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3 years ago

Which equation has the solutions x = 5+/- 2 the squire root of 7 over 3

1 answer:

5 0

If you look at the quadratic formula:

x=(-b±√(b^2-4ac))/(2a)

We can say that -b=5, b=-5

If I read it right you had ±2√7 which is equal to ±√28, and we know b=-5 so:

±√(25-4ac)=±√28

And since 3=2a, a=1.5 making the above:

±√(25-6c)=±√28

25-6c=28

-6c=3

c=-0.5

So a=1.5, b=-5, and c=-0.5 thus

y=1.5x^2-5x-0.5

We can check this by reapplying the quadratic formula to the equation above

x=(5±√(25-4(1.5)(-0.5))/3

x=(5±√(25+3))/3

x=(5±√28)/3

x=(5±√(4*7))/3

x=(5±2√7)/3

You might be interested in

Use integration by parts to find the integrals in Exercise. ∫^3_0 3-x/3e^x dx.

Viefleur [7K]

Answer:

8.733046.

Step-by-step explanation:

We have been given a definite integral $\int _0^3\:3-\frac{x}{3e^x}dx$ . We are asked to find the value of the given integral using integration by parts.

Using sum rule of integrals, we will get:

$\int _0^3\:3dx-\int _0^3\frac{x}{3e^x}dx$

We will use Integration by parts formula to solve our given problem.

$\int\ vdv=uv-\int\ vdu$

Let $u=x$ and $v'=\frac{1}{e^x}$ .

Now, we need to find du and v using these values as shown below:

$\frac{du}{dx}=\frac{d}{dx}(x)$

$\frac{du}{dx}=1$

$du=1dx$

$du=dx$

$v'=\frac{1}{e^x}$

$v=-\frac{1}{e^x}$

Substituting our given values in integration by parts formula, we will get:

$\frac{1}{3}\int _0^3\frac{x}{e^x}dx=\frac{1}{3}(x*(-\frac{1}{e^x})-\int _0^3(-\frac{1}{e^x})dx)$

$\frac{1}{3}\int _0^3\frac{x}{e^x}dx=\frac{1}{3}(-\frac{x}{e^x}- (\frac{1}{e^x}))$

$\int _0^3\:3dx-\int _0^3\frac{x}{3e^x}dx=3x-\frac{1}{3}(-\frac{x}{e^x}- (\frac{1}{e^x}))$

Compute the boundaries:

$3(3)-\frac{1}{3}(-\frac{3}{e^3}- (\frac{1}{e^3}))=9+\frac{4}{3e^3}=9.06638$

$3(0)-\frac{1}{3}(-\frac{0}{e^0}- (\frac{1}{e^0}))=0-(-\frac{1}{3})=\frac{1}{3}$

$9.06638-\frac{1}{3}=8.733046$

Therefore, the value of the given integral would be 8.733046.

6 0

3 years ago

mr. devito was ordering pencils for apl fourth grade students. There are 118 fourth graders. Pencils come in packs of 6. He need

sweet [91]

Answer:

Mr. devito needs to buy 40 packs

Step-by-step explanation:

Given:

Total number of fourth grade students = 118

The number of pencils in one pack = 6

The number pencil each student gets =2

To find:

The number of packs of pencils that is to be bought = ?

Solution:

To give all the 118 fourth grade students each 2 pencils , the total number of pencils needed will be

= > $118 \times 2$

=> 236 Pencils

We know that a pack contains 6 pencils, so the number of pack that has to be bought , in order to have 236 pencils will be

=> $\frac{ \text{total number of pencils}}{\text{ number of pencils per pack }}$

On substituting the values

=> $\frac{236}{6}$

=> 39.33

=> 40

3 0

3 years ago

Find the range for this data set {3,5,7, 3, 6, 4, 8, 6, 9, 5) *

OverLord2011 [107]

Answer:

Step-by-step explanation:

9-3=6

3 0

3 years ago

Read 2 more answers

7nadin3 [17]

Answer: A. 4p + $17.50 = $53.50; p = $9.00

1: 17.50+4p= Nothing further can be done with this topic. Please check the expression entered or try another topic.

17.5 + 4 p

2: 4p=53.50-17.50= 4p=53.50-17.50

Step-by-step explanation: 9

The total bill: $53.50

17.50 + 4 x = 53.50

4 x = 53.50 - 17.50

4 x = 36

x = 36 : 4

x = $9

Answer: The price of each shrub is $9.

...............................................................................................................................................

Answer:

Step-by-step explanation:

Let be the price of each shrub.

4 shrubs at each costs dollars

Potting soil is $17.50

Hence, total cost is the expression

We know that total bill is $53.50, so we can equate it to the expression:

This equation can be solved for to find cost of each shrub.

Solving for gives us:

So price of each shrub is $9

3 0

3 years ago

Delicious77 [7]

You can solve for the velocity and position functions by integrating using the fundamental theorem of calculus:

a(t) = 40 ft/s²

v(t) = v (0) + ∫₀ᵗ a(u) du

v(t) = -20 ft/s + ∫₀ᵗ (40 ft/s²) du

v(t) = -20 ft/s + (40 ft/s²) t

s(t) = s (0) + ∫₀ᵗ v(u) du

s(t) = 10 ft + ∫₀ᵗ (-20 ft/s + (40 ft/s²) u ) du

s(t) = 10 ft + (-20 ft/s) t + 1/2 (40 ft/s²) t ²

s(t) = 10 ft - (20 ft/s) t + (20 ft/s²) t ²

6 0

2 years ago