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3 years ago

11

Barbara has a bunny that weighs 5 pounds and gains 3 pounds per year. Her cat weighs 19 pounds and gains 1 pound per year. When

will the bunny and the cat weigh the same amount? pls help

1 answer:

N76 [4]3 years ago

4 0

Answer:

it will take 6 years

Step-by-step explanation:

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Enter the correct answer in the box. What is this expression in simplified form? <br> 3\sqrt{88}

Ugo [173]

Answer: $12\sqrt{5}$

Step-by-step explanation:

$3\sqrt{88}$

88 can be rewritten as 8 x 10 and 8 can be rewritten as $2^2*2$ making a 2 come out of the root.

$3\sqrt{2^2*2*10}\\ 3*2\sqrt{2*10}\\6\sqrt{2*10}$

10 can be rewritten as 2 x 5.

$6\sqrt{2*2*5}$

2*2 can also be rewritten as $2^2$ making a 2 come out of the root.

$6\sqrt{2^2*5}\\ 6*2\sqrt{5}\\ 12\sqrt{5}$

We can't keep simplifying the root.

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3 years ago

What is the 75% of 1119

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Step-by-step explanation:

75 x 1119 / 100 = 839.25

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Understand Mean and MAD - Quiz - Level F

pshichka [43]

Answer:

The smaller the MAD the less variability, meaning a bigger MAD means more distance between data.

Step-by-step explanation:

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Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x2/3 + y2/3 = 4 (−3 3 , 1)

vovikov84 [41]

Answer with Step-by-step explanation:

We are given that an equation of curve

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=4$

We have to find the equation of tangent line to the given curve at point $(-3\sqrt3,1)$

By using implicit differentiation, differentiate w.r.t x

$\frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=0$

Using formula : $\frac{dx^n}{dx}=nx^{n-1}$

$\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=-\frac{2}{3}x^{-\frac{1}{3}}$

$\frac{dy}{dx}=\frac{-\frac{2}{3}x^{-\frac{1}{3}}}{\frac{2}{3}y^{-\frac{1}{3}}}$

$\frac{dy}{dx}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}$

Substitute the value x= $-3\sqrt3,y=1$

Then, we get

$\frac{dy}{dx}=-\frac{(-3\sqrt3)^{-\frac{1}{3}}}{1}$

$\frac{dy}{dx}=-(-3^{\frac{3}{2}})^{-\frac{1}{3}}=-\frac{1}{-(3)^{\frac{3}{2}\times \frac{1}{3}}}=\frac{1}{\sqrt3}$

Slope of tangent=m= $\frac{1}{\sqrt3}$

Equation of tangent line with slope m and passing through the point $(x_1,y_1)$ is given by

$y-y_1=m(x-x_1)$

Substitute the values then we get

The equation of tangent line is given by

$y-1=\frac{1}{\sqrt3}(x+3\sqrt3)$

$y-1=\frac{x}{\sqrt3}+3$

$y=\frac{x}{\sqrt3}+3+1$

$y=\frac{x}{\sqrt3}+4$

This is required equation of tangent line to the given curve at given point.

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Lelechka [254]

Answer: c

Step-by-step explanation:

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