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kozerog [31]
3 years ago
10

Is (-3,11) a solution to the system of equation shown below.

Mathematics
1 answer:
natita [175]3 years ago
8 0
No, when you substitute it into the first equation for x and y it does not equal 36 this it's not a solution

(-3)-(3)(11)=36
-3-33=36
-36 =/ 36
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Find the reciprocal
Sav [38]

Answer:

Depends on what number(s)/fraction(s) your working with.

Step-by-step explanation:

Find the reciprocal of a fraction by flipping it. The definition of "reciprocal" is simple. To find the reciprocal of any number, just calculate "1 ÷ (that number)." For a fraction, the reciprocal is just a different fraction, with the numbers "flipped" upside down (inverted).

3 0
3 years ago
2x+29 find the value of x​
NikAS [45]

Answer:

The answer would be x=13.5

Step-by-step explanation: 13.5 times 2 equals 27. 27 plus 2 equals 29.

7 0
3 years ago
I don't know what the answer is.
Ksenya-84 [330]

Answer:

C.) 87

Step-by-step explanation:

if you look at it from the other side, its counting down from 91 from the left.

6 0
3 years ago
Read 2 more answers
A dog is leashed to the corner of a house with a 20 ft long leash. How much running area does the dog have? Round your answer to
11111nata11111 [884]

Answer:

The dog's running area is approximately 942 feet².

Step-by-step explanation:

The dog is leashed to a fixed point, the leash has a length of 20 ft, therefore he can rotate around that point at the maximum distance equal to the length of the leash. This pattern forms a circle, but there is an obstruction, which is the corner of the house. This obstruction takes an arc of the original circle, so the running area of the dog is the area of the whole circle minus the area of the arc formed by the corner of the house.

\text{dog's area} = \text{circle's area} - \text{arc's area}\\\\\text{dog's area} = pi*(20^2) - \frac{90}{360}*pi*(20^2)\\\\\text{dog's area} = \frac{270}{360}*pi*(400)\\\\\text{dog's area} = 942.477\text{ feet}^2

The dog's running area is approximately 942 feet².

3 0
3 years ago
The volume
Sedaia [141]
\bf \begin{array}{cccccclllll}
\textit{something}&&\textit{varies directly to}&&\textit{something else}\\ \quad \\
\textit{something}&=&{{ \textit{some value}}}&\cdot &\textit{something else}\\ \quad \\
y&=&{{ k}}&\cdot&x
&&  y={{ k }}x
\end{array}\\ \quad \\


and also

\bf \begin{array}{llllll}
\textit{something}&&\textit{varies inversely to}&\textit{something else}\\ \quad \\
\textit{something}&=&\cfrac{{{\textit{some value}}}}{}&\cfrac{}{\textit{something else}}\\ \quad \\
y&=&\cfrac{{{\textit{k}}}}{}&\cfrac{}{x}
&&y=\cfrac{{{  k}}}{x}
\end{array}


now, we know that V varies directly to T and inversely to P simultaneously
thus\bf V=T\cdot \cfrac{k}{P}

so     \bf V=T\cdot \cfrac{k}{P}\qquad 
\begin{cases}
V=42\\
T=84\\
P=8
\end{cases}\implies 42=\cfrac{84k}{8}\implies 4=k
\\\\\\
V=\cfrac{4T}{P}\qquad now\quad 
\begin{cases}
V=74\\
P=10
\end{cases}\implies 74=\cfrac{4T}{10}\implies 185=T
7 0
3 years ago
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