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3 years ago

6

Raj builds a side table in the shape of a cube. Each edge of the cube measures 30inches. Raj wants to cover the top and four sid

es of the table with ceramic tiles. Each tile has an edge length of 6inches. How many tiles will we need?

1 answer:

Lelu [443]3 years ago

5 0

Answer:

deportation

Step-by-step explanation:

deport raj

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Elijah can spend up to $23 on groceries. She wants to buy 3 pounds of tomatoes, 4 pounds of oranges, and x pounds of lean ground

Colt1911 [192]

Answer: 2 pounds

Step-by-step explanation:

Hope this helps!!! ; )

7 0

3 years ago

The polynomial of degree 4, P ( x ) , has a root of multiplicity 2 at x = 1 and roots of multiplicity 1 at x = 0 and x = − 2 . I

ICE Princess25 [194]

We want to find a polynomial given that we know its roots and a point on the graph.

We will find the polynomial:

p(x) = (183/280)*(x - 1)*(x - 1)*(x + 2)*x

We know that for a polynomial with roots {x₁, x₂, ..., xₙ} and a leading coefficient a, we can write the polynomial equation as:

p(x) = a*(x - x₁)*(x - x₂)...*(x - xₙ)

Here we know that the roots are:

x = 1 (two times)
x = 0
x = -2

Then the roots are: {1, 1, 0, -2}

We can write the polynomial as:

p(x) = a*(x - 1)*(x - 1)(x - 0)*(x - (-2))

p(x) = a*(x - 1)*(x - 1)*(x + 2)*x

We also know that this polynomial goes through the point (5, 336).

This means that:

p(5) = 336

Then we can solve:

336 = a*(5 - 1)*(5 - 1)*(5 + 2)*5

336 = a*(4)*(4)*(7)*5

336 = a*560

366/560 = a = 183/280

Then the polynomial is:

p(x) = (183/280)*(x - 1)*(x - 1)*(x + 2)*x

If you want to learn more, you can read:

brainly.com/question/11536910

5 0

3 years ago

a regular rectangular pyramid has a base and lateral faces that are congruent equilateral triangles. it has a lateral surface ar

Margarita [4]

A pyramid is regular if its base is a regular polygon, that is a polygon with equal sides and angle measures.
(and the lateral edges of the pyramid are also equal to each other)

Thus a regular rectangular pyramid is a regular pyramid with a square base, of side length say x.

The lateral faces are equilateral triangles of side length x.

The lateral surface area is 72 cm^2, thus the area of one face is 72/4=36/2=18 cm^2.

now we need to find x. Consider the picture attached, showing one lateral face of the pyramid.

by the Pythagorean theorem:

$h= \sqrt{ x^{2} - (x/2)^{2}}= \sqrt{ x^{2}- x^{2}/4}= \sqrt{3x^2/4}= \frac{ \sqrt{3} }{2}x$

thus,

$Area_{triangle}= \frac{1}{2}\cdot base \cdot height\\\\18= \frac{1}{2}\cdot x \cdot \frac{ \sqrt{3} }{2}x\\\\ \frac{18 \cdot 4}{ \sqrt{3}}=x^2$

thus:

$x^2 =\frac{18 \cdot 4}{ \sqrt{3}}= \frac{18 \cdot 4 \cdot\ \sqrt{3} }{3}=24 \sqrt{3}$ (cm^2)

but $x^{2}$ is exactly the base area, since the base is a square of sidelength = x cm.

So, the total surface area = base area + lateral area = $24 \sqrt{3}+72$ cm^2

Answer: $24 \sqrt{3}+72$ cm^2

4 0

3 years ago

How many positive integers $n$ from 1 to 5000 satisfy the congruence $n \equiv 5 \pmod{12}$?

irga5000 [103]

The equivalence $n \equiv 5 \pmod{12}$

means that n-5 is a multiple of 12.

that is

n-5=12k, for some integer k

and so

n=12k+5

for k=-1, n=-12+5=-7

for k= 0, n=0+5=5 (the first positive integer n, is for k=0)

we solve 5000=12k+5 to find the last k

12k=5000-5=4995

k=4995/12=416.25

so check k = 415, 416, 417 to be sure we have the right k:

n=12k+5=12*415+5=4985

n=12k+5=12*416+5=4997

n=12k+5=12*417+5=5009

The last k which produces n<5000 is 416

For all k∈{0, 1, 2, 3, ....416}, n is a positive integer from 1 to 5000,

thus there are 417 integers n satisfying the congruence.

Answer: 417

6 0

3 years ago

Help on this please.

olga55 [171]

Answer:

total - given, 360 -139 = 221

4 0

2 years ago