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laila [671]
3 years ago
14

Consider the following quadratic equation: 25x2=36 Using the standard form ax2+bx+c=0 of the given quadratic equation, factor th

e left hand side of the equation into two linear factors.
Mathematics
1 answer:
Stells [14]3 years ago
6 0

Answer:

  (5x -6)(5x +6) = 0

Step-by-step explanation:

Subtract 36 to put the equation in standard form. In this form, it looks like the difference of squares, so can be factored as such.

  25x^2 -36 = 0

  (5x)^2 -6^2 = 0

  (5x -6)(5x +6) = 0

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arlik [135]
Given:
Equilateral triangle: height = 2.6 inches ; base or side length = 3 inches
Rectangle: length = 6 inches ; width = 3 inches

1 name plate has 2 equilateral triangle and 3 rectangles.
Surface area of an equilateral triangle = √3/4 * a² = √3/4 * 3² = 3.9 in²

3.9 in² x 2 = 7.8 in²

Surface area of a rectangle = 6 in * 3 in = 18 in²

18 in² x 3 = 54 in²

7.8 in² + 54 in² = 61.8 in²

61.8 in² x 30 nameplates = 1,854 in²  Choice A.
7 0
3 years ago
Factor 140c + 28 - 14a to identify the equivalent expressions.
Feliz [49]

Answer:

The answer to your question are letters A, C and D.

Step-by-step explanation:

To factor this polynomial we need to look for the common factor of the three terms.

To find it, get the greatest common factor

                        140    28   14     2

                          70    14     7     2

                          35     7     7     5

                            7     7      7     7

                             1     1       1    

Greatest common factor = 2 x 7 = 14  

Factor the polynomial

140c + 28 - 14a = 14 ( 10c + 2 - a)

Factor only 7          7(20c + 4 - 2a)

Factor only 2          2 (70c + 14 - 7a)

8 0
3 years ago
Solve 6 + 5 √ 2 4 9 − 2 x = 7
NISA [10]

6+5\sqrt{249}-2x=7 \\-2x=7-6-5\sqrt{249} \\-2x\approx-77.9 \\x\approx\frac{-77.9}{2}\approx38.95

Hope this helps.

7 0
3 years ago
Two cones are similar. The surface area of the larger cone is 65π square inches. The surface area of the smaller cone is 41. 6π
kow [346]

Answer: 145

Step-by-step explanation: I say this because it has 14. 4 then

4 0
3 years ago
The number of bacteria after t hours is given by N(t)=250 e^0.15t a) Find the initial number of bacteria and the rate of growth
Art [367]

Answer:

a) N_0=250\; k=0.15

b) 334,858 bacteria

c) 4.67 hours

d) 2 hours

Step-by-step explanation:

a) Initial number of bacteria is the coefficient, that is, 250. And the growth rate is the coefficient besides “t”: 0.15. It’s rate of growth because of its positive sign; when it’s negative, it’s taken as rate of decay.

Another way to see that is the following:

Initial number of bacteria is N(0), which implies t=0. And N(0)=N_0. The process is:

N(t)=250 e^{0.15t}\\N(0)=250 e^{0.15(0)}\\ N_0=250e^{0}\\N_0=250\cdot1\\ N_0=250

b) After 2 days means t=48. So, we just replace and operate:

N(t)=250 e^{0.15t}\\N(48)=250 e^{0.15(48)}\\ N(48)=250e^{7.2}\\N(48)=334,858\;\text{bacteria}

c) N(t_1)=4000; \;t_1=?

N(t)=250 e^{0.15t}\\4000=250 e^{0.15t_1}\\ \dfrac{4000}{250}= e^{0.15t_1}\\16= e^{0.15t_1}\\ \ln{16}= \ln{e^{0.15t_1}} \\  \ln{16}=0.15t_1 \\ \dfrac{\ln{16}}{0.15}=t_1=4.67\approx 5\;h

d) t_2=?\; (N_0→3N_0 \Longrightarrow 250 → 3\cdot250 =750)

N(t)=250 e^{0.15t}\\ 750=250 e^{0.15t_2} \\ \ln{3} =\ln{e^{0.15t_2}}\\ t_2=\dfrac{\ln{3}}{0.15} = 2.99 \approx 3\;h

6 0
3 years ago
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