All categories

2 years ago

Help guyssssssssssssss

Mathematics

1 answer:

n200080 [17]2 years ago

5 0

Answer:

Step-by-step explanation:

You might be interested in

Complex zeros of 4x^4-13x^3+31x^2+293x-75

AnnZ [28]

Sorry could you elaborate a bit further on the question

3 0

2 years ago

PLEASE HELP I HATE THIS SM I WANNA KMSS

dedylja [7]

Answer:

you want to flip the original coordinates and make them both negative so (x,y) would become (-y,-x) so in your problem here Q' would be (0,-7), R' would be (-5,-8), S' would be (-10,-7) and T' would be (-5,-2). Hope that helped :>

Step-by-step explanation:

6 0

2 years ago

Please help I will mark Brainly

Sedbober [7]

The answer for your question is 686

8 0

2 years ago

Read 2 more answers

40y^14/10y^8 simplify radicals and use properties of rational exponents

Vadim26 [7]

Answer:

4y^6

Step-by-step explanation:

Take out the constants:

40y^14/10y^8

= 40/10 * y^14 / y^8

= 4 * y^(14-8)

= 4 * y^6

= 4y^6

Hope this helped you :3

7 0

3 years ago

The length of the base edge of a pyramid with a regular hexagon base is represented as x. The height of the pyramid is 3 times l

sergij07 [2.7K]

Answer:

(a)

h=3x

(b)

$A=\frac{\sqrt{3} }{4} x^2$

(c)

$A=\frac{3\sqrt{3} }{2} x^2$

(d)

$V=\frac{3\sqrt{3} }{2} x^3$ units^3

Step-by-step explanation:

We are given a regular hexagon pyramid

Since, it is regular hexagon

so, value of edge of all sides must be same

The length of the base edge of a pyramid with a regular hexagon base is represented as x

so, edge of base =x

b=x

Let's assume each blank spaces as a , b , c, d

we will find value for each spaces

(a)

The height of the pyramid is 3 times longer than the base edge

so, height =3*edge of base

height=3x

h=3x

(b)

Since, it is in units^2

so, it is given to find area

we know that

area of equilateral triangle is

$=\frac{\sqrt{3} }{4} b^2$

h=3x

b=x

now, we can plug values

$A=\frac{\sqrt{3} }{4} x^2$

(c)

we know that

there are six such triangles in the base of hexagon

So,

Area of base of hexagon = 6* (area of triangle)

Area of base of hexagon is

$=6\times \frac{\sqrt{3} }{4} x^2$

$=\frac{3\sqrt{3} }{2} x^2$

(d)

Volume=(1/3)* (Area of hexagon)*(height of pyramid)

now, we can plug values

Volume is

$=\frac{1}{3}\times\frac{3\sqrt{3} }{2} x^2\times (3x)$

$V=\frac{3\sqrt{3} }{2} x^3$ units^3

3 0

3 years ago

Read 2 more answers