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2 years ago

Helpppppppppppppppppppp

Mathematics

1 answer:

Ganezh [65]2 years ago

8 0

Answer:

See below ~

Step-by-step explanation:

Q1

y = (x + 3i)(x - 7)
y = x² + 3ix - 7x - 21
y = x² + (3i - 7)x - 21

Q2

y = (x - √5)(x - i)
y = x² - √5x - ix + √5i
y = x² - (√5 + i)x + √5i

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(x+6)(−x+1)=0 greatest x and least x = 0

Bumek [7]

Answer:

x=-6

×=1

Step-by-step explanation:

remove parentheses

connect like terms

change the signs

calculate the discriminant

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3 years ago

The height of a stuntperson jumping off a building that is 20 m high is modeled by the equation h = 20 -57, where t is the time

cupoosta [38]

A stuntman jumping off a 20-m-high building is modeled by the equation h = 20 – 5t2, where t is the time in seconds. A high-speed camera is ready to film him between 15 m and 10 m above the ground. For which interval of time should the camera film him?

Answer:

$1\leq t\geq \sqrt{2}$

Step-by-step explanation:

Given:

A stuntman jumping off a 20-m-high building is modeled by the equation

$h =20-5t^{2}$ -----------(1)

A high-speed camera is ready to making film between 15 m and 10 m above the ground

when the stuntman is 15m above the ground.

height $h = 15m$

Put height value in equation 1

$15 =20-5t^{2}$

$5t^{2} =20-15$

$5t^{2} =5$

$t^{2} =1$

$t =\pm1$

We know that the time is always positive, therefore $t=1$

when the stuntman is 10m above the ground.

height $h = 10m$

Put height value in equation 1

$10 =20-5t^{2}$

$5t^{2} =20-10$

$5t^{2} =10$

$t^{2} =\frac{10}{5}$

$t^{2} =2$

$t=\pm\sqrt{2}$

$t=\sqrt{2}$

Therefore ,time interval of camera film him is $1\leq t\geq \sqrt{2}$

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3 years ago

Find two numbers, both smaller than 150, that have a lowest common multiple of 882 and a highest common factor of 21.

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Answer:

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Step-by-step explanation:

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1 year ago

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Answer:

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Step-by-step explanation:

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3 years ago

Determine the rate of change for thw graph: $____ per mile

Ira Lisetskai [31]

Answer:

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General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS