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andreyandreev [35.5K]
2 years ago
8

B5%29%7D%7B3%7D%3D1%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bcross-multiplying%7D%7D%7B2%28x-4%29%28x%2B5%29%3D3%7D%20%5C%5C%5C%5C%5C%5C%202%28%5Cstackrel%7B%5Cmathbb%7Bfoil%7D%7D%7Bx%5E2%2Bx-20%7D%29%3D3%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bdistributing%7D%7D%7B2x%5E2%2B2x-40%7D%3D3%5Cimplies%20%5Cunderset%7B%5Ctextit%7Bstandard%20form%7D%7D%7B%5Cstackrel%7B%5Cstackrel%7Ba%7D%7B%5Cdownarrow%20%7D%7D%7B2%7Dx%5E2%5Cstackrel%7B%5Cstackrel%7Bb%7D%7B%5Cdownarrow%20%7D%7D%7B%2B2%7Dx%5Cstackrel%7B%5Cstackrel%7Bc%7D%7B%5Cdownarrow%20%7D%7D%7B-43%7D%3D0%7D" id="TexFormula1" title="\bf \cfrac{2}{3}(x-4)(x+5)=1\implies \cfrac{2(x-4)(x+5)}{3}=1\implies \stackrel{\textit{cross-multiplying}}{2(x-4)(x+5)=3} \\\\\\ 2(\stackrel{\mathbb{foil}}{x^2+x-20})=3\implies \stackrel{\textit{distributing}}{2x^2+2x-40}=3\implies \underset{\textit{standard form}}{\stackrel{\stackrel{a}{\downarrow }}{2}x^2\stackrel{\stackrel{b}{\downarrow }}{+2}x\stackrel{\stackrel{c}{\downarrow }}{-43}=0}" alt="\bf \cfrac{2}{3}(x-4)(x+5)=1\implies \cfrac{2(x-4)(x+5)}{3}=1\implies \stackrel{\textit{cross-multiplying}}{2(x-4)(x+5)=3} \\\\\\ 2(\stackrel{\mathbb{foil}}{x^2+x-20})=3\implies \stackrel{\textit{distributing}}{2x^2+2x-40}=3\implies \underset{\textit{standard form}}{\stackrel{\stackrel{a}{\downarrow }}{2}x^2\stackrel{\stackrel{b}{\downarrow }}{+2}x\stackrel{\stackrel{c}{\downarrow }}{-43}=0}" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
adell [148]2 years ago
6 0

Answer:

x =  - 6 3/10

We need to show at least two of these in workings including one in simplified format.

x = - 63/10   (minus in front of whole fraction)  

mixed number form  x = -6 3/10   (is simplified answer)

decimal form  x  =  -6.3  (is simplified answer)

Step-by-step explanation:

Just in case you ever need to write these in ans search answers already answered in same format.

You can;

Go to upper right corner and click the three dots : another menu opens.

Click the summation sign (4th from left) : you are now in math mode until you deselect it (which sometimes happens automatically). Type a^{2} and it will look like a2 .

Note : the brackets { and } aren’t necessary when your exponent is a natural number between 0 and 9, both included.

However, when you want to type e.g. a15 , you need to place them, because otherwise you will get a15 .

it will look in same way as we are taught with a pen, its the stupid copy paste thing hit and miss if it works on messages here, Obviously it didnt work when i copied just now.

Adding a Range of Cells Using the Summation Button

The Summation button will copy a formula that calculates the sum of the selected numeric cells to the clipboard.

For alphanumeric cells, clicking the Summation button will concatenate the text of the selected cells.

Select the cells by dragging the mouse over the selected area or dragging while holding Alt.

On the Document tab, click Summation.

Position the insertion point in the total cell.

On the Home tab, click Paste.

The Equation bar displays the addition of all cells in the selection.

Note: Only Numeric cells can be summed. Alphanumeric cells will have their text concatenated.

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F(x)=x^2-7 and h(x)=7x-5, then f(h(2))= ?
ziro4ka [17]
<h3>Answer:    74</h3>

==================================================

Work Shown:

h(x) = 7x-5

h(2) = 7(2)-5

h(2) = 14-5

h(2) = 9

f(h(2)) = f(9)

f(x) = x^2-7

f(9) = 9^2-7

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f(9) = 74

f(h(2)) = 74

-----------

Here's a slightly alternative approach:

f(x) = x^2 - 7

f(h(x)) = ( h(x) )^2 - 7

f(h(x)) = ( 7x-5 )^2 - 7

f(h(2)) = ( 7*2-5 )^2 - 7

f(h(2)) = ( 14-5 )^2 - 7

f(h(2)) = ( 9 )^2 - 7

f(h(2)) = 81 - 7

f(h(2)) = 74

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2 years ago
&lt;1 and &lt;2 are supplementary, so one of the angles is acute,
Ad libitum [116K]

Answer:

make each of the supplementary angles 90, then they are equal.                                                                                                                                                                                                                                                                                                                  

Step-by-step explanation:

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3 years ago
To enlarge or reduce a shape you must:
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3 years ago
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