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aivan3 [116]
3 years ago
13

X^2 + 5x = quadratic equation =​

Mathematics
1 answer:
timama [110]3 years ago
4 0

Hi there!

The question gives us the quadratic equation , and it tells us to solve it using the quadratic formula, which goes as . However, we must first find the values of a, b, and c. The official quadratic equation goes as , which matches the format of the given quadratic equation. Hence, the value of a would be 1, the value of b would be 5, and the value of c would be 3. Now, just plug it back into the quadratic equation and simplify to get the zeros of the equation. 

x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}  

x = \frac{-(5) \pm \sqrt{(5)^2 - 4(1)(3)} }{2(1)}  

x = \frac{-5 \pm \sqrt{25 - 12} }{2}  

x = \frac{-5 \pm \sqrt{13} }{2}  

x = \frac{-5 \pm 3.61 }{2}  

x = \frac{-5 + 3.61 }{2}, x = \frac{-5 - 3.61 }{2}

x=-0.695 \ \textgreater \ \ \textgreater \  -0.7, x= -4.305 \ \textgreater \ \ \textgreater \ x=-4.31

Therefore, the solutions to the quadratic equation  are x = -0.7 and x = -4.31. Hope this helped and have a phenomenal day!

Your answer is 4.31

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Answer:

3 bulbs should be flawed

Step-by-step explanation:

2+ 18 = 20

20 bulbs were examined

2/20  = 1/10 were flawed

looking at the next 30 bulbs

30 * 1/10 = 3

3 of the next 30 bulbs should be flawed


5 0
3 years ago
In the month of June, the temperature in Johannesburg, South Africa, varies over the day in a periodic way that can be modeled a
Art [367]

Answer:

The answer is "\bold{T = - 7.5 \cos \frac{\pi}{12}( t - 4 )+ 10.5}"

Step-by-step explanation:

Given value:

Temp-maximum=18^{\circ}  

Temp. minimum = 3^{\circ}

It is halfway between 10 am and 10 pm to 4 am.  

The sinus and cosine roles could be used throughout the year to predict fluctuations in climate models. Its type of formula that can be used to model such information is:  

T = A \cos B(t-C) + D, where parameters are A, B , C, D, T is the ° C temperature and t is the time (1-24)

A = amplitude = \frac{(T_{max} - T_{min})}{2}\\\\

                       = \frac{(3 - 18)}{2}\\\\= - \frac{15}{2}\\\\ = -7.5

B = \frac{2 \pi}{24}\\\\

   = \frac{\pi}{12}

C = \text{ units translated to the right}= 4

D = y_{min} + amplitude = units \ translated \ up\\\\

D = 7.5 + 3 = 10.5

Its trigonometric function equation that model temperature T hours after midnight in Johannesburg t.

T = - 7.5 \cos \frac{\pi}{12}( t - 4 )+ 10.5

6 0
4 years ago
Question 29 of 40<br> Find f(-2) for f(x) = 5.3*.
blsea [12.9K]

Answer:

A. \frac{5}{9}

Step-by-step explanation:

To find f(-2) in f(x)=5*3^x, plug in -2 into all the x values.

f(-2)=5*3^-2, you can use calculator to calculate 3^-2 which equals 0.1, forever 1's.

f(-2)=5*0.1=0.5  Forever 5's.

f(-2)=0.5 or \frac{5}{9}

Hope this helps!

If not, I am sorry.

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2 years ago
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Answer:

d

Step-by-step explanation:

i did th etst hope it helps

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--<br> Choose the correct simplification of the expression -5x2(4x – 6x2 – 3).
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\huge \bf༆ Answer ༄

Let's simply ~

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \: - 5 {x}^{2} (4x - 6 {x}^{2}  - 3)

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:( - 5x {}^{2}  \times 4x) - ( - 5 {x}^{2}  \times 6 {x}^{2} ) - ( - 5 {x}^{2}  \times  3)

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \: - 20 {x}^{3}    - ( - 30 {x}^{4} ) - ( - 15 {x}^{2} )

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \: - 20 {x}^{3}  + 30 {x}^{4}  + 15 {x}^{2}

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:30 {x}^{4}  - 20 {x}^{3}  + 15 {x}^{2}

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