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3 years ago

8

If parallelogram RSTV has diagonals then RSTV must also be a _______. square and a rectangle rhombus and a rectangle rhombus rec

tangle square

1 answer:

andrew11 [14]3 years ago

5 0

<span>a rectangle rhombus(parallelogram) since even a parallelogram has diagonals</span>

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Examine the right triangle. A right triangle with side lengths 17 centimeters and 48 centimeters. The hypotenuse is unknown. Wha

Dimas [21]

Answer:

The answer to your question is c = $\sqrt{2593}$

Step-by-step explanation:

Data

right triangle

short leg = 17 cm

long leg = 48 cm

hypotenuse = ?

Process

-Use the Pythagorean theorem to find the answer

c² = a² + b²

c = hypotenuse

a = short leg = 17 cm

b = long leg = 48 cm

- Substitution

c² = 17² + 48²

-Simplification

c² = 289 + 2304

c² = 2593

-Result

c = $\sqrt{2593}$

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3 years ago

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A water pipe is 4 7/12 feet beneath the surface of the road. What is the elevation of the pipe, in feet, expressed as a decimal?

Andrej [43]

The answer is C. U just have to divide 7 and 12 to get your decimal

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3 years ago

10. What is an expression for the perimeter of the triangle?

Whitepunk [10]

Answer:

See explanation, and comment for more details!

Step-by-step explanation:

10. The perimeter of any shape is the sum of the length of the sides, which means that the perimeter of this triangle is x+y+z.

11. Since this is a right triangle, the two legs form the height and base. The area of this triangle is therefore x*y/2.

12.

a. x=3, y=4, z=5

b. $x=\sqrt{2}, y=\sqrt{2}, z=2$

c. $x= \sqrt{3}, y=2, z=\sqrt{7}$

Hope this helps!

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3 years ago

Y=-x^2+2x+10<br> y=x+2<br><br> Substitution <br> Please show your work<br> Need ASAP

erik [133]

Answer:

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

Step-by-step explanation:

we have

$y=-x^{2} +2x+10$ ----> equation A

$y=x+2$ ----> equation B

Solve the system by substitution

substitute equation B in equation A

$x+2=-x^{2} +2x+10$

solve for x

$-x^{2} +2x+10-x-2=0$

$-x^{2} +x+8=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-x^{2} +x+8=0$

so

$a=-1\\b=1\\c=8$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}$

$x=\frac{-1\pm\sqrt{33}} {-2}$

$x=\frac{-1+\sqrt{33}} {-2}$ -----> $x=\frac{1-\sqrt{33}} {2}$

$x=\frac{-1-\sqrt{33}} {-2}$ -----> $x=\frac{1+\sqrt{33}} {2}$

<em>Find the values of y</em>

For $x=\frac{1-\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1-\sqrt{33}} {2}+2$ ----> $y=\frac{5-\sqrt{33}} {2}$

For $x=\frac{1+\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1+\sqrt{33}} {2}+2$ ----> $y=\frac{5+\sqrt{33}} {2}$

therefore

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

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3 years ago

Helpppppppppppppppppppppppppp!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Kruka [31]

Answer:

2,4

The answer is 2,4

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3 years ago

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