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4 years ago

10

Determine whether the statement is always, sometimes, or never true:

2 answers:

kiruha [24]4 years ago

6 0

Always true.
Horizontal line (mathmatic word definition) a line that goes from left to right along the x-axis of the coordinate plane

REY [17]4 years ago

4 0

Always true because if u actually draw the chart u can see that the horizontal line is parallel to the x axis line

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Multiply (3x-1)(-x+4)

iogann1982 [59]

Answer:

$-3x^{2}+13x-4$

Step-by-step explanation:

You can use the foil technique (first outer inner last)

(3x - 1)(-x + 4) = (3x · (-x)) + (3x · 4) + ((-1) · (-x)) + ((-1) · 4)

= $-3x^{2}+12x+x-4$

= $-3x^{2}+13x-4$

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3 years ago

Trevor is having trouble understanding how the degree of a nonzero constant is 0. Use the expression 3x^0 to help explain this c

dem82 [27]

Answer:

The answer is B

Step-by-step explanation:

7 0

3 years ago

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Simplify 6log6 15<br><br> A. 4.701849846e+11<br> B. 90<br> C. 15<br> D. 6

Rama09 [41]

$\begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array} \\\\\\ \textit{Logarithm Change of Base Rule} \\\\ \log_a b\implies \cfrac{\log_c b}{\log_c a}\qquad \qquad c= \begin{array}{llll} \textit{common base for }\\ \textit{numerator and}\\ denominator \end{array} \\\\[-0.35em] ~\dotfill$

$6\log_6(15)\implies \log_6(15^6)\implies \stackrel{\textit{change of base rule}}{\cfrac{\log_{10}(15^6)}{\log_{10}(6)}}\qquad \approx \qquad 9.068\qquad \approx\qquad 9$

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2 years ago

If cos() = − 2 3 and is in Quadrant III, find tan() cot() + csc(). Incorrect: Your answer is incorrect.

nydimaria [60]

Answer:

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}$

Step-by-step explanation:

Given

$\cos(\theta) = -\frac{2}{3}$

$\theta \to$ Quadrant III

Required

Determine $\tan(\theta) \cdot \cot(\theta) + \csc(\theta)$

We have:

$\cos(\theta) = -\frac{2}{3}$

We know that:

$\sin^2(\theta) + \cos^2(\theta) = 1$

This gives:

$\sin^2(\theta) + (-\frac{2}{3})^2 = 1$

$\sin^2(\theta) + (\frac{4}{9}) = 1$

Collect like terms

$\sin^2(\theta) = 1 - \frac{4}{9}$

Take LCM and solve

$\sin^2(\theta) = \frac{9 -4}{9}$

$\sin^2(\theta) = \frac{5}{9}$

Take the square roots of both sides

$\sin(\theta) = \±\frac{\sqrt 5}{3}$

Sin is negative in quadrant III. So:

$\sin(\theta) = -\frac{\sqrt 5}{3}$

Calculate $\csc(\theta)$

$\csc(\theta) = \frac{1}{\sin(\theta)}$

We have: $\sin(\theta) = -\frac{\sqrt 5}{3}$

So:

$\csc(\theta) = \frac{1}{-\frac{\sqrt 5}{3}}$

$\csc(\theta) = \frac{-3}{\sqrt 5}$

Rationalize

$\csc(\theta) = \frac{-3}{\sqrt 5}*\frac{\sqrt 5}{\sqrt 5}$

$\csc(\theta) = \frac{-3\sqrt 5}{5}$

So, we have:

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta)$

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \tan(\theta) \cdot \frac{1}{\tan(\theta)} + \csc(\theta)$

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 + \csc(\theta)$

Substitute: $\csc(\theta) = \frac{-3\sqrt 5}{5}$

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 -\frac{3\sqrt 5}{5}$

Take LCM

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}$

6 0

3 years ago

PLS HELP NEED URGENTLY!

vovikov84 [41]

Answer:

I = V/R

Step-by-step explanation:

V = IR

Divide by R on both the sides,

I = V/R

3 0

3 years ago

Read 2 more answers