TODIGIllici 8. Solve: - 5y+30=15

Mathematics

1 answer:

nasty-shy [4]2 years ago

4 0

Answer:

y=3

Step-by-step explanation:

Subtract 30 from both sides

-5y+30=15

-5y=-15

Divide by -5 on both sides

y=3

Hope this helps! :)

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The____ function is the most basic function in a family of functions

docker41 [41]

The answer is: parent function

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2 years ago

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Bowser ate 4 1/2 pounds of dog food. That amount is 3/4 of an entire bag of dog food. How many pound of dog food were originally

GalinKa [24]

Divide number of pounds eaten by the portion of the bag eaten.

= 4 1/2 pounds ÷ 3/4
= 4.5 ÷ 0.75
= 6 pound bag

CHECK:
multiply total pounds of bag by portion eaten
= 6 pound bag * 3/4
= 4 1/2 pounds eaten

ANSWER: The bag was 6 pounds of dog food.

Hope this helps! :)

8 0

2 years ago

I will give brainiest to whoever answers correctly !!

Andreyy89

Answer:
Step-by-step explanation:
We would apply the simple interest formula which is expressed as
I = PRT/100
Where
P = principal or amount borrowed
T = time in years
R = interest rate on amount borrowed.
I = interest paid.
From the given information,
Principal = $3000
T = 3 months = 3/12 = 0.25 years
R = 6 1/2 % = 6.5%
Therefore,
a) the amount that the woman pay for the use of the money is I
I = (3000 × 6.5 × 0.25)/100 = 48.75
b) The amount she repaid to the bank on the due date of the note would be
Principal + interest
= 3000 + 48.75 = $3048.75

4 0

2 years ago

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Given the functionf ( x ) = x^2 + 7 x + 10/ x^2 + 9 x + 20

vladimir1956 [14]

<em>x = -4 is a vertical asymptote for the function.</em>

<h2>Explanation:</h2>

The graph of $y=f(x)$ is a vertical has an asymptote at $x=a$ if at least one of the following statements is true:

$1) \ \underset{x\rightarrow a^{-}}{lim}f(x)=\infty\\ \\ 2) \ \underset{x\rightarrow a^{-}}{lim}f(x)=-\infty \\ \\ 3) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty \\ \\ 4) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty$

The function is:

$f(x)=\frac{x^2+7x+10}{x^2+9x+20}$

First of all, let't factor out:

$f(x)=\frac{x^2+5x+2x+10}{x^2+5x+4x+20} \\ \\ f(x)=\frac{x(x+5)+2(x+5)}{x(x+5)+4(x+5)} \\ \\ f(x)=\frac{(x+5)(x+2)}{(x+5)(x+4)} \\ \\ f(x)=\frac{(x+2)}{(x+4)}, \ x\neq 5$

From here:

$\bullet \ When \ x \ approaches \ -4 \ on \ the \ right: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(-4^{+}+2)}{(-4^{+}+4)} \\ \\ \\ The \ numerator \ is \ negative \ and \ the \ denominator \\ is \ a \ small \ positive \ number. \ So: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=-\infty$

$\bullet \ When \ x \ approaches \ -4 \ on \ the \ left: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(-4^{-}+2)}{(-4^{-}+4)} \\ \\ \\ The \ numerator \ is \ a \ negative \ and \ the \ denominator \\ is \ a \ small \ negative \ number \ too. \ So: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=+\infty$