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3 years ago

How far would she have walked if she kept the same pace?

Mathematics

2 answers:

garik1379 [7]3 years ago

5 0

4.67 miles

I think this is the answer

EastWind [94]3 years ago

5 0

4.67 miles
I'll do the math and work and get back to u

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Juan is 1 1/4 feet shorter than Maria. Maria is 1/3 foot taller than Luis. Luis is 62 inches tall, how tall are Maria and Juan

liraira [26]

It's really easy i think u can handle yourself

7 0

3 years ago

Read 2 more answers

If a = pi +3j - 7k, b = pi - pj +4k and the angle between a and is acute then the possible values for p are given by

PIT_PIT [208]

Answer:

The family of possible values for $p$ are:

$(-\infty, -4) \,\cup \,(7, +\infty)$

Step-by-step explanation:

By Linear Algebra, we can calculate the angle by definition of dot product:

$\cos \theta = \frac{\vec a\,\bullet\,\vec b}{\|\vec a\|\cdot \|\vec b\|}$ (1)

Where:

$\theta$ - Angle between vectors, in sexagesimal degrees.

$\|\vec a\|, \|\vec b \|$ - Norms of vectors $\vec {a}$ and $\vec{b}$

If $\theta$ is acute, then the cosine function is bounded between 0 a 1 and if we know that $\vec {a} = (p, 3, -7)$ and $\vec {b} = (p, -p, 4)$ , then the possible values for $p$ are:

Minimum ( $\cos \theta = 0$ )

$\frac{p^{2}-3\cdot p -28}{\sqrt{p^{2}+58}\cdot \sqrt{2\cdot p^{2}+16}} > 0$

Maximum ( $\cos \theta = 1$ )

$\frac{p^{2}-3\cdot p -28}{\sqrt{p^{2}+58}\cdot \sqrt{2\cdot p^{2}+16}} < 1$

With the help of a graphing tool we get the family of possible values for $p$ are:

$(-\infty, -4) \,\cup \,(7, +\infty)$

7 0

3 years ago

Suppose y = 2x + 1 , where x and y are functions of t. (a) if dx/dt = 6, find dy/dt when x = 4. dy dt = (b) if dy/dt = 2, find d

evablogger [386]

$y=2x+1\\\\\dfrac{dy}{dt}=2\dfrac{dx}{dt}$

(a) For dx/dt=6, dy/dt = 2·6 = 12

(b) For dy/dt=2, 2 = 2·dx/dt ⇒ dx/dt = 1

_____

The value of x is irrelevant for this question.

=====

If your function is 2^x +1 or 2^(x +1), it needs to be written as such. The answer above applies to the function given.

7 0

3 years ago

a. How wide must a wall footing be if the load is 9,500 pounds per foot of wall length, and the footing rests on a sandy gravel?

Phantasy [73]

Answer:

a) 38"

b) L = 5.7'

Step-by-step explanation:

a) Given

P = 9,500 lb/ft

Thick = 18"

We assume an allowable bearing pressure of σ = 3000 lb/sf

If Stress = Force / Area

σ = P/A

Solving for area

A = P/σ ⇒ A = 9,500 lb/ft/3000 lb/sf

⇒ A = 3.167 sf, or 3.167 ft wide, per foot of length.

⇒ A = 38" wide

We can see the pic 1 in order to understand the answer.

b) Given

P = 65,000 pounds = 65,000 lb

Thick = 18" = 1.5'

We assume an allowable bearing pressure of σ = 2000 lb/sf

If Stress = Force / Area

σ = P/A

Solving for area

A = P/σ ⇒ A = 65,000 lb/2000 lb/sf

⇒ A = 32.5 sf

then A = L² ⇒ L = √A = √(32.5 sf) = 5.7 ft

Finally L = 5.7'

We can see the pic 2 in order to understand the answer.

3 0

3 years ago

Use technology to approximate the solution(s) to the system of equations to the nearest tenth of a unit.

kodGreya [7K]

The technology I used in helping me solve the problem is using Microsoft Excel. I made one column for the x values which were gathered from the choices, one column for f(x) whose equation is 8^(x-9), and another column for g(x) whose equation is log(3x) + 2.

Inputing these equation to the last two columns yield the values shown in the picture.The 'NUM' means that the equation at that value of x is undefined. From the values, the correct solution would be

(0,0), (9.2,1.5) and (9.6,3.5) to the f(x) equation
(9.6,3.5) to the g(x) equation

5 0

3 years ago