Ask question

Ask question

All categories

2 years ago

10

Peter has just purchased a small rectangular fish tank that holds 910 milliliters of water. The tank has a width of 7 cm and a h

eight of 10 cm.
What is the length of the fish tank?

(1 milliliter is equal to 1 cubic centimeter)
A.
13 cm
B.
14 cm
C.
15 cm

2 answers:

miv72 [106K]2 years ago

8 0

Option A) 13 cm is the length of the fish tank.

<u>Step-by-step explanation:</u>

Peter has just purchased a small rectangular fish tank that holds 910 milliliters of water.

(1 milliliter is equal to 1 cubic centimeter)

⇒ Volume of the fish tank = 910 cubic centimeter

The width of the tank = 7 cm
The height of the tank = 10 cm
The length of the fish tank = x cm

We know that, Volume of a cuboid = length × width × height

⇒ 910 = x × 7 × 10

⇒ 910 = 70x

⇒ x = 910 / 70

⇒ x = 13 cm

Therefore, Option A) 13 cm is the length of the fish tank.

Wewaii [24]2 years ago

3 0

the length of the fish tank is $length=13cm$ . Correct option is A)

<u>Step-by-step explanation:</u>

Here we have , Peter has just purchased a small rectangular fish tank that holds 910 milliliters of water. The tank has a width of 7 cm and a height of 10 cm. We need to find What is the length of the fish tank . Let's find out:

The fish tank is in the shape of cuboid , and we know that volume of cuboid :

⇒ $Volume = length(width)(height)$

Now , according to question we have following parameters :

$V=910ml=910cm^3\\w=7cm\\h=10cm$

Therefore , Putting theses values in above equation we get :

⇒ $Volume = length(width)(height)$

⇒ $910 = length(7)(10)$

⇒ $910 = length(70)$

⇒ $length=13cm$

Therefore , the length of the fish tank is $length=13cm$ . Correct option is A)

You might be interested in

3 x A = 30 + 30<br><br> it’s a 3rd grade problem but don’t know the answer

xenn [34]

Answer: A=20

Step-by-step explanation:

3×A=30+30............eqn 1

3A=60

A=60/3

A=20

Substitute A=20 into equation 1

3x20= 30+30

8 0

2 years ago

When the sample size and sample standard deviation remain the same, a 99% confidence interval for a population mean, µ will be n

motikmotik

A 99% confidence interval will be wider compared to a 95% confidence interval. This is because while you don't know what the true value of mu is, you are more confident that the parameter is in the interval.

In essence, you are casting a wider net which leads to the higher level of confidence. Imagine if there was a 100 mile stretch of straight land, and somewhere on this line was a special rock that you are looking for. You would be 100% confident if you said "the rock is somewhere on that piece of land", and less confident the more you shrunk the interval.

5 0

2 years ago

Mrac [35]

Answer:

71.6 degrees

Step-by-step explanation:

Given the vectors

u= <8,4> v = <9.-9> (5 points)

u*v = (8, 4)*(9, -9)

u*v = 8(9)+(4)(-9)

u*v = 72 - 36

u*v = 36

|u| = √8²+4²

|u| = √64+16

|u| = √80

|v| = √9²+9²

|v| = √81+81

|v| = √162

Using the formula

u*v = ||u||v| cos theta

36 = √80(√162)cos theta

36 = √12960 cos theta

cos theta = 36/√12960

cos theta = 36/113.8

cos theta = 0.3162

theta = arccos(0.3162)

theta = 71.56 degrees

Hence the angle between the given vectors to the nearest tenth of a

degree is 71.6 degrees

5 0

1 year ago

A science class is tracking the progress of plant growth. The class starts the experiment with a plant five centimeters high. Th

NISA [10]

Answer:

In this equation, the y-intercept (+ 5) represents the original height of the plant. 2x represents the amount of growth multiplied by the number of days, x.

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

T what point does the curve have maximum curvature? Y = 7ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as

Nookie1986 [14]

Formula for curvature for a well behaved curve y=f(x) is

K(x)= $\frac{|{y}''|}{[1+{y}'^2]^\frac{3}{2}}$

The given curve is y=7 $e^{x}$

${y}''=7e^{x}\\ {y}'=7e^{x}$

k(x)= $\frac{7e^{x}}{[{1+(7e^{x})^2}]^\frac{3}{2}}$

${k(x)}'=\frac{7(e^x)(1+49e^{2x})(49e^{2x}-\frac{1}{2})}{[1+49e^{2x}]^{3}}$

For Maxima or Minima

${k(x)}'=0$

$7(e^x)(1+49e^{2x})(98e^{2x}-1)=0$

→ $e^{x}=0∨ 1+49e^{2x}=0∨98e^{2x}-1=0$

$e^{x}=0 , ∧ 1+49e^{2x}=0$ [not possible ∵there exists no value of x satisfying these equation]

→ $98e^{2x}-1=0$

Solving this we get

x= $-\frac{1}{2}\ln{98}$

As you will evaluate ${k(x})}''$ <0 at x= $-\frac{1}{2}\ln98$

So this is the point of Maxima. we get y=7×1/√98=1/√2

(x,y)=[ $-\frac{1}{2}\ln98$ ,1/√2]

k(x)= $\lim_{x\to\infty } \frac{7e^{x}}{[{1+(7e^{x})^2}]^\frac{3}{2}}$

k(x)= $\frac{7}{\infty}$

k(x)=0

5 0

2 years ago