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Setler [38]
3 years ago
7

If cos(2/3x+20)°=sin(2x−10)°, find the acute angles of the corresponding right triangle.

Mathematics
1 answer:
adell [148]3 years ago
7 0

Answer:

A = 40°

B = 50°

Step-by-step explanation:

If cos A = sin B, then A + B = 90

So, 2x/3 + 20  + 2x - 10 = 90

      2x + 60 + 6x - 30 = 270

               8x + 30 = 270

                       8x = 240

                          x = 30

2x/3 + 20 = 60/3 + 20 = 20 + 20 = 40

2x - 10 = 2(30) - 10 = 60 - 10 = 50  or  90 - 40 = 50

       

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Leann is 3 cm taller than Fred. If Fred's height is f cm, then Leann's height is cm?
jek_recluse [69]
F + 3 = leann's height
6 0
3 years ago
Divide using synthetic division -3 1 8 9 -18
agasfer [191]

Answer:

Step-by-step explanation:

-3   /   1    8    9    -18

              -3    -15   18

     _______________

        1      5     -6     0

The remainder here is 0 and the quoitient is 1x^2 + 5x - 6.

3 0
3 years ago
An artists builds a sculpture out of metal and wood that weighs 14.9 kilograms.3/4 of this weight is metal,and the rest is wood.
Softa [21]

Given is a sculpture that is built up of metal and wood, and the weight of this sculpture is 14.9 kilograms.

It says that three-fourth of the total weight is metal and says to find the weight of wooden part.

We can find the weight of metal part and then subtract it from the total weight to calculate the answer.

Total weight = 14.9 kilograms.

Metal part = three-fourth of total weight = \frac{3}{4} *14.9=\frac{44.7}{4} =11.175 \;kilograms

Wooden part = 14.9 - 11.175 = 3.725 kilograms.

Hence, the wooden part is 3.725 kilograms.

7 0
2 years ago
Read 2 more answers
a set v is given, together with definitions of addition and scalar multiplication. determine which properties of a vector space
agasfer [191]

The properties of a vector space are satisfied Properties 1,2, 5(a) and 5(c) are satisfied, the relaxation of the homes aren't legitimate are ifv = x ^ 2 1× v=1^ ×x ^ 2 = 1 #V

Property three does now no longer follow: Suppose that Property three is legitimate, shall we namev = a * x ^ 2 +bx +cthe neuter of V. Since v is the neuter, then O have to be constant with the aid of using the neuted, consequently 0 = O + v = (O  x ^ 2 + Ox + O) + (a × x ^ 2 + bx + c) = c × x ^ 2 + b ^ 2 + a

= 0 If O is the neuter, then it ought to restore x², but 0+ x² = (0x²+0x+zero) + (x²+0x+zero) = 1.This is a contradiction due to the fact x² isn't 1. We finish that V doesnt have a neuter vector. This additionally method that belongings four would not observe either. A set with out 0 cant

have additive inverse

Let r= v ×2x ^ 2 + v × 1x +v0 , w= w ×2x ^ 2 + w × 1x +w0 . We have that\\v+w= (vO + wO) ^  x^ 2 +(vl^ × wl)^  x+ ( v 2^ × w2)• w+v= (wO + vO) ^x^ 2 +(wl^ × vl)x+ ( w 2^ ×v2)

Since the sum of actual numbers is commutative, we finish that v + w = w + v Therefore, belongings 5(a) is valid.

Property 5(b) isn't valid: we are able to introduce

a counter example. we could use z = 1 thenv = x ^ 2 w = x ^ 2 + 1\\(v + w) + z = (x ^ 2 + 2) + 1 = 3x ^ 2 + 1

v + (w + z) = x ^ 2 + (2x ^ 2 + 1) = x ^ 2 + 3

Since 3x ^ 2 +1 ne x^ 2 +3. then the associativity rule doesnt hold.

(1+2)^ * (x^ 2 +x)=3^ * (x ^ 2 + x) = 3x + 3\\1^ × (x^ 2 +x)+2^ × (x ^ 2 + x) = (x + 1) + (2x + 2) = 3x ^ 2 + x ( ne 3x + 3 )\\(1^ ×2)^ ×(x^ 2 +x)=2^ × (x ^ 2 + x) = 2x + 2\\1^ × (2^ × (x ^ 2 + x) )=1^ × (2x+2)=2x^ 2 +2x( ne2x+2)

Property f doesnt observe because of the switch of variables. for instance, if v = x ^ 2 1 × v=1^ × x ^ 2 = 1 #V

Properties 1,2, 5(a) and 5(c) are satisfied, the relaxation of the homes arent legitimate.

Step-with the aid of using-step explanation:

Note that each sum and scalar multiplication entails in replacing the order from that most important coefficient with the impartial time period earlier than doing the same old sum/scalar multiplication.

Property three does now no longer follow: Suppose that Property three is legitimate, shall we name v = a × x ^ 2 +bx +c the neuter of V. Since v is the neuter, then O have to be constant with the aid of using the neuted, consequently0 = O + v = (O × x ^ 2 + Ox + O) + (a × x ^ 2 + bx + c) = c × x ^ 2 + b ^ 2 + a

= zero If O is the neuter, then it ought to restore x², but zero + x² = (0x²+0x+zero) + (x²+0x+zero) = 1.This is a contradiction due to the fact x² isn't 1. We finish that V doesnt have a neuter vector. This additionally method that belongings four would not observe either. A set with out 0 cant have additive inverse

Let r= v × 2x ^ 2 + v × 1x +v0 , w= w2x ^ 2 + w × 1x +w0 . \\We have thatv+w= (vO + wO) ^ x^ 2 +(vl^ wl)^x+ ( v 2^ w2)w+v= (wO + vO) ^ x^ 2 +(wl^ vl)x+ ( w 2^v2)

Since the sum of actual numbers is commutative, we finish that v + w = w + v Therefore, belongings 5(a) is valid.

Property 5(b) isn't valid: we are able to introduce

a counter example. we could usez = 1 then v = x ^ 2 w = x ^ 2 + 1(v + w) + z = (x ^ 2 + 2) + 1 = 3x ^ 2 + 1v + (w + z) = x ^ 2 + (2x ^ 2 + 1) = x ^ 2 + 3\\Since 3x ^ 2 +1 ne x^ 2 +3.then the associativity rule doesnt hold.

Note that each expressions are same because of the distributive rule of actual numbers. Also, you could be aware that his assets holds due to the fact in each instances we 'switch variables twice.

· (1+2)^ * (x^ 2 +x)=3^ * (x ^ 2 + x) = 3x + 31^ * (x^ 2 +x)+2^ * (x ^ 2 + x) = (x + 1) + (2x + 2) = 3x ^ 2 + x ( ne 3x + 3 )(1^ * 2)^ * (x^ 2 +x)=2^ * (x ^ 2 + x) = 2x + 21^ * (2^ * (x ^ 2 + x) )=1^ ×* (2x+2)=2x^ 2 +2x( ne2x+2)

Read more about polynomials :

brainly.com/question/2833285

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8 0
10 months ago
Evaluate the expression when x = 8 and y = 1.
Rzqust [24]
The answer should be 4.

First, simplify \frac{8}{2} to 4. Your problem should look like: 4 x 1. 
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Also, 
to start the equation up, make sure to add in the numbers in replace for the letters. Your problem to start with should look like: \frac{8}{2(1)}.

6 0
3 years ago
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