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3 years ago

PLEASE HELP ASAPPPP!!!!!

Mathematics

1 answer:

dexar [7]3 years ago

8 0

Answer:

see below

Step-by-step explanation: 5 20 15 39

coin 2 possibilities 1/2

three coins (1/2) × (1/2) × (1/2) = _________ or (1/2)³

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Help me plz a lot of points

Readme [11.4K]

Answer:

A: Perimeter is 10x+2

B: I believe it is 6x^2+2x

C: Perimeter: 52 un; Area: 160 un^2

un is units

Step-by-step explanation:

This is just my work so that i can do this.

3x(2x)+1(2x)=

6x^2+2x

16+16+10+10=52

8 0

3 years ago

Read 2 more answers

Calculate the double integral. $$\iint_{R}{\color{red}4} xye^{x^{2}y}\hspace*{3pt}dA, \quad R = [0, 1] \times [0, {\color{red}7}

Lady_Fox [76]

Answer:

$\mathbf{\int \int _R \ 4xy e^{x^2 \ y} \ dA = 2 (e^7 -8)}$

Step-by-step explanation:

Given that:

$\int \int _R 4xye^{x^2 \ y} \ dA, R = [0,1]\times [0,7]$

The rectangle R = [0,1] × [0,7]

R = { (x,y): x ∈ [0,1] and y ∈ [0,7] }

R = { (x,y): 0 ≤ x ≤ 1 and 0 ≤ x ≤ 7 }

$\int \int _R \ 4xy e^{x^2 \ y} \ dA = \int^{7}_{0}\int^{1}_{0} 4xye^{x^2 \ y} \ dx dy$

$\int \int _R \ 4xy e^{x^2 \ y} \ dA = \int^{7}_{0} \begin {bmatrix} ye^{yx^2} \dfrac{4}{2y} \end {bmatrix}^1 _ 0 \ dy$

$\int \int _R \ 4xy e^{x^2 \ y} \ dA = \int^{7}_{0} \begin {bmatrix} ye^{y1^2} \dfrac{4}{2y} - ye^{y0^2} \dfrac{4}{2y} \end {bmatrix}\ dy$

$\int \int _R \ 4xy e^{x^2 \ y} \ dA = \int^{7}_{0} \dfrac{4}{2}(e^y -1) \ dy$

$\int \int _R \ 4xy e^{x^2 \ y} \ dA = \dfrac{4}{2}[e^y -1]^7_0 \ dy$

$\int \int _R \ 4xy e^{x^2 \ y} \ dA = 2 [(e^7 -7)-(e^0 -0)]$

$\int \int _R \ 4xy e^{x^2 \ y} \ dA = 2 [(e^7 -7)-1]$

$\mathbf{\int \int _R \ 4xy e^{x^2 \ y} \ dA = 2 (e^7 -8)}$

3 0

4 years ago

Which equation have no real solution?

Ad libitum [116K]

To find out which equation has no real solutions, we need to calculate the discriminant for each of these given equations.

For calculating the discriminant, we need to first compare these equations with the general formula which is ax²+bx+c.

So, let's get started.

1) x² + 4x + 16 = 0

a=1, b=4, c=16

D = b²-4ac

= (4)² - 4(1)(16)

= 16-64

= -48

√D = √-48

2) 4x² + 4x - 24 = 0

a=4, b=4, c=-24

D = b²-4ac

= (4)² - 4(4)(-24)

= 16 - 16(-24)

= 16 + 384

= 400

√D = √400 = +20 or -20

3) 5x² + 3x - 1 = 0

a=5, b=3, c=-1

D = b²-4ac

= (3)² - 4(5)(-1)

= 9 + 20

= 29

√D = √29

4) 2x² - 4x + 4 = 0

a=2, b=-4, c=4

D = b²-4ac

= (-4)² - 4(2)(4)

= 16 - 32

= -16

√D = √-16

Now from all these above calculations, we can see that discriminant was negative in first equation and in last equation.

If D<0 then roots does not exist, as the square root can not contain a negative value or the equation does not have any real solutions.

Roots in such case can be calculated but those roots are known as imaginary roots, which is a higher concept.

So Final answer is,

Equation 1 => x² + 4x + 16 = 0

and

Equation 4 => 2x² - 4x + 4 = 0

has no real solutions.

4 0

4 years ago

Helpppp helpppp helppppp helppppp

mr_godi [17]

Answer:

Do you know the cylinder volume formula?Answer is: V=Pi*r^2*h Let me explain what that means. So the Pi is a constant number equal to 3.14..... and continue and this doesn't effect the volume while radius and height are the ones who do that.The volume is more effected by the radius because if both of them are a same number the radius would be greater than the height.

If r= 2 and h=2 than V=Pi*2^2*2 V=3.14*4*2

Step-by-step explanation:

Let me solve the problem!

a) r=3 h=10 than V=Pi*3^2*10 V=Pi*9*10 V=90Pi

When we increase the raduis by 1

b)r=4 h=10

V=Pi*4^2*10 so V=Pi*16*10 so V=160Pi

c)r=3 h=11

V=Pi*3^2*11 so V=Pi*9*11 so V=99Pi

It meas that when the radius is increased by 1inc the volume is greater than when you increase the height by 1.

Thank!

6 0