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Stels [109]
3 years ago
5

Find the product. (do not use spaces in your answer.) d ^j · d ^k

Mathematics
2 answers:
olganol [36]3 years ago
6 0

Answer: The product would be d^{j+k}

Step-by-step explanation:

Since we have given that

d^j.d^k

We need to find the product.

As we know that

a^m\times a^n=a^{m+n}

So, we will use the above exponential law and it becomes,

d^j.d^k=d^{j+k}

Hence, the product would be d^{j+k}

bezimeni [28]3 years ago
3 0
d^j\cdot d^k=d^{j+k}
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B) The 4 lb peanuts and the 1 lb mixture because the 4lb added to the 1lb of mixture give the correct percentages.

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How much does the angels add up to in a rectangle
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One of the requirements of a rectangle is that it must have four right angles ( a right angle is 90 degrees) so you just multiply 90 by 4 and you get 360 degrees
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Does the point (0,0) satisfy the equation y = 9x?<br> yes<br> no
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Yes, because 0 = 9 x 0.
7 0
3 years ago
If a snowball melts so that its surface area decreases at a rate of 9 cm2/min, find the rate (in cm/min) at which the diameter d
zhuklara [117]

Answer:

The diameter decreases at a rate of 0.143cm/min when it is of 10 cm.

Step-by-step explanation:

Surface area of an snowball:

An snowball has spherical format. The surface area of an sphere is given by:

S = d^2\pi

In which d is the diameter of the sphere.

In this question:

We need to differentiate S implicitly in function of time. So

\frac{dS}{dt} = 2d\pi\frac{dd}{dt}

Surface area decreases at a rate of 9 cm2/min

This means that \frac{dS}{dt} = -9

At which the diameter decreases when the diameter is 10 cm?

This is \frac{dd}{dt} when d = 10. So

\frac{dS}{dt} = 2d\pi\frac{dd}{dt}

-9 = 2(10)\pi\frac{dd}{dt}

\frac{dd}{dt} = -\frac{9}{20\pi}

\frac{dd}{dt} = -0.143

Area in cm², so diameter in cm.

The diameter decreases at a rate of 0.143cm/min when it is of 10 cm.

8 0
3 years ago
A rectangular piece of land borders a wall. The land is to be enclosed and to be into divided 3 equal plots with 200 feet of fen
baherus [9]

Answer:

Area = 2500 square feet is the largest area enclosed

Step-by-step explanation:

A rectangular piece of land borders a wall. The land is to be enclosed and to be into divided 3 equal plots with 200 feet of fencing

Let x be the length of each box and y be the width of the box

Perimeter of the box= 3(length ) + 4(width)

200=3x+4y

solve for y

200=3x+4y

200-3x=4y

divide both sides by 4

y=50-\frac{3x}{4}

Area of the rectangle = length times width

Area = 3x \cdot y

Area = 3x \cdot (50-\frac{3x}{4})

A=150x-\frac{9x^2}{4}

Now take derivative

A'=150-\frac{9x}{2}

Set it =0 and solve for x

0=150-\frac{9x}{2}

150=\frac{9x}{2}

multiply both sides by 2/9

x=\frac{100}{3}

A''=-\frac{9}{2}

For any value of x, second derivative is negative

So maximum at x= 100/3

 A=150x-\frac{9x^2}{4} , replace the value of x

A=150(\frac{100}{3})-\frac{9(\frac{100}{3})^2}{4})

Area = 2500 square feet is the largest area enclosed

3 0
3 years ago
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