Write in words 1,080,000.

stellarik [79]

$\cos 27^{\circ} = \dfrac{KL}x\\\\\implies \cos 27^{\circ} = \dfrac{6.3}x\\\\\implies x = \dfrac{6.3}{\cos 27^{\circ}}=7.1~ \text{units.}$

6 0

2 years ago

A rectangular tank has length 4m, width 12m and height 3.5..i. If the tank is filled with water 2/3 of its capacity ,calculate t

ipn [44]

Answer:

The volume of the water in the tank is 112 m³

Step-by-step explanation:

The volume of the rectangular prism V = L × W × H, where

L is its length

W is its width

H is its height

∵ A rectangular tank has a length of 4 m, a width of 12 m, and

a height of 3.5 m

∴ L = 4 m

∴ W = 12 m

∴ H = 3.5 m

∵ V = L × W × H

∴ The volume of the tank = 4 × 12 × 3.5

∴ The volume of the tank = 168 m³

∵ The tank is filled with water $\frac{2}{3}$ of its capacity

→ That means the volume of the water is $\frac{2}{3}$ the volume of the tank

∵ The volume of the water = $\frac{2}{3}$ the volume of the tank

∴ The volume of the water = $\frac{2}{3}$ × 168

∴ The volume of the water = 112 m³

∴ The volume of the water in the tank is 112 m³

3 0

3 years ago

If the tape diagram represents a 10-pound bag of almonds, but a baker only uses 3/4 of a bag (the shaded region), how many pound

ale4655 [162]

Answer:

7.5 lbs.

Step-by-step explanation:

A fraction is a division problem so to turn it into a decimal you divide 3 by 4. 0.75. 0.75 multiplied by 10 lbs. is 7.5 lbs.

5 0

3 years ago

Read 2 more answers

In order for Brady to earn a B in his biology course, his test scores must average at least 80%. On the first 5 tests, he has an

larisa [96]

Hey there! I'm happy to help!

There were 5 tests needed to make an average of 77. These numbers have to add up to 385 (77×5) because that is what you divide by 5 to find the average which is 77.

Now, we want to find an average of 80 with 6 numbers. We multiply 80 and 6, showing us that the sum of these 6 numbers must be 480. Since 5 of these numbers are the same as with the 77 average, we know that one test score (the one Brady needs to get) is added to 385 to give us 480.

480-385=95

Therefore, Brady needs to earn a 95% on the last test to receive a B in the class.

8 0

3 years ago

Let V denote the set of ordered triples (x, y, z) and define addition in V as in

icang [17]

Answer:

a) No

b) No

c) No

d) No

Step-by-step explanation:

Remember, a set V wit the operations addition and scalar product is a vector space if the following conditions are valid for all u, v, w∈V and for all scalars c and d:

1. u+v∈V

2. u+v=v+u

3. (u+v)+w=u+(v+w).

4. Exist 0∈V such that u+0=u

5. For each u∈V exist −u∈V such that u+(−u)=0.

6. if c is an escalar and u∈V, then cu∈V

7. c(u+v)=cu+cv

8. (c+d)u=cu+du

9. c(du)=(cd)u

10. 1u=u

let's check each of the properties for the respective operations:

Let $u=(u_1,u_2,u_3), v=(v_1,v_2,v_3)$

Observe that

1. u+v∈V

2. u+v=v+u, because the adittion of reals is conmutative

3. (u+v)+w=u+(v+w). because the adittion of reals is associative

4. $(u_1,u_2,u_3)+(0,0,0)=(u_1+0,u_2+0,u_3+0)=(u_1,u_2,u_3)$

5. $(u_1,u_2,u_3)+(-u_1,-u_2,-u_3)=(0,0,0)$

then regardless of the escalar product, the first five properties are met for a), b), c) and d). Now let's verify that properties 6-10 are met.

a)

6. $c(u_1,u_2,u_3)=(cu_1,u_2,cu_3)\in V$

7.

$c(u+v)=c(u_1+v_1,u_2+v_2,u_3+v_3)=(c(u_1+v_1),u_2+v_2,c(u_3+v_3))\\=(cu_1+cv_1,u_2+v_2,cu_3+cv_3)=c(u_1,u_2,u_3)+c(v_1,v_2,v_3)=cu+cv$

8.

$(c+d)u=(c+d)(u_1,u_2,u_3)=((c+d)u_1,u_2,(c+d)u_3)=\\=(cu_1+du_1,u_2,cu_3+du_3)\neq (cu_1+du_1,2u_2,cu_3+du_3)=cu+du$

Since 8 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product $a(x,y,z)=(ax,y,az)$

b) 6. $c(u_1,u_2,u_3)=(cu_1,0,cu_3)\in V$

7.

$c(u+v)=c(u_1+v_1,u_2+v_2,u_3+v_3)=(c(u_1+v_1),0,c(u_3+v_3))\\=(cu_1+cv_1,0,cu_3+cv_3)=c(u_1,u_2,u_3)+c(v_1,v_2,v_3)=cu+cv$

8.

$(c+d)u=(c+d)(u_1,u_2,u_3)=((c+d)u_1,0,(c+d)u_3)=\\=(cu_1+du_1,0,cu_3+du_3)=(cu_1,0,cu_3)+(du_1,0,du_3) =cu+du$

9.

$c(du)=c(d(u_,u_2,u_3))=c(du_1,0,du_3)=(cdu_1,0,cdu_3)=(cd)u$

10

$1u=1(u_1,u_2,u3)=(1u_1,0,1u_3)=(u_1,0,u_3)\neq(u_1,u_2,u_3)$

Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product $a(x,y,z)=(ax,0,az)$

c) Observe that $1u=1(u_1,u_2,u3)=(0,0,0)\neq(u_1,u_2,u_3)$

Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product $a(x,y,z)=(0,0,0)$ .

d) Observe that $1u=1(u_1,u_2,u3)=(2*1u_1,2*1u_2,2*1u_3)=(2u_1,2u_2,2u_3)\neq(u_1,u_2,u_3)=u$

Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product $a(x,y,z)=(2ax,2ay,2az)$ .

8 0

3 years ago