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3 years ago

If h(x)=x2-5+7. h(2)

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Mathematics

1 answer:

Marta_Voda [28]3 years ago

3 0

The way to solve this is by re-writing the equation except with 2 as x.
h(x)= x^2-5+7
h(2)= 2^2-5+7
h(2)=4-5+7
h(2)=-1+7
h(2)= 6

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Im having a hard time on this question and I hope you can help.

Masja [62]

The answer would be x ≠ 0

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First start with the left side, doing distributive property

So... -2(x) = -2x and -2(5) = -10 Therefore on the left side you now have -2x -10

Next do the same on the right side

-2(x) = - 2x and -2(-2) = 4 so you have -2x + 4 + 5 and you add 4 and 5, leaving you with -2x + 9

Now that you have simplified both sides the problem now looks like this:

-2x - 10 = -2x + 9

Because you have equal terms on both sides (-2) those cancel out so you have -10 = 9

Just from looking at this we know that the statement is false because -1o does not equal 9

*The symbol, "≠" means not equal to"

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3 years ago

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Verona is solving the equation –3 + 4x = 9. In order to isolate the variable term using the subtraction property of equality, wh

frutty [35]

Answer:

subtract -3

Step-by-step explanation:

–3 + 4x = 9

Add 3 to each side

This is the same as subtracting -3

-3 + 4x - (-3) = 9 - (-3)

4x = 9 +3

4x = 12

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3 years ago

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What is the value of x in this simplified expression

arsen [322]

Answer:

Step-by-step explanation:

(-j)^-7 = 1/(-j)^x

Changing LHS as

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Comparing LHS and RHS

x = 7

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Solve H = fraction with p minus a in the numerator and 3 times w in the denominator for p. SOLVE FOR P

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Mama L [17]

Multiplying both sides by $y^2$ gives

$xy^2\dfrac{\mathrm dy}{\mathrm dx}+y^3=1$

so that substituting $v=y^3$ and hence $\frac{\mathrm dv}{\mathrm dv}=3y^2\frac{\mathrm dy}{\mathrm dx}$ gives the linear ODE,

$\dfrac x3\dfrac{\mathrm dv}{\mathrm dx}+v=1$

Now multiply both sides by $3x^2$ to get

$x^3\dfrac{\mathrm dv}{\mathrm dx}+3x^2v=3x^2$

so that the left side condenses into the derivative of a product.

$\dfrac{\mathrm d}{\mathrm dx}[x^3v]=3x^2$

Integrate both sides, then solve for $v$ , then for $y$ :

$x^3v=\displaystyle\int3x^2\,\mathrm dx$

$x^3v=x^3+C$

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