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luda_lava [24]
3 years ago
15

Which statement is true about angles 1 and 2? ​

Mathematics
2 answers:
lapo4ka [179]3 years ago
7 0

Answer:

Angle 1 and 2 are adjacent .

Karo-lina-s [1.5K]3 years ago
5 0

Answer:

Angles 1 & 2 are adjacent

Step-by-step explanation:

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John is 5 years younger than three times his brother’s age. If the sum of their
torisob [31]

Answer:

I think 10

Step-by-step explanation:

8 0
1 year ago
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∠C and ∠D are complementary.
Brut [27]

Answer:

x = 8

Step-by-step explanation:

The sum of the measures of two complementary angles is 90°.

m<C + m<D = 90°

5x + 7x - 6 = 90

12x = 96

x = 8

5 0
3 years ago
Differentiating Functions of Other Bases In Exercise, find the derivative of the function.
aniked [119]

Answer:

\dfrac{dy}{dx} = \frac{2x-3}{\ln 16(x^2 - 3x)}

Step-by-step explanation:

We are given the following in the question:

y = \ln {16}(x^2 - 3x)

We have to find the derivative of the given expression.

y = \ln 16(x^2 - 3x)\\\text{Using the log propert}\\\\\log_a b = \dfrac{\log b}{\log a}\\\\dfrac{d(x^n)}{dx} = nx^{n-1}\\\\\dfrac{d(\log x)}{dx} = \dfrac{1}{x}\\\\\text{\bold{Differentiating we get}}\\\\\displaystyle\frac{dy}{dx} = \frac{d(\ln 16(x^2-3x))}{dx}\\\\= \frac{1}{16(x^2-3x)})\frac{d(x^2-3x)}{dx}\\\\=\frac{1}{\log 16(x^2-3x)}(2x - 3)\\\\= \frac{2x-3}{\ln 16(x^2 - 3x)}

\dfrac{dy}{dx} = \frac{2x-3}{\ln 16(x^2 - 3x)}

7 0
4 years ago
Which expression is equivalent to 3m+ 1 - m<br> A: 2 + m- 1 + m<br> B: 1 + m<br> C: 3m- 1<br> D: 3m
Vanyuwa [196]
The answer is D .............
7 0
3 years ago
Read 2 more answers
A rectangular page is to contain 6 square inches of print. The margins at the top and bottom of the page are to be 2 inches wide
mixer [17]

Answer:

A(t)  =  27.86 in²

Dimensions of the paper:

L =  3.73 in

w = 7.47 in

Step-by-step explanation:

The total area of a rectangular page  A  = ( x + 2)* (y + 4 )         x  is the length and  y the wide. x  and  y  are the dimensions of the print area

The print area of the paper is  A =  6 in²       6 = x*y     y  =  6/x

Then print area as a function of x is

A(x)  =  ( x + 2 ) * ( 6/x + 4 )    ⇒  A(x)  =  6  +  4*x  + 12/x + 8

Taking derivatives on both sides of the equation:

A´(x)  =  4  - 12/x²        A´(x)  =  0     4  =  12 /x²

x²  = 12/4       x =√3    x  =  1.73  in  and   y  =  6 / 1.73

y  =  3.47 in

Then the dimensions of the paper  are.

Length  L =  x  + 2  = 3.73 in      and   w  =  3.47 + 4  = 7.47 in

And the least amount of paper is

A(t)  =  3.73* 7.47  =  27.86 in²

To find out if x = 1.73 is an x coordinate for a minimum we get the second derivative

A´´(x)  =  24/x³    is always positive   A´´(x) > 0  then we have a minimum for A at  x = 1.73

8 0
3 years ago
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