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krek1111 [17]
3 years ago
10

Amy's car gets 32 miles per gallon. If her car's tank holds 15 gallons of gas, and she wants to keep a minimum of 1 gallon of ga

s in the tank for safety, how far (in miles) can Amy drive after filling up before she needs to fill up again?
Mathematics
1 answer:
velikii [3]3 years ago
4 0

Answer:

448 miles

Step-by-step explanation:

Amy's car can hold 15 gallons of gas. Provided that each gallon can drive her 32 miles. This means that 15 gallons can drive her 480 miles.

15 gallons x 32 miles = 480 miles

However, she needs to keep 1 gallon of gas in the tank for safety reasons. So we have to subtract 1 gallon from 15 gallons.

15 gallons - 1 gallon = 14 gallons

Now, let's compute again.

14 gallons x 32 miles = 448 miles

Amy can drive 448 miles before she needs to fill up the tank again.

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Please help on this one? Algebra question..
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Answer:

A.

Step-by-step explanation:

When you simplify (a - b)(a^2 - ab + b^2) you end up a^3 - b^3, instead of a^3 + b^3. The rest all work out.

8 0
3 years ago
Simplify the following:<br> a. a+a+a+a+a+a<br> b. 5x-4x+10x<br> c. 9t+3t-6t<br> d. -5j+11j+j
beks73 [17]
<h3><em><u>a</u></em><em><u>)</u></em><em><u> </u></em><em><u>a+a+a+a+a+a</u></em></h3>

<em><u>=(a+a+a+a+a+a)</u></em>

<em><u>=</u></em><em><u> </u></em><em><u>6a</u></em>

<h3><em><u>b. 5x-4x+10x</u></em></h3>

<em><u>=(5x+−4x+10x)</u></em>

<em><u>=</u></em><em><u> </u></em><em><u>11x</u></em>

<h3><em><u>c. 9t+3t-6t</u></em></h3>

<em><u>=(9t+3t+−6t)</u></em>

<em><u>=</u></em><em><u> </u></em><em><u>6t</u></em>

<h3><em><u>d. -5j+11j+j</u></em></h3>

<em><u>=(−5j+11j+j)</u></em>

<em><u>=</u></em><em><u> </u></em><em><u>7j</u></em>

<h3 />
8 0
2 years ago
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What is the 15th term of the sequence A(n) = 50 - 3(n-1) ?
Alexus [3.1K]

Answer:

A(15)=8

Step-by-step explanation:

Given that,

A(n)=50 - 3(n-1)

We need to find the 15th term of the sequence. Put n = 15 in the above sequence so that,

A(15)=50 - 3(15-1)\\\\A(15)=50-3(14)\\\\A(15)=50-42\\\\A(15)=8

So, the 15th term of the sequence is 8.

7 0
3 years ago
The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find
9966 [12]

Answer:

<em>The particular integral of given differential equation</em>

<em>                  </em>y_{p} = \frac{1}{4} ( x - (\frac{-5}{4} ) (1))<em></em>

<em> General solution of given differential equation</em>

<em>      </em>y = y_{c} + y_{p}<em></em>

<em>  </em>Y (x) = C_{1} e^{x} + C_{2} e^{4x} + \frac{1}{4} ( x + (\frac{5}{4} ))<em></em>

<em></em>

Step-by-step explanation:

<u><em>Step(i)</em></u>:-

Given Differential equation  y'' − 5 y' + 4 y = x

Given equation in operator form

        D²y - 5 Dy +  4 y = x

⇒     ( D² - 5 D +  4 ) y =x

⇒    f(D) y = Q

where  f(D) = D² - 5 D +  4 and Q(x) = x

<em>The auxiliary equation  f(m) =0</em>

<em>           m²-5 m + 4 =0</em>

         m² - 4 m - m + 4 =0

        m ( m -4 ) -1 ( m-4) =0

         (m - 1) =0   and ( m-4) =0

        <em> m = 1 and m =4</em>

<em>The complementary function </em>

<em></em>Y_{c} = C_{1} e^{x} + C_{2} e^{4x}<em></em>

<u><em>Step(ii)</em></u>:-

<u><em>particular integral</em></u>

<em>Particular integral</em>

<em>     </em>y_{p} = \frac{1}{f(D)} Q(x) = \frac{1}{D^{2}  - 5 D +  4} X<em></em>

<em>taking common '4' </em>

<em>                          </em>= \frac{1}{4(1 +  (\frac{D^{2}  - 5 D}{4} ))} X<em></em>

<em>                         </em>

<em>                           </em>=\frac{1}{4}  (1 + (\frac{D^{2} -5D}{4})^{-1} )} X<em></em>

<em>applying binomial expression</em>

<em>      ( 1 + x )⁻¹    = 1 - x + x² - x³ +.....       </em>

<em>                          </em>=\frac{1}{4}  (1 - (\frac{D^{2} -5D}{4}) +((\frac{D^{2} -5D}{4})^{2} -...} )X<em></em>

<em>Now simplifying and we will use notation D = </em>\frac{dy}{dx}<em></em>

<em>                        </em>=\frac{1}{4}  (x - (\frac{D^{2} -5D}{4})x +((\frac{D^{2} -5D}{4})^{2}(x) -...}<em></em>

<em>Higher degree terms are neglected</em>

<em>                     </em>=\frac{1}{4}  (x - (\frac{ -5 D}{4}) x)<em></em>

<em>The particular integral of given differential equation</em>

<em>                  </em>y_{p} = \frac{1}{4} ( x - (\frac{-5}{4} ) (1))<em></em>

<u><em>Final answer</em></u><em>:-</em>

<em>          General solution of given differential equation</em>

<em>      </em>y = y_{c} + y_{p}<em></em>

<em>  </em>Y (x) = C_{1} e^{x} + C_{2} e^{4x} + \frac{1}{4} ( x + (\frac{5}{4} ))<em></em>

<em></em>

<em></em>

<em>         </em>

<em> </em>

     

4 0
3 years ago
Pls answer this for me, this is the remainder theorem for grade 9 so help me.
professor190 [17]

Answer:

1.  3

2. 2

Step-by-step explanation:

| |  

x | + | 1 | | x^2 | + | 4 x | - | 2

x^3 | + | 5 x^2 | + | 2 x | + | 1

x^3 | + | x^2 | | | |  

| | 4 x^2 | + | 2 x | |  

| | 4 x^2 | + | 4 x | |  

| | | | -2 x | + | 1

| | | | -2 x | - | 2

| | | | | | 3

__________________________________________

| |  

x | - | 5 | | x^2 | - | x | + | 0

x^3 | - | 6 x^2 | + | 5 x | + | 2

x^3 | - | 5 x^2 | | | |  

| | -x^2 | + | 5 x | |  

| | -x^2 | + | 5 x | |  

| | | | | | 2

| | | | | | 0

| | | | | | 2

5 0
3 years ago
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