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1 year ago

Choose the number sentence whose product is 2.015.

Mathematics

2 answers:

Mariulka [41]1 year ago

8 0

I’m not too sure. Sorry

Dmitrij [34]1 year ago

7 0

I took test seems to be correct and fine thanks good luck

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What basic trigonometric identity would you use to verify that (sin2x + cos2x)/cosx=secx

Oksana_A [137]

Answer:

Step-by-step explanation:

sin(2x) = 2 sin(x) cos(x) cos(2x) = cos2(x) – sin2(x) = 1 – 2 sin2(x) = 2 cos2(x) – 1. Now I am not sure if this is right but I remember another similar formula. Here is the correct formula cot x = cos x / sin x

2 ) ( sin² x + cos² x ) / cos x = sec x

1/cos x = sec x

sec x = sec x

cos² x + sin² x = 1

7 0

2 years ago

Read 2 more answers

What is the volume of this cone?

olchik [2.2K]

Answer:

The volume of the cone is approximately 453.0 cm³

Step-by-step explanation:

The volume of a cone is one third that of a cylinder with the same height and radius. That gives us 1/3 πr²h, where r is radius and h is height.

However, we are not given the height of the cone, but the side length. We can work out the height using the Pythagorean theorem, as we have a right triangle with the height, base radius, and length. You may recall that the Pythagorean theorem states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of it's other two sides:

$a^2 = b^2 + c^2$

So we can find the height of the cone with that:

$10.8^2 = 7.4^2 + h^2\\h^2 = 10.8^2 - 7.4^2\\h^2 = 116.64 - 54.76\\h^2 = 61.88\\h = \sqrt{61.88}\\h \approx 7.9$

Now that we have the cone's height, we can solve for its volume:

$v = \frac{1}{3} \pi r^2 h\\v = \frac{1}{3} \pi \times 7.4^2 \times 7.9\\v = \frac{1}{3} \pi \times 54.76 \times 7.9\\v = \frac{1}{3} \pi \times 432.6\\v = \pi \times 144.2\\v \approx 453.0 cm^3$

5 0

3 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 if $v = x ^ 2 1× v=1^ ×x ^ 2 = 1 #V$

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, 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 then $v = x ^ 2 w = x ^ 2 + 1\\(v + w) + z = (x ^ 2 + 2) + 1 = 3x ^ 2 + 1$

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, consequently $0 = O + v = (O × x ^ 2 + Ox + O) + (a × x ^ 2 + bx + c) = c × x ^ 2 + b ^ 2 + a$

= 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 use $z = 1 then v = x ^ 2 w = x ^ 2 + 1(v + w) + z = (x ^ 2 + 2) + 1 = 3x ^ 2 + 1v + (w + z) = x ^ 2 + (2x ^ 2 + 1) = x ^ 2 + 3\\Since 3x ^ 2 +1 ne x^ 2 +3.$ 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.

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