All categories

3 years ago

Jenny walked 2.5 miles in 50 minutes . At this rate , how many minutes did it take her to walk 1 mile ?

2 answers:

8 0

For this question, let's use algebra!

2.5 miles = 50 minutes
1 mile x minutes

x minutes = 50 x 1 ÷ 2.5
x minutes = 20

Hope this helps!

BARSIC [14]3 years ago

4 0

The answer is b.

50 / 2.5= 20 (unit rate)

if Jenny walked 2.5 miles in 50 minutes, she walked 1 in 20 minutes becuase the unit rate is 20.

hopefully this helps

You might be interested in

C=2πr, for r=8 What is c

navik [9.2K]

Substitute 8 into the formula and out comes 16π!

8 0

3 years ago

Read 2 more answers

what will fica tax (assuming a 7.65% rate) on a weekly paycheck of 407.00? a_3.11. b_5.32. c_31.14. d 53.20

Mila [183]

Answer:

c $31.14

Step-by-step explanation:

The tax is the paycheck * the rate

tax = 407 * 7.65%

Change the percent to decimal form

tax = 407 * .0765

tax = 31.1355

Rounding to the nearest cent

tx = 31.14

5 0

4 years ago

The angle 0 lies in Quadrant II . Cos 0 = <img src="https://tex.z-dn.net/?f=%20-%20%20%5Cfrac%7B2%7D%7B3%7D%20" id="TexF

mel-nik [20]

Answer:

$\tan(\theta)=\frac{-\sqrt{5}}{2}$

Step-by-step explanation:

Since we are in quadrant 2, sine is positive. Since sine is positive and cosine is negative, then tangent is negative.

Now I'm going to find the sine value of this angle given using one of the Pythagorean Identities, namely $\sin^2(\theta)+\cos^2(\theta)=1$ .

If given $\cos(\theta)=\frac{-2}{3}$ , then we have $\sin^2(\theta)+(\frac{-2}{3})^2=1$ by substitution of $\cos(\theta)=\frac{-2}{3}$ .

Let's solve:

$\sin^2(\theta)+(\frac{-2}{3})^2=1$ for $\sin(\theta)$ .

$\sin^2(\theta)+\frac{4}{9}=1$

Subtract 4/9 on both sides:

$\sin^2(\theta)=1-\frac{4}{9}$

Simplify:

$\sin^2(\theta)=\frac{5}{9}$

Square root both sides:

$\sin(\theta)=\sqrt{\frac{5}{9}}$

$\sin(\theta)=\frac{\sqrt{5}}{\sqrt{9}}$

$\sin(\theta)=\frac{\sqrt{5}}{3}$

===========

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}=\frac{\frac{\sqrt{5}}{3}}{\frac{-2}{3}}$

Multiplying top and bottom by 3 gives:

$\tan(\theta)=\frac{\sqrt{5}}{-2}$

I'm going to move the factor of -1 to the top:

$\tan(\theta)=\frac{-\sqrt{5}}{2}$

7 0

3 years ago

A waterfall in a park is located 6 miles east and 3 miles north of the visitors center. There is a picnic site 1 mile east and 2

Mazyrski [523]

Answer: Hello there!

if we define the y-axis as north and the x-axis as east, then we could use vectors (x,y) to represent the locations, where the (0,0) point is the visitors center.

The waterfall is in 6 miles east and 3 miles north of the (0,0), then we could represent it as (6,3)

the picnic site is 1 mile east and 2 miles north of the (0,0) then we could represent it as (1,2)

now we want to calculate the distance between these two points, first, we need to calculate the difference; this is:

(6,3) -(1,2) = (6 -1, 3-2) = (5,1)

if we want to calculate the total magnitude ( which is the total distance) of this vector, we need to use the Pythagorean theorem, where we considere the values of x and y as the cathetus of a triangle rectangle, and we want to obtain the hipotenuse of such triangle:

$D = \sqrt{5^2 + 1^{2} } = \sqrt{26} = 5.099$ that we could round up to 5.1 miles, then the correct answer is the d.