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3 years ago

How do you find the total sum of squared values?

Mathematics

1 answer:

tester [92]3 years ago

7 0

Answer:

Step-by-step explanation:

$\sum^{12}_{r=1} r^2 = \frac{12(12+1)(2*12+1)}{6}= A$

$\sum_{r=3}^{12}= \sum_{r=1}^{12} - (r_1+r_2)$

$r_1= 1^2$

$r_2= 2^2$

you do your math!!

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Y=2x+5 and y=3x+11 solve by substitution.

kondaur [170]

Answer:

Answer: x= -6 and y= -7

Step-by-step explanation:

Substitute 2x+5 for y in y=3x+11:

y=3x+11

2x+5=3x+11

2x+5+−3x=3x+11+−3x(Add -3x to both sides)

−x+5=11

−x+5+−5=11+−5(Add -5 to both sides)

−x=6

−x−1=6−1(Divide both sides by -1)

x=−6

Step: Substitute −6 for x in y=2x+5:

y=2x+5

y=(2)(−6)+5

y=−7(Simplify both sides of the equation)

-Source: MathPapa

6 0

3 years ago

Find the measure for each angle indicated

Lorico [155]

Step-by-step explanation:

7. 180 -92 -67 =21. 180-21 = 159

8.180-144 = 36 + 94 = 130 180-130 = 50

6 0

3 years ago

Read 2 more answers

Alexandra is scuba diving. A diving computer automatically records her depth every five minutes. The depth readings from her fir

I am Lyosha [343]

Answer:

16.38 meters represents the greatest amount of change in her depth over a single 5-minute period

Step-by-step explanation:

From time = 0 to time = 5 the amount of change in her depth was -12.45 - 0 = -12.45 meters

From time = 5 to time = 10 the amount of change in her depth was -9.7 - (-12.45) = -2.75 meters

From time = 10 to time = 15 the amount of change in her depth was -22.98 - (-9.7) = -13.28 meters

From time = 15 to time = 20 the amount of change in her depth was -6.6 - (-22.98) = 16.38 meters

7 0

3 years ago

Identify the standard form of the equation by completing the square.

OLEGan [10]

Answer:

$\dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1$

Step-by-step explanation:

<u>Given equation</u>:

$4x^2-9y^2-8x+36y-68=0$

This is an equation for a horizontal hyperbola.

<u>To complete the square for a hyperbola</u>

Arrange the equation so all the terms with variables are on the left side and the constant is on the right side.

$\implies 4x^2-8x-9y^2+36y=68$

Factor out the coefficient of the x² term and the y² term.

$\implies 4(x^2-2x)-9(y^2-4y)=68$

Add the square of half the coefficient of x and y inside the parentheses of the left side, and add the distributed values to the right side:

$\implies 4\left(x^2-2x+\left(\dfrac{-2}{2}\right)^2\right)-9\left(y^2-4y+\left(\dfrac{-4}{2}\right)^2\right)=68+4\left(\dfrac{-2}{2}\right)^2-9\left(\dfrac{-4}{2}\right)^2$

$\implies 4\left(x^2-2x+1\right)-9\left(y^2-4y+4\right)=36$

Factor the two perfect trinomials on the left side:

$\implies 4(x-1)^2-9(y-2)^2=36$

Divide both sides by the number of the right side so the right side equals 1:

$\implies \dfrac{4(x-1)^2}{36}-\dfrac{9(y-2)^2}{36}=\dfrac{36}{36}$

Simplify:

$\implies \dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1$

Therefore, this is the standard equation for a horizontal hyperbola with:

center = (1, 2)
vertices = (-2, 2) and (4, 2)
co-vertices = (1, 0) and (1, 4)
$\textsf{Asymptotes}: \quad y = -\dfrac{2}{3}x+\dfrac{8}{3} \textsf{ and }y=\dfrac{2}{3}x+\dfrac{4}{3}$
$\textsf{Foci}: \quad (1-\sqrt{13}, 2) \textsf{ and }(1+\sqrt{13}, 2)$

4 0

1 year ago

Three consecutive odd integers have a sum of 33

KATRIN_1 [288]

Three consecutive integers that add up to 33 are 2 above and lower than a middle number. This means that the average of these 3 numbers will be the same. Therefore, 33/3=11 as the average. Therefore 11 is the middle number, 9 is 2 below (a consecutive odd integer) and 13 is 2 above (a third consecutive odd integer). Therefore the 3 numbers are 9, 11, and 13.

4 0

2 years ago

Read 2 more answers