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1 year ago

your dog gets fed 2/5 of a cup of food each day, your cat gets fed 2/3 of what your dog gets fed, how much do you feed your cat?

Mathematics

1 answer:

konstantin123 [22]1 year ago

8 0

2/3 cup of food = 134g

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Express ln 32 - ln 8 as a single natural logarithm

ivanzaharov [21]

$\bf \textit{Logarithm of rationals}
\\\\
log_a\left( \frac{x}{y}\right)\implies log_a(x)-log_a(y)\\\\
-------------------------------\\\\
ln(32)-ln(8)\implies ln\left( \frac{32}{8} \right)\implies ln(4)$

8 0

3 years ago

Read 2 more answers

Someone please explain this it’s confusing me.

Nesterboy [21]

Answer:

x = 4 radical 3 / 3

Step-by-step explanation:

this is a 30-60-90 triangle

the side across from the angle 30 is L

the side across from the angle 60 is L√3

the side across from the angle 90 is 2L

we are given the side across from 60⁰ so we know that:

2 = L√3

we want to solve for L, so we must divide both sides by √3 (radical three)

2/√3 = L√3/√3

you would get 2/ √3 = L, because the radical three cancels out on the L side.

but you can't have a radical in the denominator, so you have to multiply 2/√3 by radical 3

2 times √3 and √3 times √3

you get 2√3/√9

radical nine can simplify

2radical3/3 or 2√3/3

we found L, but that would equal the thirty degree angle, not the 90, 90 = 2L

multiply the number infront of the radical by two and get

4√3/3

x=4 radical 3 divided by 3

5 0

2 years ago

Find the exact length of the curve. x=et+e−t, y=5−2t, 0≤t≤2 For a curve given by parametric equations x=f(t) and y=g(t), arc len

Rama09 [41]

The length of a curve C parameterized by a vector function r(t) = x(t) i + y(t) j over an interval a ≤ t ≤ b is

$\displaystyle\int_C\mathrm ds = \int_a^b \sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2} \,\mathrm dt$

In this case, we have

x(t) = exp(t ) + exp(-t ) ==> dx/dt = exp(t ) - exp(-t )

y(t) = 5 - 2t ==> dy/dt = -2

and [a, b] = [0, 2]. The length of the curve is then

$\displaystyle\int_0^2 \sqrt{\left(e^t-e^{-t}\right)^2+(-2)^2} \,\mathrm dt = \int_0^2 \sqrt{e^{2t}-2+e^{-2t}+4}\,\mathrm dt$

$=\displaystyle\int_0^2 \sqrt{e^{2t}+2+e^{-2t}} \,\mathrm dt$

$=\displaystyle\int_0^2\sqrt{\left(e^t+e^{-t}\right)^2} \,\mathrm dt$

$=\displaystyle\int_0^2\left(e^t+e^{-t}\right)\,\mathrm dt$

$=\left(e^t-e^{-t}\right)\bigg|_0^2 = \left(e^2-e^{-2}\right)-\left(e^0-e^{-0}\right) = \boxed{e^2-\frac1{e^2}}$

5 0

2 years ago

If the length of a rectangle is 4.3 feet and its width is 2.7 feet what is its are and its perimeter

Dvinal [7]

The area is 11.61 square feet and the perimeter is 14 feet. i hope this helps :)

Area=Length times width

Perimeter=Length+Length+width+width

7 0

3 years ago

Read 2 more answers

Mystery Boxes: Breakout Rooms

ollegr [7]

Answer:

$\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}$

Step-by-step explanation:

Given

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {[ \ ]} \\ \end{array}$

Required

Fill in the box

From the question, the range is:

$Range = 60$

Range is calculated as:

$Range = Highest - Least$

From the box, we have:

$Least = 1$

So:

$60 = Highest - 1$

$Highest = 60 +1$

$Highest = 61$

The box, becomes:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question:

$IQR = 20$ --- interquartile range

This is calculated as:

$IQR = Q_3 - Q_1$

$Q_3$ is the median of the upper half while $Q_1$ is the median of the lower half.

So, we need to split the given boxes into two equal halves (7 each)

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } \\ \end{array}$

Upper half

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

The quartile is calculated by calculating the median for each of the above halves is calculated as:

$Median = \frac{N + 1}{2}th$

Where N = 7

So, we have:

$Median = \frac{7 + 1}{2}th = \frac{8}{2}th = 4th$

So,

$Q_3$ = 4th item of the upper halves

$Q_1$ = 4th item of the lower halves

From the upper halves

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

We have:

$Q_3 = 32$

$Q_1$ can not be determined from the lower halves because the 4th item is missing.

So, we make use of:

$IQR = Q_3 - Q_1$

Where $Q_3 = 32$ and $IQR = 20$

So:

$20 = 32 - Q_1$

$Q_1 = 32 - 20$

$Q_1 = 12$

So, the lower half becomes:

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {12 } & {15} & {18}& {[ \ ] } \\ \end{array}$

From this, the updated values of the box is:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$