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3 years ago

There are 6 birdcages at the zoo. Two birds are in each birdcage. How many birds are in each birdcage?

Mathematics

2 answers:

Nikolay [14]3 years ago

8 0

Answer:12

Step-by-step explanation:

2*6

Marat540 [252]3 years ago

7 0

2 birds in each birdcage

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To subtract 78 - 35 mentally Carmen out of 5 to each number and Karen a tattoo to each number will both met at work to find the

Vinil7 [7]

215 is the answer because 78-35 =43 43×5=215

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3 years ago

HELP SOON!!!! Zahra wants the equation below to have an infinite number of solutions when the missing number is placed in the bo

marin [14]

Answer:

Your answer is -3

Step-by-step explanation:

I had this on a test and i got it

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5 0

3 years ago

Read 2 more answers

Express your answer as a polynomial in standard form.

lana [24]

9514 1404 393

Answer:

f(g(x)) = -3x^2 -2x -3

Step-by-step explanation:

Given:

f(x) = -x-7

g(x) = 3x^2 +2x -4

Find:

f(g(x))

Solution:

Substitute per the function definitions.

f(g(x)) = f(3x^2 +2x -4) = -(3x^2 +2x -4) -7

f(g(x)) = -3x^2 -2x -3

7 0

3 years ago

Jada made 14 cups for of blueberry jam and divided the jam equally amoung 8 containers how much jam went in each container

N76 [4]

Uhm I’m not completely sure but 1.75

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2 years ago

Questions attached as screenshot below:Please help me I need good explanations before final testI pay attention

Nikitich [7]

The acceleration of the particle is given by the formula mentioned below:

$a=\frac{d^2s}{dt^2}$

Differentiate the position vector with respect to t.

$\begin{gathered} \frac{ds(t)}{dt}=\frac{d}{dt}\sqrt[]{\mleft(t^3+1\mright)} \\ =-\frac{1}{2}(t^3+1)^{-\frac{1}{2}}\times3t^2 \\ =\frac{3}{2}\frac{t^2}{\sqrt{(t^3+1)}} \end{gathered}$

Differentiate both sides of the obtained equation with respect to t.

$\begin{gathered} \frac{d^2s(t)}{dx^2}=\frac{3}{2}(\frac{2t}{\sqrt[]{(t^3+1)}}+t^2(-\frac{3}{2})\times\frac{1}{(t^3+1)^{\frac{3}{2}}}) \\ =\frac{3t}{\sqrt[]{(t^3+1)}}-\frac{9}{4}\frac{t^2}{(t^3+1)^{\frac{3}{2}}} \end{gathered}$

Substitute t=2 in the above equation to obtain the acceleration of the particle at 2 seconds.

$\begin{gathered} a(t=1)=\frac{3}{\sqrt[]{2}}-\frac{9}{4\times2^{\frac{3}{2}}} \\ =1.32ft/sec^2 \end{gathered}$

The initial position is obtained at t=0. Substitute t=0 in the given position function.

$\begin{gathered} s(0)=-23\times0+65 \\ =65 \end{gathered}$

8 0

1 year ago