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3 years ago

Tyler used 17.04 gallons of gas. Maria used 6.2 times as many gallons of gas as Tyler. How many gallons of gas did Maria use?

1 answer:

8 0

In that problem you will have to multiply, we know that Tyler used 17.04 gallons of gas, and that Maria used 6.2 times as many gallons of gas. So we are going to multiply 17.04 x 6.2 and the answer is 105.648, so Maria used 105.648 gallons of gas.

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If the ending time is 10:08, and the elasped is 30 minutes

yanalaym [24]

If your question is "what time was it before the elapsed time", then the answer will be 9:38. All you have to do is to deduct the time by 30 minutes based on your clock, and you will be able to get the answer right away. Also, if you add 9:38 by 30 minutes, you will get 10:08 as well.

3 0

3 years ago

Read 2 more answers

Y = 6x + 8 <br> y = –4x – 2

lara31 [8.8K]

Answer:

x= 2

y = 20

Step-by-step explanation:

y = 6x + 8

y = –4x – 2

y = 4 (6x + 8 )

y = 6 (–4x – 2)

y = 24x + 32

y = –24 -12

___________

y = 20

input that back into the equation to find x

20 = 6x + 8

20 - 8 = 6x

12 = 6x

x= 2

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3 years ago

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Ben uses a compass and straightedge to bisect segment PQ, as shown: segment PQ with compass open to greater than half of segment

il63 [147K]

Answer: To be able to create two intersecting arcs above and below the line segment

Step-by-step explanation:

That is one of the steps to bisecting a line segment lol

7 0

3 years ago

Evaluate the expression 4•9+6/-3

kykrilka [37]

You have to multiplying first, then adding on 6 over negative 3 as a fraction. The parentheses it will be multiplying.

(4)(9)+6/-3

= 34.

It will help you.

Have a great day!

-Charlie

8 0

3 years ago

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You have D dollars to buy fence to enclose a rectangular plot of land (see figure at right). The fence for the top and bottom co

alex41 [277]

The perimeter of the rectangular plot of land is given by the expression below

$P=2x+2y$

On the other hand, since the available money to buy fence is D dollars,

$\begin{gathered} D=4(2x)+3(2y) \\ \Rightarrow D=8x+6y \\ D\rightarrow\text{ constant} \end{gathered}$

Furthermore, the area of the enclosed land is given by

$A=xy$

Solving the second equation for x,

$\begin{gathered} D=8x+6y \\ \Rightarrow x=\frac{D-6y}{8} \end{gathered}$

Substituting into the equation for the area,

$\begin{gathered} A=(\frac{D-6y}{8})y \\ \Rightarrow A=\frac{D}{8}y-\frac{3}{4}y^2 \end{gathered}$

To find the maximum possible area, solve A'(y)=0, as shown below

$\begin{gathered} A^{\prime}(y)=0 \\ \Rightarrow\frac{D}{8}-\frac{3}{2}y=0 \\ \Rightarrow\frac{3}{2}y=\frac{D}{8} \\ \Rightarrow y=\frac{D}{12} \end{gathered}$

Therefore, the corresponding value of x is

$\begin{gathered} y=\frac{D}{12} \\ \Rightarrow x=\frac{D-6(\frac{D}{12})}{8}=\frac{D-\frac{D}{2}}{8}=\frac{D}{16} \end{gathered}$ <h2>Thus, the dimensions of the fence that maximize the area are x=D/16 and y=D/12.</h2><h2>As for the used money,</h2> $\begin{gathered} top,bottom:\frac{8D}{16}=\frac{D}{2} \\ Sides:\frac{6D}{12}=\frac{D}{2} \end{gathered}$ <h2>Half the money was used for the top and the bottom, while the other half was used for the sides.</h2>

7 0

1 year ago