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3 years ago

If joe runs 1,600 meters in 5 minutes, what is joes speed?

Mathematics

1 answer:

ruslelena [56]3 years ago

8 0

Answer:

320 meters per minute, or 19,200 meters per hour

Step-by-step explanation:

please comment what the answer options are or what unit you need for the answer, i will change my answer accordingly. hope this helps :)

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A worm walks 40cm in 8 minutes. Give both unit rates. Make sure you include the unit for each. (Total 8 points: 2 points for eac

vodka [1.7K]

40/8 = 5 cm per minute

7 0

3 years ago

Determine whether the forces in the pair are pulling at right angles to each other.

prohojiy [21]

Answer:

C) 3.0

Step-by-step explanation:

Let the length of the other side be x.

Using Pythagoras theorem for right-angled triangle,

4.2^2=2.9^2+x^2

or,x^2=4.2^2-2.9^2

or,x^2=9.23

or,x=√9.23=3.0 units

7 0

3 years ago

According to the Rational Root Theorem, which number is a potential root of f(x) = 9x8 + 9x6 – 12x + 7?

bogdanovich [222]

Answer:

Seven-thirds

Step-by-step explanation:

The give equation is $f(x)=9x^8+9x^6-12x+7$ .

The leading coefficient is 9

The constant term is 7

According to the rational root theorem, the ratio of factors the constant terms to that of the coefficient of the leading term are all possible rational roots of the given polynomial.

Base on this theorem, $\frac{7}{3}$ or seven thirds is a potential root because the numeration is a factor of 7 and the denominator is a factor of 9.

5 0

3 years ago

Read 2 more answers

When a bactericide is added to a nutrient broth in which bacteria are growing, the bacteria population continues to grow for a

baherus [9]

Answer:

a) 1296 bacteria per hour

b) 0 bacteria per hour

c) -1296 bacteria per hour

Step-by-step explanation:

We are given the following information in the question:

The size of the population at time t is given by:

$b(t) = 9^6 + 6^4t-6^3t^2$

We differentiate the given function.

Thus, the growth rate is given by:

$\displaystyle\frac{db(t)}{dt} = \frac{d}{dt}(9^6 + 6^4t-6^3t^2)\\\\= 6^4-2(6^3)t$

a) Growth rates at t = 0 hours

$\displaystyle\frac{db(t)}{dt} \bigg|_{t=0}= 6^4-2(6^3)(0) = 1296\text{ bacteria per hour}$

b) Growth rates at t = 3 hours

$\displaystyle\frac{db(t)}{dt} \bigg|_{t=3}= 6^4-2(6^3)(3) = 0\text{ bacteria per hour}$

c) Growth rates at t = 6 hours

$\displaystyle\frac{db(t)}{dt} \bigg|_{t=6}= 6^4-2(6^3)(6) = -1296\text{ bacteria per hour}$

3 0

3 years ago

What is the absolute value how do you do this problem?<br> f(x)=|-4-3|

Rainbow [258]

You subtract normally inside the absolute value (the two bars):
-4-3 is -7
Since the negative is INSIDE the absolute, the answer will turn to a positive. Think of the bars as a jail cell for the negatives. Once the negatives are in there, only positive numbers will come out. So your answer is 7.

5 0

2 years ago