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wlad13 [49]
3 years ago
14

You want to frame a picture that is 5 in by 7 in with a 1 in wide frame. What is the perimeter of the outside edge of the frame?

Mathematics
1 answer:
r-ruslan [8.4K]3 years ago
4 0
94 if your using the whole frame the math is 5×7=35 we have 2 sides do times 35 by 2 35×2=70 then we do 1×7=7 for the outside of the game then do 7 times 2 b/c you have 2 sides the do your final side by doing 5×1=5 then times that by 2 and you get 10 .70+14+10=94
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Solve the equation for x.
ivann1987 [24]

Answer:

x = 44

Step-by-step explanation:

sqrt(x+5) -3 =4

Add 3 to each side

sqrt(x+5) -3+3 =4+3

sqrt(x+5)  =7

Square each side

(sqrt(x+5))^2  =7^2

x+5 = 49

Subtract 5 from each side

x+5-5 = 49-5

x=44

3 0
3 years ago
Read 2 more answers
P(x)= 3x^3-5x^2-14x-4
nexus9112 [7]
   
\displaystyle\\
P(x)=3x^3-5x^2-14x-4\\\\
D_{-4}=\{-4;~-2;~\underline{\bf -1};~1;~2;~4\}\\\\
\text{We observe that } \frac{-1}{3} \text{ is a solution of the equation:}\\
3x^3-5x^2-14x-4=0\\\\


\displaystyle\\
\text{Verification}\\\\
3x^3-5x^2-14x-4=\\\\
=3\times\Big(-\frac{1}{3}\Big)^3-5\times\Big(-\frac{1}{3}\Big)^2-14\times\Big(-\frac{1}{3}\Big)-4=\\\\
=-\frac{1}{9}-\frac{5}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{42}{9}- \frac{4\times 9}{9}=\\\\
 =-\frac{6}{9}+\frac{42}{9}- \frac{36}{9}= \frac{42-6-36}{9}=\frac{42-42}{9}=\frac{0}{9}=0\\\\
\Longrightarrow~~~P(x)~\vdots~\Big(x+ \frac{1}{3}\Big)\\\\
\Longrightarrow~~~P(x)~\vdots~(3x+1)


\displaystyle\\
3x^3-5x^2-14x-4=0\\
~~~~~-5x^2 = x^2 - 6x^2\\
~~~~~-14x =-2x-12x \\
3x^3+x^2 - 6x^2-2x-12x-4=0\\
x^2(3x+1)-2x(3x+1) -4(3x+1)=0\\
(3x+1)(x^2-2x -4)=0\\\\
\text{Solve: } x^2-2x -4=0\\\\
x_{12}= \frac{-b\pm  \sqrt{b^2-4ac}}{2a}=\\\\=\frac{2\pm  \sqrt{4+16}}{2}=\frac{2\pm  \sqrt{20}}{2}=\frac{2\pm  2\sqrt{5}}{2}=1\pm\sqrt{5}\\\\
x_1 =1+\sqrt{5}\\
x_2 =1-\sqrt{5}\\
\Longrightarrow P(x)= 3x^3-5x^2-14x-4 =\boxed{(3x+1)(x-1-\sqrt{5})(x-1+\sqrt{5})}



7 0
2 years ago
Please help me 5-7g-3=
Archy [21]

Answer: −7g+2

If I Helped, Please Mark Me As Brainliest, Have A Great Day :D

5 0
3 years ago
Read 2 more answers
Alternate exterior angles ​
exis [7]

Answer:

When  \: two  \: lines  \: are \:  crossed \:  by  \: another \:  line \:  (called  \: the \:  Transversal \: ):  \: Alternate \:  Interior  \: Angles \:  are  \: a \:  pair  \: of  \: angles  \: on \:  the \:  inner  \: side  \: of  \: each  \: of \:  those  \\  \\  \\  \\  \\ two \:  lines  \: but  \: on \:  opposite  \: sides  \: of  \: the  \: transversal.

hope it helps if it helps don'tforget to like and mark

8 0
3 years ago
I dont get it plzzz help me
cricket20 [7]

Given Equation: \frac{1}{3}h-4(\frac{2}{3}h-3)=\frac{2}{3}h-6

Let us remove parenthesis first.

In order to remove parenthesis, we need to distribute -4 over parenthesis (2/3 h -3).

Distributing -4 over (2/3 h -3), we get,

-4*2/3 h - 4*-3 = -8/3 h + 12.

Substituting this value in original equation, we get

\frac{1}{3}h-\frac{8}{3}h+12 =\frac{2}{3}h-6

Now, in order to make the equation easy to solve, we always remove fractions. In order to remove fraction, we need to find the lowest common denominator of all terms(lcd).

The fraction terms has 3 in denominators, so we could multiply each and every term by 3 to remove 3's from denominators.

Multiplying each equation by 3, we get,

3*\frac{1}{3}h-3*\frac{8}{3}h+3*12 =3*\frac{2}{3}h-3*6

On simplfying this step, we get

1h -8h+36=2h-18

Combining like terms on left side, we get

-7h +36 = 2h -18.

Subtracting 36 from both sides, we get

-7h +36 -36= 2h -18-36.

-7h=2h-54

Subtracting 2h from both sides, we get

-7h-2h=2h-54-2h

-9h = -54

Dividing both sides by -9.

-9h/-9 = -54/-9

h = 6.

Therefore, final answer is h=6.


5 0
3 years ago
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