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Illusion [34]
3 years ago
5

Please HELP!! ASAP!!

Mathematics
2 answers:
ELEN [110]3 years ago
6 0

Answer:

20in

Step-by-step explanation: If the smaller frame is 3 by 5in long then the larger frame will by 12 by 20in becuase 3*4=12  so  5*4=20

3/4=12/40

serious [3.7K]3 years ago
4 0

Answer: 20 inches long.

Step-by-step explanation:

The Rectangular frame which Tamara already has is 3 inches by 5 inches. The larger one happens to be 12 inches by X inches. We can find the scale of the new Frame by dividing the new value we are provided with by the initial one. In this case, 12 (large frame) divided by 3, or 4. This means that the dimensions of the large Frame are 4 times greater than the smaller one. If this is to be the case, the length is 4 times larger than the original, and 5 x 4 = 20.

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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
2 years ago
Water tank A has a capacity of 550 gallons and is 66% full. Water tank B is 53% full. The ratio of the capacity of T. How much w
Anna71 [15]

Tank A capacity = 550 gallons       66% x 550 gallons = 363 gallons in Tank A

How about the water in tank B?

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4 0
3 years ago
What’s the domain and range for these two?
user100 [1]

Step-by-step explanation:

1) the domain is from -3 to 5 D [-3,5]

the range is from -5 to -2. R [-5,-2]

2)the domain is from -3 to 6 but not including 6 D [-3,6)

the range is from -5 but not including -5 to 2. R (-5,2]

4 0
3 years ago
Mina's kitchen sink holds up to 103.468 liters of water. Round this amount to the nearest liter
irina [24]
103 liters I think .....
7 0
3 years ago
Read 2 more answers
Given the functionf ( x ) = x^2 + 7 x + 10/ x^2 + 9 x + 20
vladimir1956 [14]

<em>x = -4 is a vertical asymptote for the function.</em>

<h2>Explanation:</h2>

The graph of y=f(x) is a vertical has an asymptote at x=a if at least one of the following statements is true:

1) \ \underset{x\rightarrow a^{-}}{lim}f(x)=\infty\\ \\ 2) \ \underset{x\rightarrow a^{-}}{lim}f(x)=-\infty \\ \\ 3) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty \\ \\ 4) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty

The function is:

f(x)=\frac{x^2+7x+10}{x^2+9x+20}

First of all, let't factor out:

f(x)=\frac{x^2+5x+2x+10}{x^2+5x+4x+20} \\ \\ f(x)=\frac{x(x+5)+2(x+5)}{x(x+5)+4(x+5)} \\ \\ f(x)=\frac{(x+5)(x+2)}{(x+5)(x+4)} \\ \\ f(x)=\frac{(x+2)}{(x+4)}, \ x\neq  5

From here:

\bullet \ When \ x \ approaches \ -4 \ on \ the \ right: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(-4^{+}+2)}{(-4^{+}+4)} \\ \\ \\ The \ numerator \ is \ negative \ and \ the \ denominator \\ is \ a \ small \ positive \ number. \ So: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=-\infty

\bullet \ When \ x \ approaches \ -4 \ on \ the \ left: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(-4^{-}+2)}{(-4^{-}+4)} \\ \\ \\ The \ numerator \ is \ a \ negative \ and \ the \ denominator \\ is \ a \ small \ negative \ number \ too. \ So: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=+\infty

Accordingly:

x=-4 \ is \ a \ vertical \ asymptote \ for \\ \\ f(x)=\frac{x^2+5x+2x+10}{x^2+5x+4x+20}

<h2>Learn more:</h2>

Vertical and horizontal asymptotes: brainly.com/question/10254973

#LearnWithBrainly

5 0
4 years ago
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