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3 years ago

You can drive 60 miles on 1/2 of a tank of gas in 1 hour? How far can you drive on one tank of gas

Mathematics

2 answers:

blondinia [14]3 years ago

8 0

Answer:

120 miles for 2 hrs

Step-by-step explanation:

REY [17]3 years ago

5 0

Answer:

120 miles in 2 hours

Step-by-step explanation:

120 because half of a tank is 60 miles so a whole tank means double half, which gives 120 and you also double the hour to 2 hours.

You might be interested in

If m = 27 and n = 4, what is the value of the following expression?

wel

Answer:

Step-by-step explanation:

m - 5 • n + 5

plug it in

27- 5 * 4+5

well, we gotta follow PEMDAS (there are no parentheses)

5*4

now we have

27-20+5

20+5

27-25

4 0

2 years ago

In right △ABC with right angle B, m∠A=(3x−8)° and m∠C=(x−2)°.

julsineya [31]

Answer:

<A = 67

Step-by-step explanation:

The three angles of a triangle add up to 180 degrees. We know a right angle is 90 degrees.

<A + <B + <C = 180

3x-8 + 90 + x-2 =180

Combine like terms

4x-10+90 = 180

4x+80 =180

Subtract 80 from each side

4x+80-80=180-80

4x=100

Divide each side by 4

4x/4 = 100/4

x=25

But we want to know <A

<A = 3x-8

Substitute x=25

=3(25) -8

=75-8

= 67

6 0

3 years ago

Read 2 more answers

-4(6-b)=4 how to solve this problem?

Kaylis [27]

You distribute -4 to each in the parentheses then you get -24 - 4b = 4 then you add 24 on both sides and get -4b = 28 then you divide each side by -4 and it equals -7

3 0

3 years ago

Read 2 more answers

A particle moves on the hyperbola xy=18 for time t≥0 seconds. At a certain instant, y=6 and dydt=8. What is x that this instant?

professor190 [17]

Answer:

The value of x at this instant is 3.

Step-by-step explanation:

Let $x\cdot y = 18$ , we get an additional equation by implicit differentiation:

$x\cdot \frac{dy}{dt}+y\cdot \frac{dx}{dt} = 0$ (1)

From the first equation we find that:

$x = \frac{18}{y}$ (2)

By applying (2) in (1), we get the resulting expression:

$\frac{18}{y}\cdot \frac{dy}{dt}+y\cdot \frac{dx}{dt} = 0$ (3)

$y\cdot \frac{dx}{dt}=-\frac{18}{y}\cdot \frac{dy}{dt}$

$\frac{dx}{dt} = -\frac{18}{y^{2}} \cdot \frac{dy}{dt}$

If we know that $y = 6$ and $\frac{dy}{dt} = 8$ , then the first derivative of x in time is:

$\frac{dx}{dt} = -\frac{18}{6^{2}} \cdot (8)$

$\frac{dx}{dt} = -4$

From (1) we determine the value of x at this instant:

$x\cdot \frac{dy}{dt} = -y\cdot \frac{dx}{dt}$

$x = -y\cdot \left(\frac{\frac{dx}{dt} }{\frac{dy}{dt} } \right)$

$x = -6\cdot \left(\frac{-4}{8} \right)$

$x = 3$

The value of x at this instant is 3.

4 0

2 years ago

Meg found one treasure at − 3 4 foot. Which of these location are higher than − 3 4 foot? Check all that apply. Negative StartFr

Neporo4naja [7]

Answer:

$\frac{-1}{4}$ and $\frac{-5}{8}$ are the locations higher than $\frac{-3}{4}$ foot

Step-by-step explanation:

We are given that Meg found one treasure at $\frac{-3}{4}$ foot

We are supposed to find Which of these location are higher than $\frac{-3}{4}$ foot?

i) $-\frac{7}{8}$

So, this location is not higher than $\frac{-3}{4}$ foot

ii) $-\frac{1}{4}$

So, $-\frac{1}{4}>\frac{-3}{4}$

So, this location is higher than $\frac{-3}{4}$ foot

iii) - 1 foot

$-1 < \frac{-3}{4}$

So,this location is not higher than $\frac{-3}{4}$ foot

iv)-2 feet

So,this location is not higher than $\frac{-3}{4}$ foot

v) $\frac{-5}{8}$

So, this location is higher than $\frac{-3}{4}$ foot

5 0

3 years ago