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AnnZ [28]
3 years ago
7

Can I get some help on this page please

Mathematics
1 answer:
Brums [2.3K]3 years ago
6 0
for first box area is 38.5 and the perimeter is 24
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Alex can plant of his garden in hour. If he continues working at this rate, what fraction of Alex's garden can he plant in one h
Nookie1986 [14]

Answer:

89

Step-by-step explanation:

2 PLUS 3456 PLUS 687=32534

4 0
3 years ago
If g (x) = 1/x^2 then g (x+h) - g (x)/h
nignag [31]

\bf g(x)=\cfrac{1}{x^2}~\hspace{5em}\cfrac{g(x+h)-g(x)}{h}\implies \cfrac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h} \\\\\\ \textit{using the LCD of }(x+h)^2(x^2)\qquad \cfrac{\frac{x^2-(x+h)^2}{(x+h)^2(x^2)}}{h}\implies \cfrac{x^2-(x+h)^2}{h(x+h)^2(x^2)} \\\\\\ \cfrac{x^2-(x^2+2xh+h^2)}{h(x+h)^2(x^2)}\implies \cfrac{\underline{x^2-x^2}-2xh-h^2}{h(x+h)^2(x^2)}\implies \cfrac{-2xh-h^2}{h(x+h)^2(x^2)}


\bf \cfrac{\underline{h}(-2x-h)}{\underline{h}(x+h)^2(x^2)}\implies \cfrac{-2x-h}{(x+h)^2(x^2)}\implies \cfrac{-2x-h}{(x^2+2xh+h^2)(x^2)} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \cfrac{-2x-h}{x^4+2x^3h+x^2h^2}~\hfill

4 0
3 years ago
What numbers can give me 8 when added and 30 when multiplied
Paul [167]
In this problem, it is important to take note that the number of numbers to be utilized isn't specified so it can be up to a thousand numbers. It wasn't also specified if repeating of numbers is allowed or not. So with those taken into consideration and the condition presented in mind, the numbers that can give you 8 when added and 30 when multiplied are 2, 3, 5, -1, and another -1. The derivation from this is mainly from factorization and a little bit of logic.

here is the solution.

2 x 3 x 5 x -1 x -1 = 30
6 x 5 x -1 x -1 = 30
30 x -1 x -1 = 30
-30 x -1 = 30
30 = 30

2 + 3 + 5 + -1 + -1 = 8
5 + 5 + -1 + -1 = 8
10 + -1 + -1 = 8
9 + -1 = 8
8 = 8
7 0
3 years ago
Select the correct answer.
shtirl [24]

Answer:

145 . this is because angles on a straight line add up to 180 degrees

5 0
4 years ago
Please help ASAP brainliest.
lisov135 [29]

Answer:

\left[\begin{array}{cc}2&8\\5&1\end{array}\right]

Step-by-step explanation:

The <em>transpose of a matrix </em>M^T is one where you swap the column and row index for every entry of some original matrix M. Let's go through our first matrix row by row and swap the indices to construct this new matrix. Note that entries with the same index for row and column will stay fixed. Here I'll use the notation p_{i,j} and p^T_{i,j} to refer to the entry in the i-th row and the j-th column of the matrices P and P^T respectively:

p_{1,1}=p^T_{1,1}=2\\p_{1,2}=p^T_{2,1}=5\\p_{2,1}=p^T_{1,2}=8\\p_{2,2}=p^T_{2,2}=1\\

Constructing the matrix P^T from those entries gives us

P^T=\left[\begin{array}{cc}2&8\\5&1\end{array}\right]

which is option a. from the list.

Another interesting quality of the transpose is that we can geometrically represent it as a reflection over the line traced out by all of the entries where the row and column index are equal. In this example, reflecting over the line traced from 2 to 1 gives us our transpose. For another example of this, see the attached image!

7 0
3 years ago
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