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3 years ago

Ashley needed to get her computer fixed. She took it to the repair store. The

Mathematics

1 answer:

ankoles [38]3 years ago

7 0

Answer:

Step-by-step explanation:

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Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s). Consider the eq

wlad13 [49]

The value of x in terms of b is x = $\frac{-3}{b}$ . Therefore the value of x when b = 3 is x = $\frac{-3}{3}$ = -1.

We can find both answers by rearranging the equation to get "x =", and then substituting in 3 for b:

-2(bx - 5) = 16

-2bx + 10 = 16

- 10

-2bx = 6

÷ -2

bx = -3

÷ b

x = -3/b, which is the answer to the first part.

To get the second answer, we just substitute b = 3 into this equation and we get:

x = -3/b = -3/3 = -1

I hope this helps!

4 0

3 years ago

hi! so, yes i do pay attention in class, although i do not understand this equation. May someone please break it down and help m

swat32

$\bf \left[ (3^2)^2 \right]^2\implies 3^{2\cdot 2\cdot 2}\implies 3^8\implies 6461$

5 0

3 years ago

Find the range and the midrange for the following data set: 7.25, 6.75, 6, 7.5, 7, 7, 7, 7.5, 6, 8.5, 8.5, 6.5, 7, 5.5, 8.5, 7.2

Korolek [52]

Answer:

Step-by-step explanation:

The min and max numbers are 5.5 and 8.5. To find range, you subtract the bigger number by the smaller number. To find midrange, you find the median of the 2 min and max numbers.

8.5-5.5=3

5.5 6 6.5 7 7.5 8 8.5

1 2 3 4 5 6 7

So the range is 3 and the midrange is 7.

---

hope it helps

3 0

3 years ago

Prove that: (Sec A- cosec A)(1+ tan A+cot A) = tan A× sec A - cot A × cosec A

mash [69]

Answer:

Step-by-step explanation:

$(sec A-cosecA)(1+tanA+cotA)=tanAsecA-cotAcosecA$

we take the LHS so here goes,

$(sec A-cosecA)(1+tanA+cotA)=tanAsecA-cotAcosecA\\(\frac{1}{cosA} -\frac{1}{sinA})(1+\frac{sinA}{cosA}+\frac{cosA}{sinA})\\\\(\frac{sinA-cosA}{sinAcosA})(\frac{sinAcosA+sin^2A+cos^2A}{sinAcosA})\\$

since , $sin^2A+cos^2A=1$

the identity becomes,

$(\frac{sinA-cosA}{sinAcosA})(\frac{1+sinAcosA}{sinAcosA})\\\\(\frac{sinA+sin^2AcosA-cosA-cos^2AsinA}{sin^2Acos^2A})\\\\$

now, we know,

$sin^2A=1-cos^2A$ and $cos^2A=1-sin^2A$

the identity becomes,

$(\frac{sinA+(1-cos^2A)cosA-cosA-(1-sin^2A)sinA}{sin^2Acos^2A} )\\\\$

$(\frac{sinA+cosA-cos^3A-cosA-sinA+sin^3A}{sin^2Acos^2A})$

sin A and cos A cancel out it becomes zero

$\frac{sin^3A-cos^3A}{sin^2Acos^2A} \\\\$

Splitting the denominator the identity becomes

$\frac{sin^3A}{sin^2Acos^2A}-\frac{cos^3A}{sin^2Acos^2A} \\\\\frac{sinA}{cos^2A} - \frac{cosA}{sin^2A} \\\\\frac{sinA}{cosA}(\frac{1}{cosA})-\frac{cosA}{sinA}(\frac{1}{sinA})\\\\$

Hence,

$tanAsecA-cotAcosecA$

3 0

3 years ago

Determime the intercepts of the line that passes through the following points. (-6,5);(-3,3);(0,1)

butalik [34]

well for a line, to get its slope all we need is two points, so let's use (-6, 5) and (0, 1), and get the equation of it.

$\bf (\stackrel{x_1}{-6}~,~\stackrel{y_1}{5})\qquad
(\stackrel{x_2}{0}~,~\stackrel{y_2}{1})
\\\\\\
slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1-5}{0-(-6)}\implies \cfrac{1-5}{0+6}\implies -\cfrac{5}{6}$

$\bf \begin{array}{|c|ll}
\cline{1-1}
\textit{point-slope form}\\
\cline{1-1}
\\
y-y_1=m(x-x_1)
\\\\
\cline{1-1}
\end{array}\implies y-5=-\cfrac{5}{6}[x-(-6)]\implies y-5=-\cfrac{5}{6}(x+6)
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
\textit{to get the x-intercept, we set y = 0, solve for \underline{x}}
\\\\\\
0-5=-\cfrac{5}{6}(x+6)\implies -30=5x+30\implies -60=5x
\\\\\\
\cfrac{-60}{5}=x\implies -12=x~\hfill \boxed{\stackrel{x-intercept}{(-12,0)}}$

now, where's the y-intercept of that line? well, to get the y-intercept, we set x = 0 and solve for "y"....hmmmm wait a second, notice (0, 1), x = 0, y = 1, that's the y-intercept already.

7 0

4 years ago