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3 years ago

Solve the equation12E − 4 = 0

1 answer:

7 0

This could get complicated. Please try to follow my steps carefully:

Begin by writing the equation to be solved: <u>12E - 4 = 0</u>

Add 4 to each side of the equation: 12E = 4

Divide each side by 12: E = 4/12 = 1/3 .

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Complete the equation of the graphed linear function in point-slope form.

vazorg [7]

The point-slope form:
$y-y_1=m(x-x_1)$
m - slope
x₁, y₁ - the coordinates of a point

It passes through the points (1,-2) and (2,2).
$(1,-2) \\
x_1=1 \\ y_1=-2 \\ \\
(2,2) \\
x_2=2 \\ y_2=2 \\ \\
m=\frac{y_2-y_1}{x_2-x_1}=\frac{2-(-2)}{2-1}=\frac{2+2}{1}=4$

$\boxed{y-(-2)=4(x-1)}$

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3 years ago

What is the value of the expression (9+6)×3 when simplified?<br> PLEASE HELP

SIZIF [17.4K]

Answer:

The answer would be 45

8 0

3 years ago

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The millers drove 150 miles in 3 hours at what rate how long will it take them to drive 400 miles

valentina_108 [34]

Rate = (150 miles)/(3 hrs) = 50 mph

so . . .

(50 mph)*(time) = 400 miles

*divide both sides by 50 mph

time = (400 miles)/(50 mph) = 8 hrs

time = 8 hrs

8 0

3 years ago

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Which function has a phase shift of pi/2 to the right?

Bess [88]

If it helps you any

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}sin({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\

\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\
\end{array}$

$\bf \begin{array}{llll}

\bullet \textit{vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\
\bullet \textit{function period}\\
\qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\
\qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)
\end{array}$

so.. check coefficients C and B for the horizontal shift

7 0

3 years ago

Read 2 more answers

OC bisects AOB, OD bisects AOC, OE bisects AOD, OF bisects AOE, OG bisects BOF. if BOF =120°, find DOE

LiRa [457]

Answer:

∠DOE = 16°

Step-by-step explanation:

The given parameters are;

∠BOF = 120°

∠AOB = 2×∠AOC ${}$ Given

∠AOC = 2×∠AOD ${}$ Given

∠AOD = 2×∠AOE ${}$ Given

∠AOE = 2×∠AOF ${}$ Given

Therefore;

∠AOB = 16×∠AOF ${}$ Angle addition postulate

∠BOF = ∠AOB - ∠AOF = 16×∠AOF - ∠AOF = 15×∠AOF ${}$ Transitive property

15×∠AOF = 120°

∠AOF = 120°/15 = 8°

Given that OE bisects ∠AOD, we have;

∠AOE ≅ ∠DOE ${}$ Angles bisected by a line

From;

∠AOE = 2×∠AOF, we have; ${}$ Given

Therefore;

∠AOE = ∠DOE = 2×∠AOF = 2×8° = 16°

∠DOE = 16°.

8 0

3 years ago