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1 year ago

Sal exercised by stretching and jogging 5 days last week.

1 answer:

4 0

So sal stretched for about 5 min. each day in 5 days and he jogged for about 43 min ..how??
I divided 215 to 5 and I got 43 min. and how did I get 215 well 240 minus 25 min is 215

and elena she stretches for 75 min in the whole 5 days total and 300 minus 75 is 225. then 225 divided by 15 is 15

Sam jogged for 43 min a day and elena jogged for 15 min a day
I hope this helps :)

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Alia wants to write the equation of the graphed line in point-slope form. These are the steps she plans to use:

baherus [9]

(2,5)(1,3)
slope(m) = (3 - 5) / (1 - 2) = -2/-1 = 2 (2 units up, one unit to the right)

y - y1 = m(x - x1)....using (1,3)
y - 3 = 2(x - 1)....talia's last step is incorrect because she didn't sub in her slope of 2

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1 year ago

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If f(x) = 1 – x, which value is equivalent to |f(i)|? options are:

g100num [7]

We have to determine which value is equivalent to | f ( i ) | if the function is: f ( x ) = 1 - x. We know that for the complex number: z = a + b i , the absolute value is: | z | = sqrt( a^2 + b^2 ). In this case: | f ( i )| = | 1 - i |. So: a = 1, b = - 1. | f ( i ) | = sqrt ( 1^2 + ( - 1 )^2) = sqrt ( 1 + 1 ) = sqrt ( 2 ). Answer: <span>C. sqrt( 2 )</span>

5 0

2 years ago

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Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i

jenyasd209 [6]

Answer:

If the limit that you want to find is $\lim_{x\to \infty}\dfrac{P(x)}{Q(x)}$ then you can use the following proof.

Step-by-step explanation:

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ and $Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0}$ be the given polinomials. Then

$\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}$

Observe that

$\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}$

and

$\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}$

Then

$\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}$

3 0

2 years ago

ΔABC is dilated using a scale factor of 12 to produce ΔA'B'C'. Select all of the statements that apply to the transformation.

GalinKa [24]

Answer:

Options (3) and (6)

Step-by-step explanation:

ΔABC is a dilated using a scale factor of $\frac{1}{2}$ to produce image triangle ΔA'B'C'.

Since, dilation is a rigid transformation,

Angles of both the triangles will be unchanged or congruent.

m∠A = m∠A' and m∠B = m∠B'

Since, sides of ΔA'B'C' = $\frac{1}{2}$ of the sides of ΔABC

Area of ΔA'B'C' = $\frac{1}{2}(\text{Area of triangle ABC})$

Area of ΔABC > Area of ΔA'B'C'

Since, angles of ΔABC and ΔA'B'C' are congruent, both the triangles will be similar.

ΔABC ~ ΔA'B'C'

Therefore, Option (3) and Option (6) are the correct options.

3 0

1 year ago

A football field is 120 yards long by 53 yards wide if a player runs diagonally from one corner to the opposite corner. How far

qaws [65]

Answer: They will travel about 131.18 yards.

Step-by-step explanation:

Given : A football field is 120 yards long by 53 yards wide.

We know that a football field is rectangular in shape.

Each interior angle in a rectangle is a right angle.

Then, by Pythagoras theorem, we have

(Diagonal)² = (Length)² + (Width)²

If a player runs diagonally from one corner to the opposite corner, then the length of the diagonal is given by :-

$(\text{Diagonal})=\sqrt{(120)^2+(53)^2}\\\\\Rightarrow\ (\text{Diagonal})=\sqrt{14400+2809}\\\\\Rightarrow\ (\text{Diagonal})=\sqrt{17209}\\\\\Rightarrow\ (\text{Diagonal})=131.183078177\approx131.18\text{ yards}$

Hence, they will travel about 131.18 yards.

7 0

1 year ago