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2 years ago

Quadrilateral LMNO is similar to quadrilateral PQRS. Find the measure of side QR.

Mathematics

2 answers:

Sonbull [250]2 years ago

7 0

The answer would be 43

oksano4ka [1.4K]2 years ago

6 0

13/56 = 10/x
13x = 560
X = 43.077
X =43.1

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Hi. I need help with these See image for question Answer no 8 ,9 and 10

Nezavi [6.7K]

Answer:

8) 4 + 2q²/p² - 4r/p + r²/p²
9) (3/4, -9/4)
10) (3/8, 41/16)

Step-by-step explanation:

8. ============

Given

α and β are roots of px² + qx + r = 0

The sum of the roots is α + β = -q/p, the product of then roots αβ = r/p

(2 + α²)(2 + β²) =
4 + 2(α² + β²) + (αβ)² =
4 + 2((α + β)² -2αβ) + (αβ)² =
4 + 2((-q/p)² - 2r/p) + (r/p)² =
4 + 2q²/p² - 4r/p + r²/p²

------------------------------

9. ============

Given function

y = 2x² - 3x - 1

The minimum point is reached at vertex

The vertex is:

x = -b/2a
x = -(-3)/2*2 = 3/4

The corresponding y-coordinate is:

y = 2(3/4)² - 3(3/4) - 1 = 9/8 - 9/4 - 1 = 1/8(9 - 18 - 9) = - 18/8 = - 9/4

So the point is:

(3/4, -9/4)

---------------

10. ============

Given function

y = 2 - 3x - 4x²

The maximum is reached at vertex

The vertex is:

x = -b/2a
x = -(-3)/2(-4) = -3/8

The corresponding y-coordinate is:

y = 2 - 3(-3/8) -4(-3/8)² = 2 + 9/8 - 9/16 = 1/16(32 + 18 - 9) = 41/16

So the maximum point is:

(3/8, 41/16)

4 0

2 years ago

8. Temperatures in degrees Celsius (C) may be changed to degrees Fahrenheit (F) by using the formula F = 1.8C + 32º. Find the te

pentagon [3]

Answer:

-40 °F = -40 °C

Step-by-step explanation:

(1) F = 1.8 C + 32

If the Fahrenheit and Celsius temperatures are the same, then

(2) F = C Substitute (2) into (1)

C = 1.8C + 32 Subtract 1.8 C from each side

C – 1.8 C = 32 Combine like terms

-0.8C = 32 Divide each side by -0.8

C = -32/0.8

(3) C = -40 Substitute (3) into (1)

F = 1.8(-40) + 32

F = -72 + 32

F = -40

So, -40 °F = -40 °C

The thermometer below shows that -40 °F = -40 °C.

8 0

3 years ago

Please help with this. 13 points and brainly PLEASE

Katyanochek1 [597]

Answer:

100 im pretty sure because 10 x 10 = 100

7 0

3 years ago

Need help with my homework

Volgvan

Answer:

$\displaystyle y=\frac{16-9x^3}{2x^3 - 3}$

$\displaystyle y=-\frac{56}{13}$

Step-by-step explanation:

Equation Solving

We are given the equation:

$\displaystyle x=\sqrt[3]{\frac{3y+16}{2y+9}}$

To make y as a subject, we need to isolate y, that is, leaving it alone in the left side of the equation, and an expression with no y's to the right side.

We have to make it in steps like follows.

Cube both sides:

$\displaystyle x^3=\left(\sqrt[3]{\frac{3y+16}{2y+9}}\right)^3$

Simplify the radical with the cube:

$\displaystyle x^3=\frac{3y+16}{2y+9}$

Multiply by 2y+9

$\displaystyle x^3(2y+9)=\frac{3y+16}{2y+9}(2y+9)$

Simplify:

$\displaystyle x^3(2y+9)=3y+16$

Operate the parentheses:

$\displaystyle x^3(2y)+x^3(9)=3y+16$

$\displaystyle 2x^3y+9x^3=3y+16$

Subtract 3y and $9x^3$ :

$\displaystyle 2x^3y - 3y=16-9x^3$

Factor y out of the left side:

$\displaystyle y(2x^3 - 3)=16-9x^3$

Divide by $2x^3 - 3$ :

$\mathbf{\displaystyle y=\frac{16-9x^3}{2x^3 - 3}}$

ii) To find y when x=2, substitute:

$\displaystyle y=\frac{16-9\cdot 2^3}{2\cdot 2^3 - 3}$

$\displaystyle y=\frac{16-9\cdot 8}{2\cdot 8 - 3}$

$\displaystyle y=\frac{16-72}{16- 3}$

$\displaystyle y=\frac{-56}{13}$

$\mathbf{\displaystyle y=-\frac{56}{13}}$

8 0

2 years ago

Help!!! Here is a screenshot of the problem.

siniylev [52]

Answer:

Step-by-step explanation:

5 0

2 years ago