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laila [671]
3 years ago
10

Line CH contains points (-2, 6) and H(5, -3) What is the slope of G?

Mathematics
1 answer:
Xelga [282]3 years ago
4 0

Answer:

The slope of line G is \dfrac{- 9}{7}  .

Step-by-step explanation:

Given as :

The points of the line CH contains

C = x_1 ,  y_1 = - 2 , 6

H = x_2 ,  y_2 = 5 , -3

Let The slop of line CH = G

<u>Now, slope of line is </u>

G = \dfrac{y_2 - y_1}{x_2-x_1}

i.e G = \dfrac{- 3 - 6}{5 -(-2)}

Or, G = \dfrac{- 9}{5 +2}

Or,G = \dfrac{- 9}{7}

So, The slope of the line CH = G = \dfrac{- 9}{7}

Hence, The slope of line G is \dfrac{- 9}{7}  . Answer

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sp2606 [1]

Answer: Smaller

Step-by-step explanation:

Multiplying a number by 1.25 increases it by a quarter

Multiplying a number by 0.75 decreases it by a quarter

1.25 / 0.75 = 0.9375

Thus, increasing something by a quarter then decreasing it by a quarter, you are effectively multiplying that number by 0.9375, which is less than 1, so the number will decrease.

4 0
3 years ago
1. The ordered pair (-1, -1) is a solution of the system -4x + 2y = 2 and x + y = -2.
Nastasia [14]
1: (-1,-1) is (x, y) to see if it is a solution, you would just plug in x and y and see if the equation is true.
-4 (-1) + 2(-1) = 2
4 + -2 = 2
2 = 2 CORRECT
So... plug in x and y in the second equation to Make sure it works for that one too.
-1 + -1 = -2
-2 = -2 CORRECT
So, yes. (-1,-1) is a solution to both equations.
5 0
3 years ago
Kristen invests $ 5745 in a bank. The bank 6.5% interest compounded monthly. How long must she leave the money in the bank for i
podryga [215]

Answer:

1. 10.7 years

2. 17.0 years

3. 2nd option

Step-by-step explanation:

Use formula for compounded interest

A=P\cdot \left(1+\dfrac{r}{n}\right)^{nt},

where

A is final value, P is initial value, r is interest rate (as decimal), n is number of periods and t is number of years.

In your case,

1. P=$5745, A=2P=$11490, n=12 (compounded monthly), r=0.065 (6.5%) and t is unknown. Then

11490=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12t},\\ \\2=(1.0054)^{12t},\\ \\12t=\log_{1.0054}2,\\ \\t=\dfrac{1}{12}\log_ {1.0054}2\approx 10.7\ years.

2. P=$5745, A=3P=$17235, n=12 (compounded monthly), r=0.065 (6.5%) and t is unknown. Then


17235=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12t},

3=(1.0054)^{12t},

12t=\log_{1.0054}3,

t=\dfrac{1}{12}\log_{1.0054}3\approx 17.0\ years.

3. <u>1 choice:</u> P=$5745, n=12 (compounded monthly), r=0.065 (6.5%), t=5 years and A is unknown. Then

A=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12\cdot 5},\\ \\A=5745\cdot (1.0054)^{60}\approx \$7936.39.

<u>2 choice:</u> P=$5745, n=4 (compounded monthly), r=0.0675 (6.75%), t=5 years and A is unknown. Then

A=5745\cdot \left(1+\dfrac{0.0675}{4}\right)^{4\cdot 5},\\ \\A=5745\cdot (1.0169)^{20}\approx \$8032.58.

The best will be 2nd option, because $8032.58>$7936.39

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Step-by-step explanation:

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Step-by-step explanation:

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