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galben [10]
3 years ago
8

If f(x+5)=x^+kx+30,k=

Mathematics
1 answer:
Vilka [71]3 years ago
5 0

30 divided by 5 is 6, so the other term is (x+6). (x+5)(x+6) becomes x^2+11x+30, so k is equal to 11.


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Choose the smallest number 3 1/8 or 10/3
Aliun [14]

Answer:

10/3

Step-by-step explanation:

31/8= 3.8

10/3= 3.3

8 0
3 years ago
Scientists plan to release a space probe that will enter the atmosphere of a gaseous planet. The temperature of the gaseous plan
vladimir1956 [14]

Answer:

its 450 k

Step-by-step explanation:

It is given that the temperature of the gaseous planet is linearly related with height of the atmosphere. we can write this in the mathematical equation:

y = m x +c

where y is the temperature values, x is height, m is the slope and c is the y-intercept. we have been given two sets of value in the image, using which we can find the value of slope in y-intercept.

at x = 18.40 km, y = 147.54 K

⇒147.54 = 18.40 m + c      

⇒c = 147.54 - 18.40 m  ..(1)

at x =78.11 km, y = 567.00 K

⇒567.00 =78.11 m + c       ..(2)

Put equation 1 in 2 and solve:

⇒567.00 =78.11 m + 147.54 - 18.40 m

⇒419.46 = 59.71 m

⇒ m =419.46 ÷59.71 = 7.025 K/km

c = 147.54 - 18.40 × 7.025 = 18.28 K

At height, x = 61.5 km the approximate temperature is :

y = 7.025 K/km ×  61.5 km + 18.28 K = 450.3 K

Thus, the approximate temperature at altitude 61.5 km is 450 K.

7 0
4 years ago
F(x) = 3x + x3
____ [38]

Answer:

Please check the explanation.

Step-by-step explanation:

Given

f(x) = 3x + x³

Taking differentiate

\frac{d}{dx}\left(3x+x^3\right)

\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'

=\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(x^3\right)

solving

\frac{d}{dx}\left(3x\right)

\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'

=3\frac{d}{dx}\left(x\right)

\mathrm{Apply\:the\:common\:derivative}:\quad \frac{d}{dx}\left(x\right)=1

=3\cdot \:1

=3

now solving

\frac{d}{dx}\left(x^3\right)

\mathrm{Apply\:the\:Power\:Rule}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}

=3x^{3-1}

=3x^2

Thus, the expression becomes

\frac{d}{dx}\left(3x+x^3\right)=\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(x^3\right)

                    =3+3x^2

Thus,

f'(x) = 3 + 3x²

Given that f'(x) = 15

substituting the value  f'(x) = 15 in f'(x) = 3 + 3x²

f'(x) = 3 + 3x²

15 =  3 + 3x²

switch sides

3 + 3x² = 15

3x² = 15-3

3x² = 12

Divide both sides by 3

x² = 4

\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}

x=\sqrt{4},\:x=-\sqrt{4}

x=2,\:x=-2

Thus, the value of x​ will be:

x=2,\:x=-2

5 0
3 years ago
Please help me with this
malfutka [58]
43 I added and multiplied everything I’m sure
4 0
3 years ago
Plwees help:)
drek231 [11]

Dan takes about 8 minutes to run a mile

Yoko payed around $1.75

8 0
3 years ago
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