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8_murik_8 [283]
3 years ago
7

Solving a system of linear equations algebraically:

Mathematics
1 answer:
guajiro [1.7K]3 years ago
8 0

It was so easy xD

its x=1

y=12×1+7

y=-6×1+25

=19

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Choose the function whose graph is given by:
adoni [48]

Answer: y = 0.5cos(x), choice A

The parent function y = cos(x) follows this same basic pattern of the curve shown, but it starts off at (0,1) as its highest point. This graph has its highest point to be half as high, so we multiply by 1/2 or 0.5

3 0
3 years ago
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HELP!!! URGENT!!! I WILL GIVE BRAINLEIST THING ​
lisov135 [29]

Answer:

Horizontal distance = 0 m and 6 m

Step-by-step explanation:

Height of a rider in a roller coaster has been defined by the equation,

y = \frac{1}{3}x^{2}-2x+8

Here x = rider's horizontal distance from the start of the ride

i). y=\frac{1}{3}x^{2}-2x+8

      =\frac{1}{3}(x^{2}-6x+24)

      =\frac{1}{3}[x^{2}-2(3x)+9-9+24]

      =\frac{1}{3}[(x^{2}-2(3x)+9)+15]

      =\frac{1}{3}[(x-3)^2+15]

      =\frac{1}{3}(x-3)^2+5

ii). Since, the parabolic graph for the given equation opens upwards,

    Vertex of the parabola will be the lowest point of the rider on the roller coaster.

    From the equation,

    Vertex → (3, 5)

    Therefore, minimum height of the rider will be the y-coordinate of the vertex.

    Minimum height of the rider = 5 m

iii). If h = 8 m,

    8=\frac{1}{3}(x-3)^2+5

    3=\frac{1}{3}(x-3)^2

    (x - 3)² = 9

    x = 3 ± 3

    x = 0, 6 m

    Therefore, at 8 m height of the roller coaster, horizontal distance of the rider will be x = 0 and 6 m

6 0
3 years ago
3(x+6)+2x-5=-2(x+1)+10
Leto [7]

Step-by-step explanation:

3x+18+2x-5= -2x+-2+10

combining like terms

3x+2x+2x=5-2+10-18

7x=-5

x=-5/7

8 0
4 years ago
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A small combination lock on a suitcase has 55 ​wheels, each labeled with the 10 digits 0 to 9. How many 55 digit combinations ar
den301095 [7]
Suppose a_n is the number of possible combinations for a suitcase with a lock consisting of n wheels. If you added one more wheel onto the lock, there would only be 9 allowed possible digits you can use for the new wheel. This means the number of possible combinations for n+1 wheels, or a_{n+1} is given recursively by the formula

a_{n+1}=9a_n

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For example, if the combination for a 3-wheel lock is 282, then a 4-wheel lock can be any one of 2820, 2821, 2823, ..., 2829 (nine possibilities depending on the second-to-last digit).

By substitution, you have

a_{n+1}=9a_n=9^2a_{n-1}=9^3a_{n-2}=\cdots=9^na_1=10\times9^n

This means a lock with 55 wheels will have

a_{55}=10\times9^{54}

possible combinations (a number with 53 digits).
7 0
3 years ago
Help me out, please-
olga_2 [115]

unfortunately i cannot

4 0
3 years ago
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