10 задание пожалуйста

For the training adaptation of maximum strength, what is the recommended number of repetitions per set? 1 to 5 repetitions 1 to

Slav-nsk [51]

For the training adaptation of maximum strength, 1 to 5 repetitions per set is recommended.

The one-repetition maximum test, also called a one-rep max or 1RM, is used to find out the heaviest weight you can lift just once.

If Maximal strength is desired, it is best achieved by performing repetitions at 85 to 100 % of the one-repetition maximum (1RM).

Formula to find the repetition is

weight used x (reps done x 0.33 + 1) = 1RM

so,

Reps done = {{1RM/weight used} - 1}/ 0.33

By this find the no of reps are recommended for maximum strength.

The maximum strength is achieved by doing the number of reps by 85 to 100% of the value of 1 RM.

So,

For the training adaptation of maximum strength, 1 to 5 repetitions per set is recommended.

Learn more about the MAXIMUM STRENGTH here

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6 0

2 years ago

(3.2 * 10000000) -3

andreev551 [17]

The exponents distribute over factors:

$(ab)^c = a^c \times b^c$

so, you have

$(3.2 \times 10^6)^{-3} = 3.2^{-3} \times (10^6)^{-3}$

Let's focus on each factor: applying the definition of negative exponents, you have

$3.2^{-3} = \cfrac{1}{3.2^3} \approx 0.03$

While for the power of 10, you can use the following rule

$(a^b)^c = a^{bc}$ to write

$(10^6)^{-3} = 10^{6\cdot (-3)} = 10^{-18}$

So, the expression evaluates to

$0.03\times 10^{-18} = 3\times 10^{-20}$

3 0

4 years ago

The company also discovered that it costs $25 to produce 2 camera cases, $59 to produce 4 camera cases and $305 to produce 10 ca

sasho [114]

Answer:

Cost of 5 camera cases is $85.

Step-by-step explanation:

First we need to determine what type of function we're dealing with, is it linear, quadratic or exponential? We can determine this by calculating the first difference.

We can translate the number of cameras to correspond to the x axis and cost to the y axis. By doing this we can come up with the ordered pairs of (2,25), (4,59),(10,305). Let's get the first two ordered pairs and calculate the first difference:

$\frac{59-25}{4-2}=17$

And now the second and third value:

$\frac{305-59}{10-4}=41$

Notice how these values do not match, this means that this function is not increasing at a constant rate meaning that this function is not linear. Now let's calculate the second difference:

$\frac{41-17}{10-2}=3$

This means our function is quadratic with a=3. Remember that generally speaking a quadratic function is such that:

$y=ax^2+bx+c$

Where a,b,c are coefficients and in this case a=3. Now, let's solve for b and c values by using the coordinates given. Let's use the coordinates (2,25) where x=2 and y=25 and so:

$f(x)=ax^2+bx+c\\\\25=3(2)^2+b(2)+c\\25=3(4)+2b+c\\25=12+2b+c\\13=2b+c\\\\b=\frac{13-c}{2}$

Now let's solve for "c" by using different coordinates (4,59) where x=4 and y=59 and also use the "b" value that we calculated above and so:

$f(x)=ax^2+bx+c\\\\59=3(4)^2+(\frac{13-c}{2})(4)+c\\59=48+\frac{52-4c}{2}+c\\11=\frac{52-4c+2c}{2}\\11=26-2c+c\\-15=-c\\\\c=15$

Now that we know the value of "c" we can use the equation for b to solve for b and so:

$b=\frac{13-c}{2}=\frac{13-15}{2}\\\\b=-1$

So this means in the quadratic equation a=3, b=-1 and c=15 therefore by using the general equation of a quadratic function we obtain or final equation:

$f(x)=ax^2+bx+c\\\\f(x)=3x^2-x+15$