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3 years ago

The number of cookies Natalie can bake varies directly with the time she

Mathematics

2 answers:

EastWind [94]3 years ago

4 0

The answer would be 10 dozen cookies

CaHeK987 [17]3 years ago

3 0

We already know that 0.5 of an hour, or half of an hour, is 30 minutes. We also know that 3 dozen equals 36. With this information, we can conclude that Natalie can make 72 cookies in one hour. 72 times 5 is 360. 360 divided by 12 is 30, so, Natalie can bake 30 dozen, or 360, cookies in five hours.

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Express the vector (4,-6) in terms of the standard unit vectors i and j

Aliun [14]

Four, -6 in terms of standard unit factors I NJ makes no sense

8 0

3 years ago

A 6m ladder leans against a wall. The bottom of the ladder is 1.3m from the wall at time =0sec and slides away from the wall at

icang [17]

Answer:

- 0.100

Step-by-step explanation:

Length of the ladder, H = 6 m

Distance at the bottom from the wall, B = 1.3 m

Let the distance of top of the ladder from the bottom at the wall is P

Thus,

from Pythagoras theorem,

B² + P² = H² .

B² + P² = 6² ..............(1) [Since length of the ladder remains constant]

at B = 1.3 m

1.3² + P² = 6²

P² = 36 - 1.69

P² = 34.31

P = 5.857

Now,

differentiating (1)

$2B(\frac{dB}{dt})+2P(\frac{dP}{dt})=0$

at t = 2 seconds

change in B = 0.3 × 2= 0.6 ft

Thus,

at 2 seconds

B = 1.3 + 0.6 = 1.9 m

therefore,

1.9² + P² = 6²

P = 5.69 m

on substituting the given values,

2(1.9)(0.3) + 2(5.69) × $(\frac{dP}{dt})$ = 0

1.14 + 11.38 × $(\frac{dP}{dt})$ = 0

11.38 × $(\frac{dP}{dt})$ = - 1.14

$(\frac{dP}{dt})$ = - 0.100

here, negative sign means that the velocity is in downward direction as upward is positive

5 0

4 years ago

Vicki has a 50 inch roll of ribbon. She used 3 feet of the ribbon to wrap a gift, How many inches of ribbon does she have left?

taurus [48]

She has 14 inches left. (1 ft and 2 in)

5 0

3 years ago

Read 2 more answers

Find all solutions of each equation on the interval 0 ≤ x < 2π.

Korvikt [17]

Answer:

$x = 0$ or $x = \pi$ .

Step-by-step explanation:

How are tangents and secants related to sines and cosines?

$\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}$ .

$\displaystyle \sec{x} = \frac{1}{\cos{x}}$ .

Sticking to either cosine or sine might help simplify the calculation. By the Pythagorean Theorem, $\sin^{2}{x} = 1 - \cos^{2}{x}$ . Therefore, for the square of tangents,

$\displaystyle \tan^{2}{x} = \frac{\sin^{2}{x}}{\cos^{2}{x}} = \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}$ .

This equation will thus become:

$\displaystyle \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} \cdot \frac{1}{\cos^{2}{x}} + \frac{2}{\cos^{2}{x}} - \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} = 2$ .

To simplify the calculations, replace all $\cos^{2}{x}$ with another variable. For example, let $u = \cos^{2}{x}$ . Keep in mind that $0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1$ .

$\displaystyle \frac{1 - u}{u^{2}} + \frac{2}{u} - \frac{1 - u}{u} = 2$ .

$\displaystyle \frac{(1 - u) + u - u \cdot (1- u)}{u^{2}} = 2$ .

Solve this equation for $u$ :

$\displaystyle \frac{u^{2} + 1}{u^{2}} = 2$ .

$u^{2} + 1 = 2 u^{2}$ .

$u^{2} = 1$ .

Given that $0 \le u \le 1$ , $u = 1$ is the only possible solution.

$\cos^{2}{x} = 1$ ,

$x = k \pi$ , where $k\in \mathbb{Z}$ (i.e., $k$ is an integer.)

Given that $0 \le x < 2\pi$ ,

$0 \le k$ .

$k = 0$ or $k = 1$ . Accordingly,

$x = 0$ or $x = \pi$ .

8 0

3 years ago

Read 2 more answers

What is the value of the expression |a + b| + |c| when a = –3, b = 7, and c = –15?

nikdorinn [45]

We know:
$|a|= \left\{\begin{array}{ccc}a&if\ a \ \textgreater \ 0\\0&if\ a=0\\-a&if\ a \ \textless \ 0\end{array}\right\\\\|2|=2\\|0|=0\\|-3|=3$

$|a+b|+|c|$
substitute a = -3; b = 7; c = -15
$|-3+7|+|-15|=|4|+15=4+15=19$

3 0

3 years ago