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gulaghasi [49]
3 years ago
8

Please help quick! Write the quadratic function f(x) = 3(x + 1)(x – 7) in standard form. (Answers attached below)

Mathematics
1 answer:
Musya8 [376]3 years ago
3 0

Answer:

hope it will help you ...

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All four arms of a mechanical jack are the same​ length, and they form a parallelogram. Turning the crank pulls the arms​ togeth
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10 inches

Step-by-step explanation:

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12.7 is it<br> Irrational<br> Rational<br> Integers<br> Whole<br> Natural
marissa [1.9K]

Answer:

Rational number

Step-by-step explanation:

It is a rational number because it terminates. However, it is NOT an integer because there is a decimal making it unable to be a whole and natural number.

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3 years ago
Need some help with this!
Morgarella [4.7K]

Answer:

substitution

reflexive

Step-by-step explanation:

8 0
4 years ago
I am trying to understand how to solve for X, then solve the equation
Alina [70]

Answer:

x = 16

m<Y = 34°

Step-by-step explanation:

∆XYZ is an isosceles ∆. An isosceles ∆ has two equal sides, as well as the bases of the isosceles triangle are congruent. In this case, therefore:

<X = <Z

(6x - 23)° = (4x + 9)

Solve for x

6x - 23 = 4x + 9

Collect like terms

6x - 4x = 23 + 9

2x = 32

Divide both sides by 2

x = 16

m<Y = 180° - (m<X + m<Z) (sum of ∆)

m<Y = 180 - ((6x - 23) + (4x + 9))

Plug in the value of x

m<Y = 180 - ((6(16) - 23) + (4(16) + 9))

m<Y = 180 - (73 + 73)

m<Y = 34°

5 0
3 years ago
Solve to find x <br> log (3x+5) to base 5= 2log (1-3x) to base 25
Eduardwww [97]

Answer:

x = -2/ 3

Step-by-step explanation:

in order to cancel out the logs they should have common bases

\mathsf{ log_{5}(3x + 5) = 2 log_{25}(1 - 3x)  }

we can write 25 as 5²

\mathsf{ \implies  log_{5}(3x + 5) =  2 \: log_{ {5}^{2} }(1 - 3x)  }

we know that the reciprocal of the exponents of the bases are multiplied to the log

\mathsf{ \implies  log_{5}(3x + 5) =   \frac{1}{\cancel{2} } \times \cancel{2} \: log_{ 5}(1 - 3x)  }

and now since the logs have common bases

\mathsf{ \implies  \cancel { log_{5}}(3x + 5) = \cancel{ log_{ 5}}(1 - 3x)  }

we're left with

\mathsf{ \implies 3x + 5 = 1 - 3x}

\mathsf{ \implies 9x = -4}

<u>x = -2/ 3</u>

8 0
3 years ago
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