Which table represents the graph of a logarithmic function in the form y-log, when :

A treehouse has an inside room and a porch, as shown below. Which equation

Nonamiya [84]

Answer:

B

Step-by-step explanation:

It tells you

6 0

2 years ago

To find the product of 42.12 and 10³, move the decimal point in 42.12 __ places to the right because 10³ has __ zeros.

Yakvenalex [24]

<h3>To find the product of 42.12 and 10^3, move the decimal point in 42.12 3 places to the right because 10^3 has 3 zeros</h3>

<em><u>Solution:</u></em>

Given that,

$\text{ product of } 42.12 \text{ and } 10^3$

Which means,

$42.12 \times 10^3$

Here, the exponent of 10 is positive ( which is 3)

When the exponent is positive, we have to move the decimal point to right

When you multiply a number by a power of 10, ( 10!, 10^2, and so on ) move the decimal point of the number to the right the same number of places as the number of zeros in the power of 10

Here, exponent is 3 , therefore move the decimal point right 3 places in 42.12

Therefore,

$42.12 \times 10^3 = 42120$

7 0

3 years ago

PLEASE HELP<br><br> Find the surface area of the triangular prism.

ASHA 777 [7]

Answer:

139

Step-by-step explanation:

First, find all the faces that you must find the areas of. In this case, you need to find the areas of two triangles and three rectangles.

In this triangular prism, the base is one triangle. If the area of the base is 7.75, we can assume that that holds true for both bases, and multiply 7.75 by 2 in order to find the area of both triangles.

Now find the area of each of the three rectangles by multiplying their individual heights and bases by each other. You should get 38, 47.5, and 38 as the areas of the rectangles.

Now add all the individual area's together. (They're bolded for clarity).

4 0

3 years ago

Determine the zeroes of the polynomial

Anna71 [15]

Answer:

3,7/6

Step-by-step explanation:

$(\sqrt{x^2-4x+3} )+(\sqrt{x^{2} -9} )-(\sqrt{4x^2-14x+6} )\\=(\sqrt{x^2-x-3x+3} )+(\sqrt{(x^2-3^2})-(\sqrt{4x^2-2x-12x+6})\\ =(\sqrt{x(x-1)-3(x-1)} )+\sqrt{(x+3)(x-3)}-\sqrt{2x(2x-1)-6(2x-1)} \\=\sqrt{(x-1)(x-3)}+\sqrt{(x+3)(x-3)} -\sqrt{2(2x-1)(x-3)} \\=\sqrt{x-3} (\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} )\\$

$\sqrt{x-3} =0~gives~x=3\\or~\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} =0\\or~ \sqrt{x-1} +\sqrt{x+3} =\sqrt{2(2x-1)} \\squaring\\x-1+x+3+2\sqrt{x-1} \sqrt{x+3} =2(2x-1)\\2x+2+2\sqrt{(x-1)(x+3)} =4x-2\\2\sqrt{x^2-x+3x-3} =2x-4\\\sqrt{x^2+2x-3} =x-2\\again ~squaring\\x^2+2x-3=x^2-4x+4\\\\2x+4x=4+3\\6x=7\\x=\frac{7}{6}$