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Arturiano [62]
3 years ago
5

A customer went to a garden shop and bought some potting soil for $17.50 and 4 shrubs. The total bill was $53.50. Choose the equ

ation that models this situation and solve to find the price p of each shrub.
Mathematics
1 answer:
olga2289 [7]3 years ago
6 0
4x + 17.50 = 53.50

the prices of each shrub would be 9 dollars each
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Find the similarity ratio of a cube with volume 729 m3 to a cube with volume 3,375 m3.iR²
irinina [24]

The similarity ratio of a cube with volume 729 m3 to a cube with a volume of 3,375 m3.iR² is 3:5. The correct answer between all the choices given is the second choice or letter B. I am hoping that this answer has satisfied your query and it will be able to help you, and if you would like, feel free to ask another question.

8 0
4 years ago
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One leg of a 45 - 45 - 90 triangle measures 10 inches. What is the length of the hypotenuse
djyliett [7]

Answer:

Therefore the length of the hypotenuse is 14.14 inches.

Step-by-step explanation:

Given:

The Triangle is of 45 -45 - 90

Length of one leg = 10 inches

To Find:

Length of the hypotenuse = ?

Solution:

The Triangle is of 45 -45 - 90 ..................Given

So For 45 - 45 - 90 Both the Legs of Triangle is SAME

i.e Longer leg = Shorter leg = 10 inches

Now in Right angle Triangle By Pythagoras Theorem we have

(\textrm{Hypotenuse})^{2} = (\textrm{Shorter leg})^{2}+(\textrm{Longer leg})^{2}

On substituting the values we get

Hypotenuse^{2}= 10^{2}+ 10^{2} \\\\Hypotenuse^{2}=200\\Square\ Rooting\ we\ get\\Hypotenuse=\sqrt{200}=10\sqrt{2}=10\times 1.414=14.14\ inches

Therefore the length of the hypotenuse is 14.14 inches.

5 0
3 years ago
How many square feet of outdoor carpet will we need for this hole?
Lina20 [59]
24 square feet + 12 square feet = total square feet?

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5 0
3 years ago
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If w = 5 cos (xy) − sin (xz) and x = 1/t , y = t, z = t^3 ; then find dw/dt
Scrat [10]

In this question, we find the derivatives, using the chain's rule.

Doing this, the derivative is:

\frac{dw}{dt} = \frac{5}{t}(\sin{1} - \cos{1}) - 2t\cos{t^2}

Chain Rule:

Suppose we have a function w(x,y,z), x = x(t), y = y(t), z = z(t), and want to find it's derivative as function of t. It will be given by:

\frac{dw}{dt} = \frac{dw}{dx}\frac{dx}{dt} + \frac{dw}{dy}\frac{dy}{dt} + \frac{dw}{dz}\frac{dz}{dt}

Thus, we have to find the desired derivatives, which are:

  • w of x:

\frac{dw}{dx} = -5y\sin{(xy)} - z\cos{(xz)}

Considering x = \frac{1}{t}, y = t, z = t^3

\frac{dw}{dx} = -5t\sin{(1)} - t^3\cos{(t^2)}

  • w of y:

\frac{dw}{dy} = -5x\cos{(xy)}

Considering x = \frac{1}{t}, y = t

\frac{dw}{dy} = -\frac{5}{t}\cos{1}

  • w of z:

\frac{dw}{dz} = -x\cos{(xz)}

Considering x = \frac{1}{t}, z = t^3

\frac{dw}{dz} = -\frac{1}{t}\cos{(t^2)}

  • Derivatives of x, y and z as functions of t:

\frac{dx}{dt} = -\frac{1}{t^2}

\frac{dy}{dt} = 1

\frac{dz}{dt} = 3t^2

  • Derivative of w as function of t.

Now, we just replace what we found into the formula. So

\frac{dw}{dt} = \frac{dw}{dx}\frac{dx}{dt} + \frac{dw}{dy}\frac{dy}{dt} + \frac{dw}{dz}\frac{dz}{dt}

\frac{dw}{dt} = (-5t\sin{(1)} - t^3\cos{(t^2)})(-\frac{1}{t^2}) - (\frac{5}{t}\cos{1}) - (\frac{1}{t}\cos{(t^2)})3t^2

Applying the multiplications:

\frac{dw}{dt} = \frac{5}{t}\sin{1} + t\cos{t^2} - \frac{5}{t}\cos{1} - 3t\cos{t^2}

Applying the simplifications:

\frac{dw}{dt} = \frac{5}{t}(\sin{1} - \cos{1}) - 2t\cos{t^2}

Which is the derivative.

For more on the chain rule, you can check brainly.com/question/12795383

8 0
3 years ago
Does (3/4,radica 7/4) lie on the circumference of the unit circle
galina1969 [7]
We have (3/4)^2 + (\sqrt{7}/4)^2 = 9/16 + 7/16 = (9+7)/16 = 16/16 = 1.
The answer is yes.
3 0
4 years ago
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