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3 years ago

A prism of height 12" has a rhombus with diagonals 6" and 8" for a base. Find the volume. 240 cu. in. 288 cu. in. 576 cu. in.

Mathematics

2 answers:

weqwewe [10]3 years ago

6 0

Its not 576 because i got it wrong

oee [108]3 years ago

6 0

Answer:

Option 2nd is correct

volume of prism is, 288 cu. in.

Step-by-step explanation:

Volume of prism(V) is given by:

$V = B \cdot h$ .....[1]

where,

B is the Base area

h is the height of the prism.

Area of rhombus(A) is given by:

$A = \frac{1}{2}d_1 \cdot d_2$

where,

$d_1, d_2$ are the diagonals of the rhombus.

As per the statement:

A prism of height 12" has a rhombus with diagonals 6" and 8" for a base.

⇒ $d_1 = 6"$ , $d_2= 8"$ and h = 12"

Find the base area i.,e Area of rhombus.

$B = \frac{1}{2}(6 \cdot 8)$

Simplify:

$B = 24$ square inches.

Substitute the given values in [1] we have;

$V = 24 \cdot 12 = 288$ cubic inches.

Therefore, the volume of prism is, 288 cu. in.

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6.31 is a rational number.

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2 years ago

8. If the discriminant is positive, there will be ______ real solutions.

Deffense [45]

Answer:

Step-by-step explanation:

The discriminant is the quantity under the radical in the quadratic formula ...

[tex]x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[tex]

When that quantity is positive, the square root is real, and the formula gives two real values for x.

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2 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.

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

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

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

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

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

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

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

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

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

What is the answer to 12➗3624

jeka57 [31]

Solutions

$12/3624
$

$=0.00331125827$

3 0

3 years ago

Read 2 more answers

A bag contains eleven equally sized marbles, which are numbered. Two marbles are chosen at random and replaced after each select

Digiron [165]

Answer:

B. $\frac{24}{121}$ .

Step-by-step explanation:

Total number of marbles = 11.

Since, after choosing the first marble, we are putting it back and then the second marble is chosen.

As, there are 4 shaded marbles.

So, the probability of getting the first marble shaded = $\frac{\binom{4}{1}}{\binom{11}{1}}$ = $\frac{4}{11}$

Also, there are 6 odd labeled marbles.

So, the probability of getting the second marble being odd labeled = $\frac{\binom{6}{1}}{\binom{11}{1}}$ = $\frac{6}{11}$

So, the probability of getting the first marble shaded and second marble labeled odd = $\frac{4}{11}\times \frac{6}{11}$ = $\frac{24}{121}$ .

Hence, the required probability is $\frac{24}{121}$ .

4 0

2 years ago

Read 2 more answers