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2 years ago

In triangle IJK, JK=8, KI=14, IJ=11. Which statement about the angles of triangle IJK must be true?

Mathematics

1 answer:

lora16 [44]2 years ago

3 0

Answer:true

Step-by-step explanation: It may be true but jus ask an adult ty

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Original price: $50; markdown: 2.2%; retail price:

larisa86 [58]

$50x2.2%
50x0.022=1.1
$50-$1.1=$48.9

4 0

3 years ago

if you were to solve the following system by substitution, what would be the best variable to solve for and from what equation?

Romashka-Z-Leto [24]

Answer:

see below

Step-by-step explanation:

2x+8y=12 3x-8y=11

If we have to solve by substitution, Take the first equation and divide by 2

2x/2 + 8y/2 =12/2

x+4y = 6

Then subtract 4y from each side

x = 6 -4y

Then substitute this into the second equation

This is best solved by elimination

2x+8y=12

3x-8y=11

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

5x = 36

x = 36/5

6 0

2 years ago

Solve the following quadratic equation<br> 9m2 +9= 11

jeka94

Answer: m=

√2 or m=

−1

√2

Step-by-step explanation:

4 0

3 years ago

Jorge watches cars that pass in front of his house. He counts 25 red cars and 40 cars that are not red. What is the approximate

jenyasd209 [6]

Step-by-step explanation:

firstly you will add

the red cars and non red cars together and which is 65

probability of the red car = 25/65

to the lowest form is the 5/13(D)

8 0

3 years ago

Show that d^2y/dx^2=-2x/y^5, if x^3 + y^3=1

aalyn [17]

Answer:

y³ + x³ = 1

First, differentiate the first time, term by term:

${3y^{2}.\frac{dy}{dx} + 3x^{2}} = 0 \\\\{3y^{2}.\frac{dy}{dx} = -3x^{2}} \\\\\frac{dy}{dx} = \frac{-3x^{2}}{3y^{2}} \\\\\frac{dy}{dx} = \frac{-x^{2}}{y^{2}}$

↑ we'll substitute this later (4th step onwards)

Differentiate the second time:

$3y^{2}.\frac{dy}{dx} + 3x^{2} = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} + 6x = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{-x^{2} }{y^{2} })^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{x^{4} }{y^{4} }) = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + \frac{6x^{4} }{y^{3} } = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} = - 6x - \frac{6x^{4} }{y^{3} } \\\\$

$3y^{2}.\frac{d^{2} y}{dx^{2}} = - \frac{- 6xy^{3} - 6x^{4} }{y^{3}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{- 6xy^{3} - 6x^{4} }{3y^{2}. y^{3}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{- 2xy^{3} - 2x^{4} }{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x (y^{3} + x^{3})}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x (1)}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x}{y^{5}}$

3 0

2 years ago