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3 years ago

What would a proper answer be to this question

Mathematics

1 answer:

Musya8 [376]3 years ago

5 0

I cant help you there m8 the I'm not able to see the problem I'm sorry.

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Work out the perimeter of the triangle when the sides are : 3x-5 2x 19-x

mr Goodwill [35]

Answer:

4x + 14

Step-by-step explanation:

The perimeter of the triangle is equal to the sum of its three sides.

The sides of the triangle are 3x-5, 2x, and 19-x.

Calculate their sum.

3x - 5 + 2x + 19 - x

Rearrange.

3x + 2x - x + 19 - 5

Add or subtract like terms.

4x + 14

The perimeter of the triangle is 4x + 14.

5 0

3 years ago

Read 2 more answers

Please answer ASAP! Consistently wash your hands for a good minute.

8_murik_8 [283]

Answer:

20 inches

Step-by-step explanation:

x/2=10

20/2=10

8 0

3 years ago

Solve the system of equations. -5x+2y=9. y=7x

il63 [147K]

$- 5x + 2y = 9 \\ y = 7x \\ \\ - 5x + 2 \times 7x = 9 \\ - 5x + 14x = 9 \\ 9x = 9 \\ x = 1 \\ \\ y = 7 \times 1 = 7 \\$

Answer: (1; 7)

7 0

3 years ago

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

A safety regulation states that the height hh of a guardrail should be 106 centimeters with an absolute deviation of no more tha

iren [92.7K]

Let us say that h is the height of the guardrail. Therefore the inequality equation that we can generate from this scenario is:

| h – 106 | = ± 7

There are two ways to solve this, either the equation is positive or negative.

When the equation is positive, therefore:

| h – 106 | = 7

h = 7 + 106 = 113 cm

When the equation is negative, therefore:

| h – 106 | = - 7

h = -7 + 106 = 99 cm

So the height must be 99 cm to 113 cm

6 0

3 years ago