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3 years ago

Find all possible values of x in the similar triangles pqr stu

1 answer:

5 0

If triangles PQR and STU are similar then PQ corresponds to ST and PR corresponds to SU. Therefore, PQ/ST=PR/SU
Considering that, PQ= 7-x, ST= 13-x, PR= x²+5 and SU= x² +20
therefore, (7-x)/(13-x)= (x²+5)/(x²+20)
cross multiplying,
7x² +140-x³+20x =13x²+65-x³-5x
combining the like terms,
6x² +15x -75=0
solving for x,
x = 5/2 or -5

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16(4 1/4) distrubuted

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Answer:

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Is each line parallel, perpendicular, or neither parallel nor perpendicular to the line −3x+5y=−15? Drag and drop each choice in

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The first line is perpendicular, the second line is not perpendicular and not parallel , the third line is parallel to the above line and the forth line not parallel or perpendicular to the line.

<h3>How can the slope of the line be calculate?</h3>

The slope of the line is been given as −3x+5y=−15

Then $y= \frac{3}{5x} -3$ , hence the slope is $\frac{3}{5}$

Slope of the first line 5x + 3y = 15 after calculation is

the slope is $y= \frac{5}{-3x} +5\\\\\frac{5}{-3}$

this means that the line is perpendicular to the given line

Slope of the second line 3x + 5y = 15 after calculation is

$y= \frac{3}{-5x} +15\\\\\frac{3}{-5}$

This means that the line is not perpendicular and not parallel

Slope of the Third line -3x + 5y = 15 can be calculated as

$y= \frac{3}{5x} +15\\\\\frac{3}{5}$

This means that the line is parallel to the above line.

Slope of the first line 3x + 5y = 15 after calculation is

$y= \frac{-3}{5x} +3\\\\\frac{-3}{5}$

This means that the line is not parallel or perpendicular to the line

Learn more about slope of line at:

brainly.com/question/3493733

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2 years ago

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3 years ago

"If a ball is thrown into the air with a velocity of 44 ft/s, its height in feet t seconds later is given by y = 44t − 16t2. (a)

Nataly_w [17]

Answer:

i) -28 ft/s

ii) -21.6 ft/s

iii) -20.8 ft/s

iv) -20.16 ft/s

b) -20 ft/s

Step-by-step explanation:

if v represents vertical velocity then if the height y is

y = 44*t − 16*t²

and the instant velocity v is the derivative with respect to the time

v= dy/dt = 44*(1) - 16*(2*t) = 44- 32*t

while the average velocity is va= (y-y₀)/(t-t₀)

where t₀ = 2 and y₀=y(t₀) = 44*2 − 16*2² = 24

then

va= (y-y₀)/(t-t₀) = (44*t − 16*t² - 24)/(t-2)

for

i) t= 0.5s + 2 s= 2.5 s

va = (44* 2.5 − 16* 2.5 ² - 24)/( 2.5 -2) = -28 ft/s

ii) t= 0.1s + 2 s= 2.1 s

va = (44* 2.1 − 16* 2.1 ² - 24)/( 2.1 -2) = -21.6 ft/s

iii) t= 0.05 s + 2 s= 2.05 s

va = (44* 2.05 − 16* 2.05 ² - 24)/( 2.05 -2) = -20.8 ft/s

iv) t= 0.01s + 2 s= 2.01 s

va = (44* 2.01 − 16* 2.01 ² - 24)/( 2.01 -2) = -20.16 ft/s

b) the instantaneous velocity when t=2

v (t=2) = 44- 32*(2) = -20 ft/s

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3 years ago