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4 years ago

Find [h(x) +m(x)]-[h(x)-m(x)] H(x)=1/3x3/3 3x2+9x-2 M(x)=13 3/5x+6x3-2x2-4x

Mathematics

1 answer:

Andre45 [30]4 years ago

7 0

Answer:

\frac{276}{5}-8x)

Step-by-step explanation:

We are given with

H(x)=1/3x3/3 3x2+9x-2

And

M(x)=13 3/5x+6x3-2x2-4x

And asked to determine

[h(x) +m(x)]-[h(x)-m(x)]

Opening the second parenthesis

h(x) +m(x)-h(x)+m(x)

As we subtracted the second bracket, the sign get reversed.

As h(x) has opposite signs , they get cancelled

h(x) +m(x)-h(x)+m(x)

=2m(x)

=2(13\tfrac{3}/{5}x+6x3-2x2-4x)

=2(\frac{13 \times 5 + 3}{5}+18-4-4x)

=2(\frac{68}{5}+14-4x)

=2(\frac{68+70}{5}-4x)

=2(\frac{138}{5}-4x)

=\frac{276}{5}-8x)

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2. If y varies inversely as x and<br> y = 3 when x = 5, find x when y = 2.5

4vir4ik [10]

Answer:

Step-by-step explanation:

From the law of variation,

y <> 1/x, where <> is the constant of proportionality. Therefore

y = k/x where k is a constant

3 = k/5 and k = 15

To find x when y = 2.5(5/2)

Go back to the second equation

y = k/x

5/2 = 15/x

When you cross multiply

5x = 30.

Divide through by 5,

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8 0

4 years ago

For each value of y, determine whether it is a solution to -2y+75-5.<br> Is it a solution?

ser-zykov [4K]

Answer:

• No

• Yes

• No

Step-by-step explanation:

To determine if the 4 given values of y are solutions to the inequality, start by solving the inequality. Solving an inequality is just like that of an equation, except that the direction of the sign changes when the inequality is divided by a negative number.

-2y +7≤ -5

Subtract 7 on both sides:

-2y≤ -5 -7

-2y≤ -12

Divide by -2 on both sides:

y≥ 6

This means that the solution can be 6 or greater than 6.

-10 and 3 are smaller than 6 and are not a solutions, while 7 and 6 satisfies the inequality and are therefore solutions.

_______

Alternatively, we can also substitute each value of y into the inequality and check if the value is less than or equal to -5.

Here's an example to check if -10 is a solution.

-2y +7≤ -5

When y= -10,

-2y +7

= -2(-10) +7

= 20 +7

= 27

Since 27 is greater than 5, it is <u>not</u> a solution to the inequality.

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2 years ago

Evaluate the integral e^xy w region d xy=1, xy=4, x/y=1, x/y=2

LUCKY_DIMON [66]

Make a change of coordinates:

$u(x,y)=xy$
$v(x,y)=\dfrac xy$

The Jacobian for this transformation is

$\mathbf J=\begin{bmatrix}\dfrac{\partial u}{\partial x}&\dfrac{\partial v}{\partial x}\\\\\dfrac{\partial u}{\partial y}&\dfrac{\partial v}{\partial y}\end{bmatrix}=\begin{bmatrix}y&x\\\\\dfrac1y&-\dfrac x{y^2}\end{bmatrix}$

and has a determinant of

$\det\mathbf J=-\dfrac{2x}y$

Note that we need to use the Jacobian in the other direction; that is, we've computed

$\mathbf J=\dfrac{\partial(u,v)}{\partial(x,y)}$

but we need the Jacobian determinant for the reverse transformation (from $(x,y)$ to $(u,v)$ . To do this, notice that

$\dfrac{\partial(x,y)}{\partial(u,v)}=\dfrac1{\dfrac{\partial(u,v)}{\partial(x,y)}}=\dfrac1{\mathbf J}$

we need to take the reciprocal of the Jacobian above.

The integral then changes to

$\displaystyle\iint_{\mathcal W_{(x,y)}}e^{xy}\,\mathrm dx\,\mathrm dy=\iint_{\mathcal W_{(u,v)}}\dfrac{e^u}{|\det\mathbf J|}\,\mathrm du\,\mathrm dv$
$=\displaystyle\frac12\int_{v=}^{v=}\int_{u=}^{u=}\frac{e^u}v\,\mathrm du\,\mathrm dv=\frac{(e^4-e)\ln2}2$

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3 years ago