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Irina-Kira [14]
3 years ago
13

Can someone help with question 51 please I need to break it down

Mathematics
1 answer:
dlinn [17]3 years ago
3 0
\dfrac{2}{3}t-3\leq1|\cdot3\\
2t-9\leq3\\
2t\leq12\\
t\leq6\\\\
\dfrac{3}{4}t-2>7|\cdot4\\
3t-8>28\\
3t>36\\
t>12\\\\
t\leq6 \vee t>12\\
\boxed{t\in(-\infty,6]\cup(12,\infty)}

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50 points
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(680 * d) - (960 * d) + 52000 =

is the equation, i’m not sure what the total answer is though.

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6. Dawn lingua bought three yards of cloth to make some curtains. The cloth was on sale for $2.25 per yard. How much did dawn pa
jekas [21]

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8 0
3 years ago
Lim x approaches 0 (1+2x)3/sinx
jok3333 [9.3K]

Interpreting your expression as

\dfrac{3(1+2x)}{\sin(x)}

when x approaches zero, the numerator approaches 3:

3(1+2x) \to 3(1+2\cdot 0) = 3(1+0) = 3\cdot 1 = 3

The denominator approaches 0, because \sin(0)=0

Moreover, we have

\displaystyle \lim_{x\to 0^-} \sin(x) = 0^-,\quad \displaystyle \lim_{x\to 0^+} \sin(x) = 0^+

So, the limit does not exist, because left and right limits are different:

\displaystyle \lim_{x\to 0^-} \dfrac{3(1+2x)}{\sin(x)}= \dfrac{3}{0^-} = -\infty,\quad \displaystyle \lim_{x\to 0^+}\dfrac{3(1+2x)}{\sin(x)}= \dfrac{3}{0^+} = +\infty

8 0
3 years ago
Evalute costheta if sintheta = (sqrt5)/3
torisob [31]

Answer:

\large\boxed{\cos\theta=\pm\dfrac{2}{3}}

Step-by-step explanation:

Use \sin^2x+\cos^2x=1.

We have

\sin\theta=\dfrac{\sqrt5}{3}

Substitute:

\left(\dfrac{\sqrt5}{3}\right)^2+\cos^2\theta=1\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\\dfrac{(\sqrt5)^2}{3^2}+\cos^2\theta=1\qquad\text{use}\ (\sqrt{a})^2=a\\\\\dfrac{5}{9}+\cos^2\theta=1\qquad\text{subtract}\ \dfrac{5}{9}\ \text{from both sides}\\\\\cos^2\theta=\dfrac{9}{9}-\dfrac{5}{9}\\\\\cos^2\theta=\dfrac{4}{9}\to \cos\theta=\pm\sqrt{\dfrac{4}{9}}\\\\\cos\theta=\pm\dfrac{\sqrt4}{\sqrt9}\\\\\cos\theta=\pm\dfrac{2}{3}

3 0
3 years ago
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