If 2 inches are equivalent to 5 centimeters, how many square centimeters are in one square inch?

Which equation can be solved to find one of the missing side lengths in the triangle?

algol13

There are several different equations that can be used to find missing sides, these can be trigonometric functions or the distance formula. The trigonometric functions consist of sine (opposite/hypotenuse), cosine (adjacent/hypotenuse), and tangent (opposite/adjacent. The adjacent side is represented by the side next to the given angle measure, the opposite is the side that is connected to adjacent side and across from the given angle, and the hypotenuse is the diagonal that connects the opposite side to the given angle- most notable because its line isn't straight like the other sides.
The distance formula is used to find the measurement of missing side lengths in all quadrilaterals, and it's: D = sqrt(x2 - x1)^2 + (y2 - y1)^2 where x are the x-coordinates of two given points and y are the y-coordinates of the same two given points.

7 0

3 years ago

If the point (a, b) is equidistant from the points (-a, 2) and (2, -b), then prove that 3(a+b)+4=0

erastova [34]

Answer: see proof below

Step-by-step explanation:

Since (a, b) is equidistant from (-a, 2) and (2, -b), then it is the midpoint of the those two points. Use Midpoint formula to find (a, b).

$M_x=\dfrac{x_1+x_2}{2}\qquad \qquad \qquad M_y=\dfrac{y_1+y_2}{2}\\\\\\a=\dfrac{-a+2}{2}\qquad \qquad \qquad \quad b=\dfrac{2-b}{2}\\\\\\2a=-a+2\qquad \qquad \qquad \quad 2b=2-b\\\\\\3a=2\qquad \qquad \qquad \qquad \qquad 3b=2\\\\\\a=\dfrac{2}{3}\qquad \qquad \qquad \qquad \qquad b=\dfrac{2}{3}$

3(a + b) - 4 = 0

$3\bigg(\dfrac{2}{3}+\dfrac{2}{3}\bigg)-4=0\\\\\\3\bigg(\dfrac{4}{3}\bigg)-4=0\\\\\\4-4=0\\\\0=0\qquad \text{TRUE!}$

Notice that I changed the equation to "negative 4" because the equation you provided did not make a true statement.

4 0

3 years ago

On the first of each month, Shelly runs a 5k race. She keeps track of her times to track her progress. Her time in minutes is re

Vikki [24]

Answer:

C . 0.76

Step-by-step explanation:

We are given that on the first of each month , Shelly runs a 5 k race. She keeps track of her times to track her progress. Her time in minutes is recorded in the table:

Jan 40.55 Feb 41.51

March 42.01 Apr 38.76

May 36.32 June 34.28

July 35.38 Aug 37.48

Sept 40.87 Oct 48.32

Nov 41.59 Dec 42.71

We have to find the difference between the mean of the data , including the outlier and excluding the outlier

Outlier: That observation which is different from other observations.

The outlier in the given observations is 48.32 because is different from other observations.

Mean of the data including the outlier

Mean = $\frac{Sum \;of\;observations}{Total\;number\;of\;observations}$

Mean= $\frac{40.55+41.51+42.01+38.76+36.32+34.28+35.38+37.48+40.87+48.32+41.59+42.71}{12}$

Mean= $\frac{479.78}{12}$

Mean=39.982

Mean of the data excluding the outlier

Mean= $\frac{40.55+41.51+42.01+38.76+36.32+34.28+35.38+37.48+40.87+41.59+42.71}{11}Mean=[tex]\frac{431.46}{11}$

Mean=39.224

Difference between mean of the data including the outlier and excluding the outlier=39.982-39.224=0.758

Difference between mean of the data including the outlier and excluding the outlier=0.76

Answer Answer:

Step-by-step explanation:

We are given that on the first of each month , Shelly runs a 5 k race. She keeps track of her times to track her progress. Her time in minutes is recorded in the table:

Jan 40.55 Feb 41.51

March 42.01 Apr 38.76

May 36.32 June 34.28

July 35.38 Aug 37.48

Sept 40.87 Oct 48.32

Nov 41.59 Dec 42.71

We have to find the difference between the mean of the data , including the outlier and excluding the outlier

Outlier: That observation which is different from other observations.

The outlier in the given observations is 48.32 because is different from other observations.

Mean of the data including the outlier

Mean = $\frac{Sum \;of\;observations}{Total\;number\;of\;observations}$

Mean= $\frac{40.55+41.51+42.01+38.76+36.32+34.28+35.38+37.48+40.87+48.32+41.59+42.71}{12}$

Mean= $\frac{479.78}{12}$

Mean=39.982

Mean of the data excluding the outlier

Mean= $\frac{40.55+41.51+42.01+38.76+36.32+34.28+35.38+37.48+40.87+41.59+42.71}{11}Mean=[tex]\frac{431.46}{11}$

Mean=39.224

Difference between mean of the data including the outlier and excluding the outlier=39.982-39.224=0.758

Difference between mean of the data including the outlier and excluding the outlier=0.76

Answer :C . 0.76

5 0

3 years ago

Read 2 more answers

Please to these click on the paper clip. No fake stuff please

Akimi4 [234]

You just fill the variables with the numbers

5 0

4 years ago

I will give brainlist

Katen [24]

I think is -4 ( y + 2) Srry if I’m wrong

8 0

3 years ago