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3 years ago

X-5y=2 and 9x+8=15 solve by substitution

1 answer:

7 0

$x-5y=2$
Hence $x=2+5y$ (Equation I)
$9x+8=15$ (Equation II)
Substituting Equation I in Equation II
$9(2+5y)+8=15$
$18+45y+8=15$
$26+45y=15$
$45y=-11$
Hence $y= - \frac{11}{45}$

Substituting this in Equation in I to get X

$x-5(- \frac{11}{45} )=2$
$x+ \frac{11}{9}=2$
$x= 2- \frac{11}{9}$
$x= \frac{17}{9}$

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3 years ago

Read 2 more answers

Anyone know where I can get these answers? <br> (There’s also a back but I can’t send it)

Nutka1998 [239]

Answer:

See below

Step-by-step explanation:

<u>Covert equations in standard form into slope-intercept form of y = mx + b</u>

x - y = 3 ⇒ y = x - 3

5x + 3y = 6 ⇒ 3y = -5x + 6 ⇒ y = -5/3x + 2

x - 2y = 2 ⇒ 2y = x - 2 ⇒ y = 1/2x - 1

3x + 2y = -8 ⇒ 2y = -3x - 8 ⇒ y = -3/2x - 4

4 0

3 years ago

Write an equation of the line that passes through (1,2) and is parallel to the line y = -5x+4

ArbitrLikvidat [17]

Answer:

y =-5x + 7

Step-by-step explanation:

y = -5x + 4

Slope = -5

Parallel lines have same slope

Slope of the parallel line to this line = -5

(1 , 2)

y -y1 =m(x - x1)

y - 2 = -5(x - 1)

y -2 = -5x + 5

y = -5x + 5 + 2

y = -5x + 7

8 0

3 years ago

1) work out the value of (1.7x10^4)x (8.5x10^-2) give your answer in standard form.

Mashutka [201]

ANSWER TO QUESTION 1

$(1.7\times 10^4)\times(8.5\times 10^{-2})$

We rewrite to obtain;

$(1.7\times 10^4)\times(8.5\times 10^{-2})=(1.7\times 8.5)\times(10^4\times 10^{-2})$

Recall this product law of indices

$a^m\times a^n=a^{m+n}$

we apply this law to obtain,

$(1.7\times 10^4)\times(8.5\times 10^{-2})=14.45\times10^{4+-2}$

This simplifies to

$(1.7\times 10^4)\times(8.5\times 10^{-2})=14.45\times10^{2}$

We need to rewrite this in standard form;

$(1.7\times 10^4)\times(8.5\times 10^{-2})=1.445\times 10^1\times10^{2}$

We apply the product law again to get

$(1.7\times 10^4)\times(8.5\times 10^{-2})=1.445\times 10^{1+2}$

This simplifies to

$(1.7\times 10^4)\times(8.5\times 10^{-2})=1.445\times 10^{3}$

ANSWER TO QUESTION 2

$(6.8\times 10^2)\times(1.3\times 10^{-3})$

We rewrite to obtain;

$(6.8\times 10^2)\times(1.3\times 10^-3)=(6.8\times 1.3)\times(10^2\times 10^{-3})$

Recall this product law of indices

$a^m\times a^n=a^{m+n}$

we apply this law to obtain,

$(6.8\times 10^2)\times(1.3\times 10^-3)=8.84\times10^{2+-3}$

This simplifies to

$(6.8\times 10^2)\times(1.3\times 10^-3)=8.84\times10^{-1}$

This is already in standard form.

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3 years ago

hayley wants to attend bellhaven collage after high school. she knows that the yearly cost is around $17000 per year. her parent

KengaRu [80]

I think the answer is 13,840

to start off use the formula IPRT which stands for interest, principal, rate, and time. so then you do 2,000 times 6% (or 0.06) then times that by 18 which is 2,160 so she earned 2,160 in her college account, next you add the 1,000 and that would make it 3,160. next you subtract 3,160 from 17000 to see how much she still has to pay which i think is 13,840

if this is wrong lmk and i can try to fix it

6 0

3 years ago