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3 years ago

Given h(x)=2x^3+x^2+7, what is the value of h(-3)?

Mathematics

1 answer:

Vika [28.1K]3 years ago

6 0

The answer to h(-3) would be -38

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Can someone please help me !

Kay [80]

Answer: 14

Step-by-step explanation:

1.75x -2 + 1.75x -2 + 1.75x -2 =67.5

5.25x-6= 67.5

5.25x =73 .5

X = 14

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3 years ago

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Which equation below has no solution?

soldier1979 [14.2K]

Answer:

Step-by-step explanation:

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3 years ago

if you could generate energy by fusing the hydrogen in solution a, how much of the solution would you need to generate the elect

iogann1982 [59]

Answer:

We know that:

Energy released by fusion of hydrogen in 1 liter of solution A = 7.6x10^10 J

Energy used daily in a certain family home = 3x10^4 J

(you did not write the units, so i suppose that are the same in both cases)

Then, if x is the number of liters of solution A used, the energy produced will be:

E(x) = x*7.6x10^10 J

And we want this equal to 3x10^4

then:

E(x) = x*7.6x10^10 J = 3x10^4 j

now we solve this for x.

x = (3x10^4 j)/(7.6x10^10 j) = 3.9x10^-7

Then you need to use 3.9x10^-7 L of solution a.

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3 years ago

I need help with this one

Aleks [24]

Answer:

Step-by-step explanation:

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3 years ago

Measure the lengths of the sides of ∆ABC in GeoGebra, and compute the sine and the cosine of ∠A and ∠B. Verify your calculations

marusya05 [52]

Answer:

$Sin \angle A =0.80$

$Cos \angle A=0.60$

$Sin \angle B =0.60$

$Cos \angle B=0.80$

Step-by-step explanation:

Given

I will answer this question using the attached triangle

Solving (a): Sine and Cosine A

In trigonometry:

$Sin \theta =\frac{Opposite}{Hypotenuse}$ and

$Cos \theta =\frac{Adjacent}{Hypotenuse}$

So:

$Sin \angle A =\frac{BC}{BA}$

Substitute values for BC and BA

$Sin \angle A =\frac{8cm}{10cm}$

$Sin \angle A =\frac{8}{10}$

$Sin \angle A =0.80$

$Cos \angle A=\frac{AC}{BA}$

Substitute values for AC and BA

$Cos \angle A=\frac{6cm}{10cm}$

$Cos \angle A=\frac{6}{10}$

$Cos \angle A=0.60$

Solving (b): Sine and Cosine B

In trigonometry:

$Sin \theta =\frac{Opposite}{Hypotenuse}$ and

$Cos \theta =\frac{Adjacent}{Hypotenuse}$

So:

$Sin \angle B =\frac{AC}{BA}$

Substitute values for AC and BA

$Sin \angle B =\frac{6cm}{10cm}$

$Sin \angle B =\frac{6}{10}$

$Sin \angle B =0.60$

$Cos \angle B=\frac{BC}{BA}$

Substitute values for BC and BA

$Cos \angle B=\frac{8cm}{10cm}$

$Cos \angle B=\frac{8}{10}$

$Cos \angle B=0.80$

Using a calculator:

$A = 53^{\circ}$

So:

$Sin(53^{\circ}) =0.7986$

$Sin(53^{\circ}) =0.80$ -- approximated

$Cos(53^{\circ}) = 0.6018$

$Cos(53^{\circ}) = 0.60$ -- approximated

$B = 37^{\circ}$

So:

$Sin(37^{\circ}) = 0.6018$

$Sin(37^{\circ}) = 0.60$ --- approximated

$Cos(37^{\circ}) = 0.7986$

$Cos(37^{\circ}) = 0.80$ --- approximated

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3 years ago

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