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3 years ago

15

Which values of x and y will satisfy y = 5x and 12x + 9y = 60?. . A. . . x = 1, y = 19 . . B. . . x = 35 y= 20/19. . C. . . x =

1/20 y = 1/90. . D.. . x = 20/19 y = 100/19. . E.. . x = 19 y = 50/19

2 answers:

Alinara [238K]3 years ago

8 0

The answer is D=> x=20/19 and y=<span>100/19

</span>

grigory [225]3 years ago

3 0

The answer is D, (x, y)=(20/19, 100/19)

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Simplify this expression <br> 7z+5+6z-9

Simora [160]

Answer:

7z + 5 + 6z - 9 (Add 7z and 6z, subtract 9 from 5)

<em>13z - 4 </em>(Can't be simplified further)

8 0

2 years ago

You are saving money to buy an electric guitar. You deposit $1000 in an account that earns interest compounded annually. The exp

Katena32 [7]

Let's move like a crab, backwards some.

after 2 years?

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$1000\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &2
\end{cases}
\\\\\\
A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 2}\implies A=1000(1.03)^2$

after 3 years?

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$1000\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &3
\end{cases}
\\\\\\
A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 3}\implies A=1000(1.03)^3$

is that enough to pay the $1100?

now, let's write 1000(1+r)² in standard form

1000( 1² + 2r + r²)

1000(1 + 2r + r²)

1000 + 2000r + 1000r²

1000r² + 2000r + 1000 <---- standard form.

8 0

2 years ago

Jenny bought a 5 notebooks and 2 pencils, and the total was 9 dollars.

weeeeeb [17]

You would solve this with simultaneous equations, so if we write it as:
5n + 2p = 9
3n + 2p = 6
(subtract)
2n = 3
÷ 2
notebooks = 1.5

Now you would substitute it in:
(3 × 1.5) + 2p = 6
4.5 + 2p = 6
- 4.5
2p = 1.5
÷ 2
pens = 0.75

So your final answer is notebooks are $1.50 and pens are $0.75, I hope this helps!

6 0

3 years ago

Sami made 23,400 last year in income from his job. He worked 30 hours per week and worked all 52 weeks of the year. What was his

xxTIMURxx [149]

Answer:

$15 per hour

Step-by-step explanation:

30 hours per week x 52 weeks in a year = 1560 hours per year

23,400 / 1560 = $15 per hour

7 0

3 years ago

Factorize 2a^2+7a-15

loris [4]

$\huge\bf \pink {\underline{Solution :-}}$

$2 {a}^{2} + 7a - 15$

$= 2 {a}^{2} + 10a - 3a - 15$

$= 2a(a + 5) - 3(a + 5)$

$= (a + 5)(2a - 3)$

$\sf \red{Hence, Answer \: is \: (a + 5)(2a - 3).}$

$\\$

$\bf \purple{ \underline{ Important \: Formulas \: for \: Factorization :-}}$

$• \: {a}^{2} + 2ab + {b}^{2} = {(a + b)}^{2}$

$• \: (a + b)(a - b) = {a}^{2} - {b}^{2}$

$• \: {a}^{2} - 2 ab + {b}^{2} = {(a - b)}^{2}$

$• \: {a}^{3} + 3 {a}^{2} b + 3a {b}^{2} + {b}^{3} = {(a + b)}^{3}$

$• \: {a}^{3} - 3 {a}^{2} b + 3a {b}^{2} - {b}^{3} = {(a - b)}^{3}$

$• \: {(a + b)}^{2} + {(a - b)}^{2} = 2( {a}^{2} + {b}^{2} )$

$• \: {(a + b)}^{2} - {(a - b)}^{2} = 4ab$

$• \: (a + b)( {a}^{2} -ab + {b}^{2} ) = {a}^{3} + {b}^{3}$

$• \: (a - b)( {a}^{2} + ab + {b}^{2} ) = {a}^{3} - {b}^{3}$

$• \: {( \frac{a + b}{2} )}^{2} - ( {\frac{a - b}{2} } )^{2} = ab$

5 0

3 years ago

Read 2 more answers