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2 years ago

Is (–3, –2) a solution to the equation y = x + –1?

Mathematics

1 answer:

natka813 [3]2 years ago

6 0

Answer:

No.

Step-by-step explanation:

We plug in the coordinates and that gives us -2=-3+-1, that is not true, therefore your answer is no.

Brainliest would be appreciated! :-)

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What times 0.2 minus 7 equals 3

Murrr4er [49]

X*0.2-7=3
+7 +7
x*0.2=10
divide by 0.2 on both sides
x=50

6 0

3 years ago

Read 2 more answers

Find the value of y in the equation attached to this question

dlinn [17]

Answer:

y = 4

Step-by-step explanation:

1) Simplify $3^5$ to 243.

${9}^{y}=\frac{243\times {9}^{6}}{{27}^{3}}$

2) Simplify ${9}^{6}$ to 531441.

${9}^{y}=\frac{243\times 531441}{{27}^{3}}$

3) Simplify $243\times 531441$ to 129140163.

${9}^{y}=\frac{129140163}{{27}^{3}}$

4) Simplify ${27}^{3}$ to 19683.

${9}^{y}=\frac{129140163}{19683}$

5) Simplify $\frac{129140163}{19683}$ to 6561.

${9}^{y}=6561$

6) Convert both sides to the same base.

$9^y=9^4$

7) Cancel the base of 9 on both sides.

$y=4$

Cheers,

ROR

3 0

2 years ago

Read 2 more answers

Write an equation for the translation of x^2+y^2=49 by 3 units left and 4 units up

ludmilkaskok [199]

$\bf \textit{equation of a circle}\\\\ 
(x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2
\qquad 
\begin{array}{lllll}
center\ (&{{ h}},&{{ k}})\qquad 
radius=&{{ r}}
\end{array}\\\\
-----------------------------\\\\
x^2+y^2=49\iff (x-0)^2+(y-0)^2=7^2\quad 
\begin{cases}
h=0\\
k=0
\end{cases}
\\\\\\
(x-\underline{(-3)})^2+(y-\underline{(+4)})^2=7^2\quad 
\begin{cases}
h=-3\\
k=4
\end{cases}
\\\\\\
\boxed{(x+3)^2+(y-4)^2=49}$