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2 years ago

9

fertilizer you put 2 qts for every 50 gal of water I only want to use 15gal of water how many qts,pts,cups??? do i put

1 answer:

xxMikexx [17]2 years ago

7 0

Answer is 1.2 pints fertilizer
Step by step
First I’m going to change 2 quarts to 4 pints because we are going for a smaller portion of water to fertilizer
Next you find your rate
50 gal water to 4 pints fertilizer = 12.5 rate
We want to solve for x, 15 gallons water to x fertilizer
50/2 to 15/x
Substitute your rate for x and divide
15/12.5 = 1.2 pints

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Their are 5280 feet in 1 mile. How many inches are there in 4 miles

schepotkina [342]

Answer:

21120

Step-by-step explanation:Because you take 5280 and multiply it by 4

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3 years ago

I really don’t get it

posledela

<em>you just add 8 + 8 + 8 it equals </em><u><em>24. rounded it will be 20</em></u>

<u><em>--------------------------------------------------</em></u>

<u><em>im pretty sure thats how you do it if not sorry</em></u>

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2 years ago

Canada has a population that is about 1/10 as large as United States. If Canada population is about 32 million, about how many p

Sladkaya [172]

Do 32,000,000 divided by 10 to get your answer.

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3 years ago

Suppose that a function f(x) is defined for all real values of x except at xequals=c. can anything be said about the existence

Margaret [11]

we are given that

f(x) is defined for all values of x except at x=c

Limit may or may not exist

case-1:

If there is hole at x=c , then limit exist

case-2:

If there is vertical asymptote at x=c , then limit does not exist

Examples:

case-1:

$\lim_{x \to c} \frac{x^2-cx}{(x-c)}$

We can simplify it

$\lim_{x \to c} \frac{x(x-c)}{(x-c)}$

$=\lim_{x \to c} x$

$=c$

so, we can see that limit exist and it's value defined

case-2:

$\lim_{x \to c} \frac{1}{(x-c)}$

Left limit is

$\lim_{x \to c-} \frac{1}{(x-c)}$

$=-\infty$

Right Limit is

$\lim_{x \to c+} \frac{1}{(x-c)}$

$=+\infty$

so, we can see that left limit is not equal to right limit

so, limit does not exist

4 0

3 years ago

Which equations represent the asymptotes of the hyperbola (x-1)^2/36-(y-2)^2/64=1 ?

allochka39001 [22]

Answer:

$4x+3y=10$ and $4x-3y=-2$

Step-by-step explanation:

The asymptote equation of a translate hyperbola with equation $\frac{(x-h)^2}{b^2}-\frac{(y-k)^2}{a^2}=1$ is $y=\pm\frac{b}{a}(x-h)+k$

The given hyperbola has equation: $\frac{(x-1)^2}{36}-\frac{(y-2)^2}{64}=1$

Or $\frac{(x-1)^2}{6^2}-\frac{(y-2)^2}{8^2}=1$

Comparing to $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$

We have a=6 and b=8, h=1 and k=2.

We substitute this values to get:

$y=\pm\frac{6}{8}(x-1)+2$

$y=\pm\frac{4}{3}(x-1)+2$

$3y=\pm4(x-1)+6$

We split the plus or minus sign to get:

$3y=4x-4+6$

$3y=4x+2$

$3y-4x=2$

Or

$3y=-4(x-1)+6$

$3y=-4x+4+6$

$3y+4x=10$

5 0

3 years ago