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3 years ago

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How much is 40 gallonsMai is filling her fish tank. Water flows into the tank at a constant rate. Complete the table below and a

nswer the following questions. time (minutes) water (gallons) Please help asap!!

1 answer:

svp [43]3 years ago

5 0

I’ll need to see the table and I’ll be able to help

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Help me please help me I really need help

lyudmila [28]

Answer:

44%

0.44 is 44% of 1.

6 0

3 years ago

To rent a certain meeting room, a college charges a reservation fee of $37 and an additional fee of $5.70 per hour. The chemistr

gulaghasi [49]

Answer: t < 4 hours

Step-by-step explanation:

Renting = $37 + t x $5.70

If $37 + t$5.70 < $59.80

t$5.70 < $59.80 - $37

t$5.70 < $22.80

t < $22.80/$5.70

t < 4 hours

[tex]\textit{\textbf{Spymore}}[/tex

7 0

3 years ago

What is the math problem 1/3 divided by 2

tia_tia [17]

0.1666666666666667 that is your answer to your question

8 0

3 years ago

Read 2 more answers

Compute the area of the triangle ( express answer in cm2 <br><br> Height =6 cm<br><br> Base= 12cm

Nostrana [21]

The formula for the area of a triangle is 1/2bh
1/2(12)(6)
1/2(72)
36cm2 is the answer.

5 0

3 years ago

Read 2 more answers

Convert r = 8cos θ to rectangular form.

nikklg [1K]

Answer:

$(x-4)^2+y^2=16$

This is a circle with radius $\sqrt{16}=4$ and center $(4,0)$ .

Step-by-step explanation:

Recall the following:

$x=r\cos(\theta)$

$y=r\sin(\theta)$

$\frac{y}{x}=\tan(\theta)$

$x^2+y^2=r^2$

We are given $r=8\cos(theta)$ .

Multiply both sides by $r$ :

$r(r)=8\cos(\theta)r$

Reorder on right and simplify on left:

$r^2=8r\cos(theta)$

We can now make some replacements from our "Recall" section:

$x^2+y^2=8x$

We can probably leave like this, but sometimes we are required to put in a more identifying form.

This is a circle. I'm going to write in the form $(x-h)^2+(y-k)^2=r^2$ .

Subtract $8x$ on both sides:

$x^2-8x+y^2=0$

$x^2-8x+\text{(blank)}+y^2=0+\text{(blank)}$

We are going to fill those "blanks" in so that we can complete the square for the $x$ part.

$x^2-8x+(\frac{-8}{2})^2+y^2=0+(\frac{-8}{2})^2$

$(x+\frac{-8}{2})^2+y^2=0+(-4)^2$

$(x-4)^2+y^2=0+16$

$(x-4)^2+y^2=16$

So our equation is that of a circle with radius $\sqrt{16}=4$ and center $(4,0)$ .

6 0

3 years ago