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Vesna [10]
3 years ago
8

Given that f(x) = −x + 4 and g(x) = −2x − 3, solve for f(g(x)) when x = 2.

Mathematics
1 answer:
krok68 [10]3 years ago
5 0

Answer:

Do g(2) first:  g(2) = -2*2 - 3 = -7

Put this into f:  f(g(2)) = -(-7) + 4 = 7 + 4 = 11

Step-by-step explanation:

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Given: ∆ABC, AB = CB BD − median to AC E∈ AB ,F∈ BC AE = CF Prove: △ADE ≅ △CDF ΔBDE ≅ ΔBDF
NemiM [27]

Answer:

1) By SAS theorem, ΔADE≅ΔCDF

2) By SSS theorem, ΔBDE≅ΔBDF

Step-by-step explanation:

Consider isosceles triangle ABC (see diagram).

1. In triangles ADE and CDF:

  • AD≅DC (since BD is median, then it divides side AC in two congruent parts);
  • AE≅CF (given);
  • ∠A≅∠C (triangle ABC is isosceles, then angles adjacent to the base are congruent).

By SAS theorem, ΔADE≅ΔCDF.

2. In triangles BDE and BDF:

  • side BD is common;
  • DE≅DF (ΔADE≅ΔCDF, then congruent triangles have congruent corresponding sides);
  • BE≅FB (triangle ABC is isosceles, AB≅BC, AE≅CF, then BE=AB-AE, FB=BC-CF).

Be SSS theorem, ΔBDE≅ΔBDF.

7 0
3 years ago
What are the answers to type in the yellow boxes?
Scorpion4ik [409]

Answer: The yellow boxes in order are 64, -40, 24

Step-by-step explanation:

(-8)^2 = 64

5*-8 = -40

y = 1*64 + (-40)

y = 64 - 40

y = 24

8 0
2 years ago
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A television station allows 15 minutes of advertising each hour. How many 30 second (1/2 minute) commercials can be run in 1 day
AleksAgata [21]
(15 minutes/hour) x (1 spot / 1/2 minute) x (24 hour/day) = <u><em>720</em></u> spots/day
3 0
3 years ago
Determine the location and values of the absolute maximum and absolute minimum for given function : f(x)=(‐x+2)4,where 0&lt;×&lt
brilliants [131]

Answer:

Where 0 < x < 3

The location of the local minimum, is (2, 0)

The location of the local maximum is at (0, 16)

Step-by-step explanation:

The given function is f(x) = (x + 2)⁴

The range of the minimum = 0 < x < 3

At a local minimum/maximum values, we have;

f'(x) = \dfrac{(-x + 2)^4}{dx}  = -4 \cdot (-x + 2)^3 = 0

∴ (-x + 2)³ = 0

x = 2

f''(x) = \dfrac{ -4 \cdot (-x + 2)^3}{dx}  = -12 \cdot (-x + 2)^2

When x = 2, f''(2) = -12×(-2 + 2)² = 0 which gives a local minimum at x = 2

We have, f(2) = (-2 + 2)⁴ = 0

The location of the local minimum, is (2, 0)

Given that the minimum of the function is at x = 2, and the function is (-x + 2)⁴, the absolute local maximum will be at the maximum value of (-x + 2) for 0 < x < 3

When x = 0, -x + 2 = 0 + 2 = 2

Similarly, we have;

-x + 2 = 1, when x = 1

-x + 2 = 0, when x = 2

-x + 2 = -1, when x = 3

Therefore, the maximum value of -x + 2, is at x = 0 and the maximum value of the function where 0 < x < 3, is (0 + 2)⁴ = 16

The location of the local maximum is at (0, 16).

5 0
3 years ago
HELP ASAP 5 STARS, THANKS, 20 POINTS, BRAINLIEST,
Bingel [31]

Answer:

b.7

Step-by-step explanation:

you take the three and substitute that in for x.  2(3) + 1

2(3)=6 then add 1

5 0
3 years ago
Read 2 more answers
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