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3 years ago

Solve exponential equation with different bases: 17^j=16

1 answer:

8 0

Answer:{−17}. In Example 1, we were able to write the left and right sides of the equation with a common base. ... To solve such an equation, we can take a logarithm of the same base on ... Steps to Solve Exponential Equations by Using Logarithms ... of x is a solution to the logarithmic equation. 35. log2(x − 31) = 5 − log2 x a. x = 16.

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A rectangular school banner has a length of 36 inches and a width of 52 inches. A sign is made that is similar to the school ban

zalisa [80]

Answer:

4212 : 1573 is the ratio.

Step-by-step explanation:

Given:

Length of rectangular school banner is 36 inches and width is 52 inches.

And, the sign made is similar to banner.

The sign's length is 22 inches.

Now, we have to find the ratio of the area of the banner to the area of the sign.

So, we have dimensions of rectangular school banner:

Length = 36 inches.

Width = 52 inches.

But, we have only length of sign:

Length = 22 inches.

Now, we have to find the width of sign by using cross multiplication method:

Let the width of sign be $x.$

As, sign is similar to banner.

So, if 36 inches is equivalent to 52 inches.

Then, 22 inches is equivalent to $x.$

$\frac{36}{52} =\frac{22}{x}$

By cross multiplying we get:

$36x=1144$

Dividing both sides by 36 we get:

$x=31\frac{7}{9} .$

Hence, the width of sign is $31\frac{7}{9}$ inches.

Now, we find the area of banner and the area of sign by putting formula:

Area of banner = length × width.

$Area\ of\ banner=36\times 52$

$Area\ of\ banner=1872\ square\ inches.$

Now, area of sign:

$Area\ of\ sign=length\times width\\\\Area\ of\ sign=22\times 31\frac{7}{9} \\\\Area\ of\ sign=22\times \frac{286}{9} \\\\Area\ of\ sign=\frac{6292}{9} \ square\ inches.$

Now, to get the ratio of the area of the school banner to the area of the sign:

$1872:\frac{6292}{9}$

$=\frac{1872}{\frac{6292}{9} } \\\\=\frac{16848}{6292}$

On simplifying we get:

$=\frac{4212}{1573}$

$=4212:1573.$

Therefore, the ratio of the area of the school banner to the area of the sign is 4212:1573.

8 0

3 years ago

Emmanuel carves a cone-shaped hollow out of a solid clay cylinder. What is the approximate volume of the solid portion of the cy

bixtya [17]

Answer:

$3.14r^2(h-\frac{1}{3}h_1)$

Step-by-step explanation:

Let h be the cylinders height and r the radius.

-The volume of a cylinder is calculated as:

$V=\pi r^2h$

-Since the cone is within the cylinder, it has the same radius as the cylinder.

-Let $h_1$ be the height of the cone.

-The area of a cone is calculated as;

$V=\pi r^2 \frac{h}{3}\\\\=\frac{1}{3}\pi r^2h_1$

The volume of the solid section of the cylinder is calculated by subtracting the cone's volume from the cylinders:

$V=V_{cy}-V_{co}\\\\=\pi r^2h-\frac{1}{3}\pi r^2 h_1, \pi=3.14\\\\=3.14r^2(h-\frac{1}{3}h_1)$

Hence, the approximate area of the solid portion is $3.14r^2(h-\frac{1}{3}h_1)$

5 0

3 years ago

A comic-strip writer churns out a different number of comic strips each day. For 16 days, the writer logged the number of comic

Arte-miy333 [17]

The distribution will be skewed to the positive side because the mean is greater than the median.
Mean: 3.7
Median: 3

Answer: B

6 0

3 years ago

A) y=2(4)^x growth or decay y-intercept:

tatiyna

It's growth and the y-intercept is (0,2). ;)

5 0

3 years ago

If I have 24 jolly ranchers and 64 starbursts what is the largest amount of bags I can fill? How many starbursts will be I’m eac

Rus_ich [418]

Answer:

Largest amount of bags that can be filled: 11

Starbursts in each bag: 3

Step-by-step explanation:

Find the greastest common factor between these two (which is 8) and divide the sum of the jolly ranchers and starbursts-which will be 88-by the greatest common factor. This will leave you with 11 bags that can be filled with 8 candies. To find the starbursts, divide 24 by 8 (3.)

5 0

3 years ago