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Fiesta28 [93]
2 years ago
15

Write the equation in standard form. 8−5y=−2x

Mathematics
2 answers:
Elena-2011 [213]2 years ago
7 0

Answer:

-5y+2x=-8

Step-by-step explanation:

8-5y=-2x

move the 8 to the right..


-5y=-2x-8


move the -2x to the left..

-5y+2x=-8!


Hope if helps :D

GarryVolchara [31]2 years ago
6 0
8-5y= -2x
⇒ -5y= -2x-8
⇒ y= -2/(-5)x -8/(-5)
⇒ y= 2/5x+ 8/5

The final answer is y= 2/5x+ 8/5~
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Simplify 15(10-4-2)<br> 14<br> 015/2<br> 52-4110<br> O 105-4110
inna [77]

\qquad\qquad\huge\underline{\boxed{\sf Answer☂}}

let's simplify ~

\qquad \sf  \dashrightarrow \:  \sqrt{5} (10 - 4 \sqrt{2} )

\qquad \sf  \dashrightarrow \: ( \sqrt{5}  \times 10) - ( \sqrt{5}  \times  4\sqrt{2)}

\qquad \sf  \dashrightarrow \: 10 \sqrt{5}  -  4\sqrt{10}

3 0
1 year ago
Easy question.<br><br> Solve the inequality -5x&lt;40
laiz [17]

\huge\text{Hey there!}

\mathsf{-5x < 40}

\large\text{DIVIDE -5 to BOTH SIDES}

\mathsf{\dfrac{-5x}{-5}

\rm{CANCEL\  out: \dfrac{-5}{-5} \ because\  that \ gives\  you\  1}

\rm{KEEP: \dfrac{40}{-5}\ because \ that \ gives \  you \ what \ you \ are\ comparing}\\\\\rm{to \ the \ x-value}

\mathsf{\dfrac{40}{-5}=\boxed{\bf -8}}

\mathsf{x > -8}

\boxed{\boxed{\large\textsf{Answer: \large \bf x }\bf  > \large\textsf{\bf -8}}}\huge\checkmark

\boxed{\large\textsf{IT is an O(P)(E)(N)(E)(D)   circle shaded to the right}}

\large\text{Good luck on your assignment and enjoy your day!}

~\frak{Amphitrite1040:)}

3 0
3 years ago
PLEASE HELP ILL MARK YOU THE BRAINLIEST ASAP!!
Ratling [72]

Answer:

second 1 is right

Step-by-step explanation:

it is righttttt

4 0
2 years ago
Read 2 more answers
Solve inequality in interval notation Y^2&gt;-5y-9
nata0808 [166]
It’s going to be negative infinitely, infinitely.
4 0
2 years ago
PLS HELP ME WITH THIS!!!!!! HOW DID THEY GET 80ft^2
noname [10]

\huge \boxed{\mathfrak{Question} \downarrow}

  • Determine the surface area of the right square pyramid.

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

The formula for finding the surface area of a right square pyramid is ⇨ b² + 2bl, where

  • b = base of the right square pyramid
  • l = slant height of the right square pyramid.

In the given figure,

  • base (b) = 4 ft.
  • slant height (l) = 8 ft.

Now, let's substitute the values of b & l in the formula & solve it :-

\sf \: {b}^{2}  + 2bl \\  =   \sf \: {4}^{2}  + 2 \times 4 \times 8 \\  =  \sf \: 16 + 8 \times 8 \\  =  \sf \: 16 + 64 \\  =  \huge\boxed{\boxed{ \bf 80 \: ft ^{2} }}

So, the surface area of the right square pyramid is <u>8</u><u>0</u><u> </u><u>ft²</u><u>.</u>

7 0
2 years ago
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