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2 years ago

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yesterday, Linda traveled 128 miles to visit her uncle. Her car used for gallons of gasoline. If Linda's car uses gasoline at th

e same rate, how many gallons of gasoline will her car used today when she travels 432 miles?

1 answer:

aliya0001 [1]2 years ago

8 0

Answer:

Linda car will use 13.5 gallons

Step-by-step explanation:

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Lines 3x-2y+7=0 and 6x+ay-18=0 is perpendicular. What is the value of a?

BlackZzzverrR [31]

Answer:

$\boxed{\sf a = 9 }$

Step-by-step explanation:

Two lines are given to us which are perpendicular to each other and we need to find out the value of a . The given equations are ,

$\sf\longrightarrow 3x - 2y +7=0$

$\sf\longrightarrow 6x +ay -18 = 0$

Step 1 : <u>Conver</u><u>t</u><u> </u><u>the </u><u>equations</u><u> in</u><u> </u><u>slope</u><u> intercept</u><u> form</u><u> </u><u>of</u><u> the</u><u> line</u><u> </u><u>.</u>

$\sf\longrightarrow y = \dfrac{3x}{2} +\dfrac{ 7 }{2}$

and ,

$\sf\longrightarrow y = -\dfrac{6x }{a}+\dfrac{18}{a}$

Step 2: <u>Find </u><u>the</u><u> </u><u>slope</u><u> of</u><u> the</u><u> </u><u>lines </u><u>:</u><u>-</u>

Now we know that the product of slope of two perpendicular lines is -1. Therefore , from Slope Intercept Form of the line we can say that the slope of first line is ,

$\sf\longrightarrow Slope_1 = \dfrac{3}{2}$

And the slope of the second line is ,

$\sf\longrightarrow Slope_2 =\dfrac{-6}{a}$

Step 3: <u>Multiply</u><u> </u><u>the </u><u>slopes </u><u>:</u><u>-</u><u> </u>

$\sf\longrightarrow \dfrac{3}{2}\times \dfrac{-6}{a}= -1$

Multiply ,

$\sf\longrightarrow \dfrac{-9}{a}= -1$

Multiply both sides by a ,

$\sf\longrightarrow (-1)a = -9$

Divide both sides by -1 ,

$\sf\longrightarrow \boxed{\blue{\sf a = 9 }}$

<u>Hence </u><u>the</u><u> </u><u>value</u><u> of</u><u> a</u><u> </u><u>is </u><u>9</u><u> </u><u>.</u>

8 0

2 years ago

Evaluate 3^-3 x 3^4 x 3 x 3^-5 Answers<br> a. -27<br> b. 1/27<br> c. -81<br> d. 1/81

Sauron [17]

The answer is b. 1/27
3^-3=1/27 and 3^4=81. 1/27×81=3. 3×3=9. 3^-5=1/243. 1/243×9= 1/27. The answer is b. 1/27.

8 0

3 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

4 0

3 years ago

Follow the steps below to find the mean for the set of data. Use the work shown to help you. Data set: 6, 5, 8, 5, 9, 6, 7, 5, 1

lbvjy [14]

Answer:

7

Step-by-step explanation: 6 + 5 + 8 + 5 + 9 + 6 + 7 + 5 + 12 = 63 so now u just divide that by 9

6 0

3 years ago

Read 2 more answers

Simplify the following exponential expression. Show your work step by step and list the Properties of Exponents used to solve th

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3x^0 (2x^3y^2)^4

--------------------------

(4x^7y^4) ^2

= 3 * 1 (2x^3y^2)^4

-------------------------- Zero Exponent Property X^0 =1

(4x^7y^4) ^2

3 (2^4 *x^3*4 y^2*4)

-------------------------- power of a power property x^a ^b = x^(a*b)

4^2 x^7*2 y^4*2

3 *16 *x^12 y^8

-------------------------- simplify

16 x^14 y^8

3 *x^12 y^8

-------------------------- simplify

x^14 y^8

3 *x^(12-14) y^(8-8)

-------------------------- Quotient of Power X^a/ X^b = X^ (a-b)

3x^-2 y^0 simplify

3x^-2 *1 Zero Exponent Property X^0 =1

3 / x^2 Negative exponent property x^-a = 1/x^a

5 0

3 years ago