24 in 70 in What is the length of the hypotenuse?

Polygon F has an area of 36 square units. Aimar drew a scaled version of Polygon F and labeled it Polygon G. Polygon G has an ar

timofeeve [1]

Answer:

1/3

Step-by-step explanation:

im just smart (wink)

6 0

3 years ago

1. A right triangle LMN is given where: side MN = 8 side NL (the hypotenuse) =

inessss [21]

Step-by-step explanation:

$\underline{ \underline{ \text{Given}}} :$

Length of MN ( Base ) = 8
Length of NL ( Hypotenuse ) = 10

$\underline{ \underline{ \text{To \: find}}} :$

Length of LM ( Perpendicular )

$\underline{ \underline{ \text{Using \: pythagoras \: theorem}}} :$

$\boxed{ \sf{ {Hypotenuse}^{2} = {Perpendicular}^{2} + {Base}^{2} }}$

⤑ $\sf{ Perpendicular = \sqrt{ {(Hypotenuse)}^{2} - {(Base)}^{2} } }$

⤑ $\sf{ \sqrt{ {(10)}^{2} - {(8)}^{2} } }$

⤑ $\sf{ \sqrt{100 - 64}}$

⤑ $\sf{ \sqrt{36}}$

⤑ $\boxed{ \sf{6\: units}}$

$\pink{ \boxed{ \boxed{ \tt{Our \: final \: answer : \boxed{ \underline { \tt{6 \: units}}}}}}}$

Hope I helped ! ツ

Have a wonderful day / night ! ♡

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5 0

2 years ago

I need help with 1-3 i don’t understand that problems and its already over due

Nina [5.8K]

Step-by-step explanation:

For a quadratic equation y = ax² + bx + c, the vertex (the maximum or minimum point) is at x = -b/(2a).

1) y = -0.5t² + 2t + 38

The maximum is at:

t = -2 / (2 × -0.5)

t = 2

The maximum height is:

y = -0.5(2)² + 2(2) + 38

y = 40

The coordinates of the vertex are (2, 40). That means the missile reaches a maximum height of 40 km after 2 minutes.

2) y = -4.9t² + 12t + 1.6

The maximum is at:

t = -12 / (2 × -4.9)

t = 1.22

The maximum height is:

y = -4.9(1.22)² + 12(1.22) + 1.6

y = 8.95

The coordinates of the vertex are (1.22, 8.95). That means the missile reaches a maximum height of 8.95 m after 1.22 seconds.

3) y = -0.04x² + 0.88x

The maximum is at:

x = -0.88 / (2 × -0.04)

x = 11

The maximum height is:

y = -0.04(11)² + 0.88(11)

y = 4.84

The maximum height of the tunnel is 4.84 meters.

The maximum width is when y = 0.

0 = -0.04x² + 0.88x

0 = -0.04x (x − 22)

x = 22

The maximum width is 22 feet.

7 0

3 years ago

Help!

nignag [31]

1) 8 + 4 = -5 + 7

12 = 2

FALSE

2) y = -11x + 4

(0, -7): -7 = -11(0) + 4 ⇒ -7 = 0 + 4 ⇒ -7 = 4 False

(-1, -7): -7 = -11(-1) + 4 ⇒ -7 = 11 + 4 ⇒ -7 = 15 False

(1, -7): -7 = -11(1) + 4 ⇒ -7 = -11 + 4 ⇒ -7 = -7 True

(2, 26): 26 = -11(2) + 4 ⇒ 26 = -22 + 4 ⇒ 26 = -18 False

Answer: C

3) Input Output

0 0

1 3

2 6

3 9

4 12

5 15

6 18

Rule: input is being added by 1, output is 3 times x

4) c = 65h

5) 2x = -6

$\frac{2x}{2} = \frac{-6}{2}$

x = -3

6) 8j - 5 + j = 67

9j - 5 = 67 added like terms (8j + j)

+5 +5

9j = 72

$\frac{9j}{9} = \frac{72}{9}$

j = 8

7) y = mx + b

-b -b

y - b = mx

$\frac{y - b}{x} = \frac{mx}{x}$

$\frac{y - b}{x} = m$

5 0

3 years ago

What is the equation of the best-fitting regression line? Find this in two ways. First, using the regression analysis part of th

Lemur [1.5K]

Step-by-step explanation:

Regression analysis is used to infer about the relationship between two or more variables.

The line of best fit is a straight line representing the regression equation on a scatter plot. The may pass through either some point or all points or none of the points.

Method 1:

Using regression analysis the line of best fit is: $y=\alpha +\beta x+e$

Here α = intercept, β = slope and e = error.

The formula to compute the intercept is:

$\alpha =\bar y-\beta \bar x$

Here $\bar y$ and $\bar x$ are mean of the y and x values respectively.

$\bar y=\frac{\sum y_{i}}{n} \\\bar x=\frac{\sum x_{i}}{n}$

The formula to compute the slope is:

$\beta =\frac{\sum (x-\bar x)(y - \bar y)}{\sum (x=\bar x)^{2}}$

And the formula to compute the error is:

$e=y-\alpha -\beta x$

Method 2:

The regression line can be determined using the descriptive statistics mean, standard deviation and correlation.

The equation of the line of best fit is:

$(y-\bar y)=r\frac{\sigma_{x}}{\sigma_{y}} (x-\bar x)$

Here r = correlation coefficient = $r=\frac{Cov (x,y)}{\sqrt{\sigma^{2}_{x}\sigma^{2}_{y}} }$

$\sigma_{x}$ and $\sigma_{y}$ are standard deviation of x and y respectively.

$\sigma_{x}=\frac{1}{n}\sum (x-\bar x)^{2} \\\sigma_{y}=\frac{1}{n}\sum (y-\bar y)^{2}$

3 0

3 years ago