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4 years ago

What is the soulution to this system of linear equation 7x-2y=-6 8x+y=3

Mathematics

1 answer:

tatuchka [14]4 years ago

7 0

Answer:

(0,3)

Step-by-step explanation:

Step 1: 7x-2y=-6 Step 2: 7x-2y=-6 solve by elimination of the y value

2(8x+y=3)2 +(16x+2y=6)

Step 3: 23x=0, which means <u>x=0</u>

Step 4: subtitute 0 into the equation 8x+y=3

you get 8(0)+y=3, which comes to 0+y=3

you get <u>y=3</u>

therefore the solution to this system is (0,3)

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Which ratio correctly compares 60 mm to 45 cm?

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60 mm converts to 6 centimeters.
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2 : 15

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3 years ago

2) The cost of renting a boat for one or more

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3 years ago

I know you want to answer this question.

Alik [6]

Answer:

D. x = 3

Step-by-step explanation:

$\frac{1}{2} ^{x-4}$ - 3 = $4^{x-3}$ - 2

First, convert $4^{x-3}$ to base 2:

$4^{x-3}$ = $(2^{2})^{x-3}$

$\frac{1}{2} ^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Next, convert $\frac{1}{2} ^{x-4}$ to base 2:

$\frac{1}{2} ^{x-4}$ = $(2^{-1})^{x-4}$

$(2^{-1})^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{-1})^{x-4}$ = $2^{-1*(x-4)}$

$2^{-1*(x-4)}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{2})^{x-3}$ = $2^{2(x-3)}$

$2^{-1*(x-4)}$ - 3 = $2^{2(x-3)}$ - 2

Apply exponent rule: $a^{b+c}$ = $a^{b}$ $a^{c}$ :

$2^{-1(x-4)}$ = $2^{-1x}$ * $2^{4}$ , $2^{2(x-3)}$ = $2^{2x}$ * $2^{-6}$

$2^{-1 * x}$ * $2^{4}$ - 3 = $2^{2x}$ * $2^{-6}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$2^{-1x}$ = $(2^{x})^{-1}$ , $2^{2x}$ = $(2^{x})^{2}$

$(2^{x})^{-1}$ * $2^{4}$ - 3 = $(2^{x})^{2}$ * $2^{-6}$ - 2

Rewrite the equation with $2^{x} = u$ :

$(u)^{-1}$ * $2^{4}$ - 3 = $(u)^{2}$ * $2^{-6}$ - 2

Solve $u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2:

$u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2

Refine:

$\frac{16}{u}$ - 3 = $\frac{1}{64}$ $u^{2}$ - 2

Add 3 to both sides:

$\frac{16}{u}$ - 3 + 3 = $\frac{1}{64}$ $u^{2}$ - 2 + 3

Simplify:

$\frac{16}{u}$ = $\frac{1}{64}$ $u^{2}$ + 1

Multiply by the Least Common Multiplier (64u):

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify:

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify $\frac{16}{u}$ * 64u:

1024

Simplify $\frac{1}{64}$ $u^{2}$ * 64u:

$u^{3}$

Substitute:

1024 = $u^{3}$ + 64u

Solve for u:

u = 8

Substitute back u = $2^{x}$ :

8 = $2^{x}$

Solve for x:

x = 3

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3 years ago

What is the range of the function y=e^x+1 graphed below

REY [17]

The range of y=e^x+1 is y>1.

5 0

3 years ago

Read 2 more answers

Solve:

svlad2 [7]

Hello!

Judging on the questions you have provided I have come to the conclusion that to solve all of these you need to find the value of x by isolating it all on one side and simplifying.

For question #1 the first step would be to subtract 6 from each side.
The outcome should be -x=-6-12, and if simplified -x=-18.
Next, you want to divide each side by -1 because x needs to always be positive when finding the solution.
Your answer for #1 should be X=18, D.

For question #2 the first step you should take is to subtract 1.33 from each side to isolate x.
The outcome should be x=7.82-1.33, and if simplified x=6.49.
Thus your answer for #2 should be X=6.49, A.

For question #3 the first step you should take to approaching the equation is to isolate x by adding 15.99 towards each side of the equation.
The outcome should be x=58.06+15.99, and if simplified x=74.05.
Thus your answer for question #3 should be X=74.05, D.

Lastly, for question #4 the first step you should take towards isolating x is to divide the entire equation by 1.32.
Your outcome should be x= 6.8
Thus, your answer for the last question, #4, should be X=6.8, C.

Hope this helped!
-Blake

8 0

3 years ago

Read 2 more answers