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3 years ago

What is the volume of this square pyramid? I'm giving out brainliest for the best answer!

Mathematics

2 answers:

AlekseyPX3 years ago

3 0

Answer:

1280

Step-by-step explanation:

Paha777 [63]3 years ago

3 0

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Find m∠ABD and m∠CBD given m∠ABC = 111∘.

sineoko [7]

Answer:

m∠ABD = 88º

m∠CBD = 23º

Step-by-step explanation:

(-10x + 58) + (6x + 41) = 111

Combine like terms

-4x + 99 = 111

Subtract 99 from both sides

-4x = 12

Divide both sides by -4

x = -3

------------------------

m∠ABD = -10x + 58

m∠ABD = -10(-3) + 58

m∠ABD = = 30 + 58

m∠ABD = 88º

m∠CBD = 6x + 41

m∠CBD = 6(-3) + 41

m∠CBD = -18 + 41

m∠CBD = 23º

7 0

3 years ago

Read 2 more answers

Find the value of x.

enot [183]

X is 10 take that’s known 80 degrees and the fact that it is a right angle means the whole this is 90 degrees so therefore x is 10 90=80+_

6 0

3 years ago

Multiply and simplify: –3x2y2 • x4 –3x5y2 9x5y2 –3x6y2 –3x5y6

Contact [7]

Answer:

Multiplying and simplifying the term $-3x^2y^2 \:.\: x^4$ we get $\mathbf{-3x^6y^2}$

Option C is correct option.

Step-by-step explanation:

We need to multiply and simplify: $-3x^2y^2 \:.\: x^4$

Simplifying the term:

$-3x^2y^2 \:.\: x^4$

While multiplying, the variables with same bases are multiplied: $a^m\:.\:a^n=a^{m+n}$

In the given question only x^2 and x^4 are variables with same bases, So we get:

$=-3(x^{2+4})y^2\\=-3x^6y^2$

So, multiplying and simplifying the term $-3x^2y^2 \:.\: x^4$ we get $\mathbf{-3x^6y^2}$

Option C is correct option.

4 0

3 years ago

Can someone please help?

PIT_PIT [208]

Step-by-step explanation:

are this form 1 question???

4 0

3 years ago

Estimate 1 13 cos(x2) dx 0 using the Trapezoidal Rule and the Midpoint Rule, each with n = 4. (Round your answers to six decimal

babunello [35]

Answer:

(a) 4.152698

(b) 3.215557

Step-by-step explanation:

(a)

$\int\limits^{13}_1 {cos(x^2)} \, dx =M_n=$\sum_{n=1}^{\infty} f(m_i)\Delta x $$

n=4, so :

Each subinterval has length :

$\Delta x= \frac{b-a}{n} =\frac{13-1}{4} =\frac{12}{4} =3$

Therefore the subintervals consist of:

$[1,5], [5,9], [9,13]$

Now, the midpoints of these subintervals are:

$\frac{1+5}{2} =3\\\\\frac{5+9}{2} =7\\\\\frac{9+13}{2} =11$

Hence:

$M_4= 3*(cos(3^2))+3*(cos(7^2))+3*cos((11^2))\approx 4.152698$

(b)

$\int\limits^{13}_1 {cos(x^2)} \, dx =T_n=\frac{\Delta x}{2} (f(x_o)+2f(x_1)+2f(x_2)+...+2f(x_n_-_1)+f(x_n))$

n=4, so :

Each subinterval has length :

$\Delta x= \frac{b-a}{n} =\frac{13-1}{4} =\frac{12}{4} =3$

Therefore the subintervals consist of:

[1,5], [5,9], [9,13]

The endpoints of the subintervals consist of:

5,9

Hence:

$T_4= \frac{3}{2}(cos(1^2)+2*cos(5^2)+2*cos(9^2)+cos(13^2)) \approx 3.215557$

8 0

3 years ago