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zlopas [31]
3 years ago
12

I need help on (13.)

Mathematics
1 answer:
Trava [24]3 years ago
5 0
(-6,2),(-7,10)
slope = (10 - 2) / (-7 - (-6) = 8 / (-7 + 6) = 8/-1 = -8

y = mx + b
slope(m) = -8
use either of ur points...(-6,2)...x = -6 and y = 2
now we sub and find b, the y int
2 = -8(-6) + b
2 = 48 + b
2 - 48 = b
-46 = b

so ur equation is : y = -8x - 46....or 8x + y = -46
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2x + y = 15 x + y = 10​
Artyom0805 [142]

Answer:

2x+y=15x+y=10

Consider the first equation. Subtract 15x from both sides.

2x+y−15x=y

Combine 2x and −15x to get −13x.

−13x+y=y

Subtract y from both sides.

−13x+y−y=0

Combine y and −y to get 0.

−13x=0

Divide both sides by −13. Zero divided by any non-zero number gives zero.

x=0

Consider the second equation. Insert the known values of variables into the equation.

15×0+y=10

Multiply 15 and 0 to get 0.

0+y=10

Anything plus zero gives itself.

3 0
3 years ago
In a geometric sequence, a2=108 and as 256. Write the explicit formula for this sequence.
torisob [31]

Answer:

The first choice is correct.

Step-by-step explanation:

First work out the common ratio r:-

a2 = a1r and a5 = a1 r^4

So r^3 = a1r^4 / a1r = 256/ 108

r = cube root (256/-108) =  4/3

So the explicit formula is 108(4/3)^(n-1)

8 0
3 years ago
Read 2 more answers
2 sin^2 (x) -5 sin (x) -3=0
andreyandreev [35.5K]

we have that

2sin^{2} x-5sin x-3=0

I. Rewrite the equation by substituting the expression u in for sin x.

2u^{2} -5u-3=0

II. Factor the quadratic expression. Rewrite the equation with factors instead of the original polynomial.

2u^{2} -5u-3=0 is equal to

using a graph calculator-----> see the attached figure

(u-3)*(2u+1)=0

III. Use the zero product property to solve the quadratic equation.

(u-3)*(2u+1)=0

(u-3)=0--------------> u=3

(2u+1)=0-------- 2u=-1------> u=-1/2-----> u=-0.5

IV. Rewrite your solutions to Part III by replacing u with sin x.

sin x=3--------> is not the solution (sin x can not be greater than 1)

sin x=-0.50------>is the solution

V. Solve the remaining equations for x, giving all solutions to the equation.

sin x=-0.50

if the sine is negative

then

x belong to the III or IV quadrant

we know that

sin 30°=0.50

so

the solution for the III quadrant is

x=180°+30°-------> x=210°

the solution for the IV quadrant is

x=360°-30°------> x=330°

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Chuge%5Cbold%5Cred%7B%7BHELP%7D%7D" id="TexFormula1" title="\huge\bold\red{{HELP}}" alt="\hu
FromTheMoon [43]

9514 1404 393

Answer:

  1. (f×g)(x) = 2x^2 +2x
  2. (f×g)(x) = 6y^2 -11y -35

Step-by-step explanation:

The distributive property applies.

1. (fg)(x) = f(x)·g(x) = (2x)·(x +1)

  (f×g)(x) = 2x^2 +2x

__

2. (fg)(x) = f(x)·g(x) = (2y -7)·(3y +5) = 2y(3y +5) -7(3y +5)

  (f×g)(x) = 6y^2 -11y -35

_____

<em>Additional comment</em>

Please note in problem 2 that the function argument is x and the variable in the given expressions is y. This means the function value is exactly and only <em>6y^2 -11y -35 for any value of x</em>. It does not change when x changes. (We suspect a typo.)

7 0
3 years ago
Read 2 more answers
Please help me!! Iv been stuck!
Sloan [31]

Answer:

7

Step-by-step explanation:

Because the leftmost triangle is a 45-45-90 isosceles right triangle, a = b =\frac{7\sqrt2}{\sqrt2} = 7\\

The rightmost triangle is a 30-60-90 triangle, and the side lengths are in the proportion 1:\sqrt3:2 from shortest to longest. Therefore, c = 7\cdot\frac{\sqrt3}{1} = 7\sqrt{3} \implies \boxed{7}

7 0
3 years ago
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