (or) }(x+2)=0$

$15x=0\text{ (or) }x-3=0\text{ (or) }x+2=0$

$\frac{15x}{15}=\frac{0}{15}\text{ (or) }x-3=0\text{ (or) }x+2=0$

$x=0\text{ (or) }x=3\text{ (or) }x=-2$

Therefore, the zeros of our given function are $x=-2,0,3$ .

7 0

3 years ago