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3 years ago

Y= negative 2/3 graph it

Mathematics

1 answer:

Sindrei [870]3 years ago

5 0

Answer:I searched graph -2/3 and it came up with it on google. The answer is on there if you search that up

Step-by-step explanation:

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How many flowers, spaced every 4 inches, are needed to surround a circular garden with a 200 inch radius?

alukav5142 [94]

Answer:

Step-by-step explanation:

7 0

3 years ago

Equation of the line (-6,5) and (8,14) in standard form

blsea [12.9K]

Answer:

$\huge\boxed{9x-14y=-70}$

Step-by-step explanation:

The point-slope form of an equation of a line:

$y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}$

We have two points (-6, 5) and (8, 14).

Substitute:

$m=\dfrac{14-5}{8-(-6)}=\dfrac{9}{14}\\\\y-5=\dfrac{9}{14}(x-(-6))\\\\y-5=\dfrac{9}{14}(x+6)$

Thew standard form of an equation of a line:

$Ax+By=C$

transform:

$y-5=\dfrac{9}{14}(x+6)\qquad|\text{multiply both sides by 14}\\\\14y-14\cdot5=14\!\!\!\!\!\diagup^1\cdot\dfrac{9}{14\!\!\!\!\!\diagup_1}(x+6)\\\\14y-70=9(x+6)\\\\14y-70=9x+54\qquad|\text{subtract}\ 9x\ \text{from both sides}\\\\-9x+14y-70=54\qquad|\text{add 70 to both sides}\\\\-9x+14y=70\qquad|\text{change the signs}\\\\9x-14y=-70$

4 0

3 years ago

PLEASE ANSWER THIS ASAP I WILL MARK YOU THE BRAINLIEST

emmasim [6.3K]

Answer = 314.2

Area of cyclinder:
Area at base x height

Pi = 3.142

5^2 x 3.142 = 78.55

78.55 x 4 = 314.2

5 0

3 years ago

Find the? inverse, if it? exists, for the given matrix.<br><br> [4 3]<br><br> [3 6]

True [87]

Answer:

Therefore, the inverse of given matrix is

$=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}$

Step-by-step explanation:

The inverse of a square matrix $A$ is $A^{-1}$ such that

$A A^{-1}=I$ where I is the identity matrix.

Consider, $A = \left[\begin{array}{ccc}4&3\\3&6\end{array}\right]$

$\mathrm{Matrix\:can\:only\:be\:inverted\:if\:it\:is\:non-singular,\:that\:is:}$

$\det \begin{pmatrix}4&3 \\3&6\end{pmatrix}\ne 0$

$\mathrm{Find\:2x2\:matrix\:inverse\:according\:to\:the\:formula}:\quad \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}^{-1}=\frac{1}{\det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}}\begin{pmatrix}d\:&\:-b\:\\ -c\:&\:a\:\end{pmatrix}$