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Brilliant_brown [7]
3 years ago
7

Simplify. show work please 5 1/3+(-3 9/18)

Mathematics
1 answer:
Nikolay [14]3 years ago
8 0
51/3+-39/18

= 89/6

Answer is 14.833333
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Perimeter of triangle:20
goblinko [34]

Answer:

perimeter of rectangle= 2(4+X)

32=8+2x

32-8=2x

26=2x

26/2=X

14=X

perimeter of triangle=3×sides

20=3x

20/3=X

6.66=x

4 0
3 years ago
Jim had some dimes, quarters, and nickels. First, he counted only the dimes and quarters and found that he had 9 coins. Then, he
Olegator [25]

Answer:

$1.90

Step-by-step explanation:

Represent the numbers of nickels, dimes and quarters by n, d and q.

Then: d + q = 9, or q = 9 - d.

Also, n + d = 10

Lastly, n + d + q = 15.

Let's substitute 9 - d for q in the equation immediately above:

n + d + (9 - d) = 15, or n + 9 = 15.  Then n must be 6.

In summary, Jim has 6 nickels, (10 - 6) dimes and (9 - 4) quarters, or:

                                    6 nickels, 4 dimes and 5 quarters

Thus, he has 5($0.05) + 4($0.10) + 5($0.25), or

                         $0.25   +   $0.40   + $1.25, or          $1.90

7 0
3 years ago
Liz train for rock climbing 4 days a week
Goshia [24]
7 times 4 =28 this is cuz there is seven days in a week and 7x4
7 0
2 years ago
Solving Exponential and Logarithmic Equations In Exercise, solve for x.<br> In 2x - In(3x - 1) = 0
LenKa [72]

Answer:

\frac{-1}{3x^2-x}

Step-by-step explanation:

  1. If f(x) is in th form of f(x)=g(x)-h(x) then f'(x)=g'(x) - h'(x)
  2. When f(x)=z(g(x)) then f'(x)= z'(g(x))g'(x) (called as chain rule)

<u>using these information</u>:

g(x)=ln2x then g'(x)=\frac{(2x)'}{2x} =\frac{2}{2x}=\frac{1}{x}

h(x)=In(3x - 1) then h'(x)=\frac{(3x-1)'}{3x-1} =\frac{3}{3x-1}f'(x)=g'(x) - h'(x) =[tex]\frac{1}{x} - \frac{3}{3x-1} =\frac{-1}{3x^2-x}

7 0
4 years ago
What is X if
umka2103 [35]

Answer:

X = \begin{bmatrix} - 4 & 1 \\ - 2 & - 7\end{bmatrix}

Step-by-step explanation:

\begin{bmatrix} - 1 & - 2\\ 4 & 8\end{bmatrix} +X = \begin{bmatrix} - 5 & - 1\\ 2 & 1\end{bmatrix}

X = \begin{bmatrix} - 5 & - 1\\ 2 & 1\end{bmatrix}-\begin{bmatrix} - 1 & - 2\\ 4 & 8\end{bmatrix}

X = \begin{bmatrix} - 5-(-1) & - 1-(-2)\\ 2-4 & 1-8\end{bmatrix}

X = \begin{bmatrix} - 5+1 & - 1+2\\ 2-4 & 1-8\end{bmatrix}

X = \begin{bmatrix} - 4 & 1 \\ - 2 & - 7\end{bmatrix}

7 0
3 years ago
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