Please Help!!

Mike and Ike are twins. Mike’s age is one more than twice a certain number. Ike’s age is 10 less than triple that same number. F

Misha Larkins [42]

Mike age is 3 less then twice the age of his brother nuke.
M = Mike's age now
N = Nuke's age now
M = 2N - 3

Ten years ago, four times nuke's age was one more mike's age.
M-10 = Mike's age 10 years ago
N-10 = Muke's age 10 years ago
4(N - 10) = 1 + M -10

what are their present ages

find M and N

substitute M = 2N - 3 into 4(N - 10) = 1 + M -10:

4(N - 10) = 1 + [2N - 3] - 10
4N - 40 = 2N - 12
2N = 28
N = 14

M = 2(14) - 3 = 28 - 3 = 25

Mike is 25 and Nuke is 14.

check:
4(14 - 10) = 4(4) = 16 = 26 -10 = 1 + 25 - 10

4 0

2 years ago

What statement is true? (Please help‼️)

Olenka [21]

Answer:

The statement that is accurate is A and pls 5 stars and brainliest

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

$I = \int \int_E \int zdV$ where E is bounded by the cylinder $y^2 + z^2 = 81$ and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and $y^2 + z^2 = 81$

∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

$y^2 = 81 \\ \\ y = \sqrt{81} \\ \\ y = 9$

∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix} ^ {\sqrt {{81-y^2}}}_{0} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{(\sqrt{81 -y^2})^2 }{2}-0 \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{{81 -y^2} }{2} \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

I = 91.125

4 0

3 years ago

Ryan gets an allowance of $9 a week which he deposits into his bank account where he already has $45 he wants to buy some shoes

Stells [14]

Answer:

Step-by-step explanation:

First add the shoes and the toys: 63+54 = $117

Second subtract 45 dollars from $117 = 72

Third divide 72 by 9 = 8 weeks

Answer: 8 weeks

5 0

3 years ago

Question 5 of 10

Mashcka [7]

Answer:

B. -8

Step-by-step explanation:

F(x) = 2x

F(-4) = 2(-4) = -8

4 0

3 years ago