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andreyandreev [35.5K]
3 years ago
6

What is the volume of the rectangular pyramid shown belown?

Mathematics
1 answer:
Anna35 [415]3 years ago
7 0

9514 1404 393

Answer:

  168 cubic inches

Step-by-step explanation:

The formula for the volume of a pyramid is ...

  V = (1/3)Bh

where B is the area of the base, and h is the height.

The base is a rectangle, whose are is given by the formula ...

  A = LW

  A = (8 in)(7 in) = 56 in²

Then the pyramid volume is ...

  V = (1/3)(56 in²)(9 in) = 168 in³

The volume of the pyramid is 168 cubic inches.

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Pls help (see picture)
PSYCHO15rus [73]

Answer:

6.8

Step-by-step explanation:

Similar means the same shape but smaller. So we have to find the connection between 17 and 5. Which can be achieved by this equation

17/5=3.4

Now we apply the number 3.4 to find the length of PQ using the same side on the similar shape

2*3.4=6.8

Hope this helps and I’m sorry to ask but I would love a Brainliest

8 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 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

#SPJ4

8 0
1 year ago
A rectangle has an area of 3/2 square yards. If the width of the rectangle is 1/2 yard, which statement is true?
dezoksy [38]
The answer is in your heart
7 0
3 years ago
Will mark as Brainliest...
tatiyna

Answer:

280 ft squared

Step-by-step explanation:

To find the area of the nonshaded portion, we can find the area of the entire floor and then subtract the shaded area.

The total area is that of a rectangle: 30 * 15 = 450 ft squared.

Now, the shaded region is made up of a rectangle and a triangle.

- The rectangle has length 8 and width 10, so its area is 10 * 8 = 80 ft squared.

- The triangle has base 12 and height 15, so using the area of a triangle formula: A=(bh)/2 (where b is the base and h is the height) = (12 * 15)/2 = 180/2 = 90 ft squared.

- The total shaded region is: 80 + 90 = 170 ft squared

Subtract 110 from 450: 450 - 170 = 280 ft squared.

Thus, the answer is 280 ft squared.

Hope this helps!

8 0
3 years ago
Read 2 more answers
I need help remembering the steps to solve a polynomial equation
murzikaleks [220]
Example 1 – Solve: 3x3 = 12x

Step 1: Write the equation in the correct form. In this case, we need to set the equation equal to zero with the terms written in descending order.

 3x^3-12=0

Step 2: Use a factoring strategies to factor the problem.

 3x(x^2-4)=0
3x(x+2)(x-2)=0

Step 3: Use the Zero Product Property and set each factor containing a variable equal to zero.

 3x=0 or x+2=0 or x-2=0

Step 4: Solve each factor that was set equal to zero by getting the x on one side and the answer on the other side.

 x=0 or x=-2 or x=


I hope this helped you!


3 0
3 years ago
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