Ask question

Ask question

All categories

2 months ago

11

Mrs. Allan's car uses 8 gallons of gas for a 224-mile trip. Mrs. Owen's car uses 6 gallons of gas for a 210-mile trip. How many

gallons of gas would each car use if both cars traveled 560 miles?

1 answer:

lawyer [12.5K]2 months ago

8 0

For a 560-mile journey, Mrs. Allan's car will require 20 gallons of fuel. Meanwhile, Mrs. Owen's vehicle will need 16 gallons for the same distance.

You might be interested in

You and your friend are standing back-to-back. Your friend runs 16 feet forward and then 12 feet right. At the same time, you ru

babunello [11817]

16-12=4ft long 9-12=3ft 4+3=7ft 7 feet long

4 0

1 month ago

Read 2 more answers

Kendra is choosing bouquets of flowers for table centerpieces. She decides to buy 40 bouquets for $120 because she thinks that p

babunello [11817]

Answer:

Find below:

Step-by-step explanation:

To determine this, we will either calculate the total cost of acquiring 40 bouquets at $2.50 each or find the single bouquet’s cost at $120.

Cost of one in pack of 40 priced at $120.

120 divided by 40 equals $3

Now, we notice that $3>$2.50

This indicates Kendra has made an error by purchasing the 40 bouquet pack at $120

Hope this helps.

Good Luck

7 0

2 months ago

casey bought sandwiches and a bag of chips. Each sandwich cost three times as much as bag of chips. She bought 5 sandwiches for

AnnZ [12381]

Answer:

4

Step-by-step explanation:

Each sandwich costs 6 dollars, while chips sell for 3 dollars. She purchased 5 sandwiches at 6 dollars each, amounting to 30 dollars (5*6=30). This leaves her with 12 dollars to use since her total expenditure was 42 dollars. Given that chips are 3 dollars each, with 12 dollars left, 4 bags could be bought (4 * 3 = 12).

3 0

2 months ago

Read 2 more answers

A flat circular plate has the shape of the region x2 + y2≤1. The plate, including the boundary where x2 + y2 = 1, is heated such

Leona [12618]

Setting both partial derivatives to zero results in a single critical point at $(x,y)=\left(\dfrac12,0\right)$ , located within the unit disk.

At this given point, the derivative value of the Hessian matrix is

$|H|=\begin{vmatrix}T_{xx}&T_{xy}\\T_{yx}&T_{yy}\end{vmatrix}=\begin{vmatrix}2&0\\0&4\end{vmatrix}=8>0$

and the second-order partial derivative with respect to $x$ yields

$T_{xx}\bigg|_{(x,y)=(1/2,0)}=2>0$

This suggests that the critical point represents a local minimum, marking it as the coldest area on the plate with a temperature of $T\left(\dfrac12,0\right)=-\dfrac14$ .

To find the hottest area on the plate, it must be located along the boundary. Let $x=\cos\theta$ and $y=\sin\theta$ , so that

$T(x,y)=T(\theta)=\cos^2\theta+2\sin^2\theta-\cos\theta$
$T(\theta)=\dfrac32-\cos\theta-\dfrac12\cos2\theta$

Thus, the plate's boundary (the circle $x^2+y^2=1$ ) is treated as a single variable function $\theta$ examined over $\theta\in[0,2\pi)$ . A single differentiation gives

$T'(\theta)=\sin\theta+\sin2\theta=0$
$\implies\theta=0,\theta=\dfrac{2\pi}3,\theta=\pi,\theta=\dfrac{4\pi}3$

You will discover that $T(\theta)$ achieves three extrema on the interval $(0,2\pi)$ , with relative maxima occurring at $\theta=\dfrac{2\pi}3$ and $\theta=\dfrac{4\pi}3$ , and a relative minimum at $\theta=\pi$ (and $\theta=0$ , if you wish to include that).

Our minimum has already been identified inside the plate - which you can check to have a lower temperature than at the points noted by $T(\theta)$ - and we identify two maxima at $\theta=\dfrac{2\pi}3$ and $\theta=\dfrac{4\pi}3$ , both showing a maximum temperature of $T=\dfrac94$ .

Reverting to Cartesian coordinates, these points match up with $\left(-\dfrac12,\pm\dfrac{\sqrt3}2\right)$ .

4 0

1 month ago

Helen draws a random circle.

Leona [12618]

Answer:

Upper limit: 3.116

Lower limit: 3.125

Step-by-step breakdown:

The random circle drawn by Hellen has a circumference of C=405 mm, measured to 3 significant figures, leading to a diameter, d, of 130 mm, accurate to 2 significant figures.

We utilize the formula

$\pi = \frac{C}{d}$

Substituting in the figures provides:

$\pi = \frac{405}{130}$

$\pi = 3.11538461538$

Consequently, Hellen’s value for π is recorded as 3.115 to three decimal places.

To find both limits, we divide the specified level of precision by two.

$\frac{0.001}{2} = 0.0005$

This figure is then added to the rounded number to determine the upper limit:

3.115+0.0005=3.1155

The lower limit calculates as 3.115-0.0005=3.1145

7 0

2 months ago