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Likurg_2 [28]
4 years ago
11

7p+3q for p=2 and q=-6

Mathematics
1 answer:
ddd [48]4 years ago
4 0
The answer is -4

7×2+-6×3

14+-18=-4
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Write the equation of a line through (-1,-5) parallel to 3x-y=5, and in slope intercept form
MAXImum [283]

Answer:

y =3 x

Step-by-step explanation:

The standard equation of a line in point slope form is expressed as;

y -y0 = m(x-x0)

m is the slope

(x0, y0) is the point on the line

Given the line 3x - y = 5

Get the slope

- y = -3x + 5

y = 3x - 5

Slope m  = 3

Substitute m = 3 and (-1, -5) into the given formula

y - (-5) = 3(x-(-1))

y+5 = 3(x+1)

y + 3 = 3x+3

y = 3x + 3 - 3

y = 3x

Hence the equation in slope intercept form is y =3 x

4 0
3 years ago
mrs. miller bought 3 large pizzas for the class. each pizza had 8 slices. at the end of the day , the 5 fifth grade teachers eac
anastassius [24]

Answer:

19 Slices.

Step-by-step explanation:

The number number of Pizzas = 3

Each Pizza had = 8 Slices

<u>Determining the Total number of Slices:</u>

Total no. of slices =Pizzas × Slices per Pizza

Total no. of Slices = 24

<u>Determining the number of slices Consumed by 5th Grade Teachers :</u>

Total no. of slices Consumed by the Teachers = 5 × 1

Total no. of slices Consumed by the Teachers = 5

<u>Total number of slices Consumed by the students:</u>

Total no. of slices Consumed by the students = Total slices - Slices consumed by teachers.

Total no. of slices Consumed by the students= 24 - 5

Total no. of slices Consumed by the students= 19

3 0
4 years ago
Start with the basic function f(x)=2x. If you have an initial value of 1, then you end up with the following iterations. f(1)=2*
denpristay [2]
<span>f(x)=2x

f(1)=2*1=2

f^2(1)=2*2*1=4

f^3(1) =2*2*2*1=8

1. If you continue this pattern, what do you expect would happen to the numbers as the number of iterations grows?

I expect the numbers continue growing multiplying each time by 2.

Check your result by conducting at least 10 iterations.

f^4(1) = f^3(1) * f(1) = 8*2 = 16

f^(5)(1) = f^4(1) * f(1) = 16 * 2 = 32

f^6 (1) = f^5 (1) * f(1) = 32 * 2 = 64

f^7 (1) = f^6 (1) * f(1) = 64 * 2 = 128

f^8 (1) = f^7 (1) * f(1) = 128 * 2 = 256

f^9 (1) = f^8 (1) * f(1) = 256 * 2 = 512

f^10 (1) = f^9 (1) * f(1) = 512 * 2 = 1024

2. Repeat the process with an initial value of −1. What happens as the number of iterations grows?

f(-1) = 2(-1) = - 2

f^2 (-1) = f(-1) * f(-1) = - 2 * - 2 = 4

f^3 (-1) = f^2 (-1) * f(-1) = 4 * (-2) = - 8

f^4 (-1) = f^3 (-1) * f(-1) = - 8 * (-2) = 16

f^5 (-1) = f^4 (-1) * f(-1) = 16 * (-2) = - 32

As you see the magnitude of the number increases, being multiplied by 2 each time, and the sign is aleternated, negative positive negative positive ...
</span>
4 0
4 years ago
Solve the following equation for g.<br> Rg=m+h^2g
marshall27 [118]

Answer:

G=m/r-h^2

Step-by-step explanation:

move all the terms to the left side and set equal to zero. Then set each factor equal to zero. Hope this helps!

4 0
3 years ago
Express the prime number 31 in the form 2^p − 1 where p is a prime number and as a difference of two perfect squares using the i
stellarik [79]

Answer:

31 = 2⁵ -1

31 =  16²- 15²

Step-by-step explanation:

As we know that

2¹=2

2²= 4

2³= 8

2⁴= 16

2⁵=32

So

32 = 2⁵

By subtracting 1 in bot sides

32-1 = 2⁵ -1

31 = 2⁵ -1

So

2⁵ -1 = 31

By comparing with

2^p-1

p= 5

As we know that

15²= 225

16² = 256

16²- 15² = 256-225 = 31

16²- 15²  = (16+15)(16-15)

31 =  (16+15)(16-15)

5 0
4 years ago
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