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4 years ago

Pls help me on this 1 sum 1 its important

Mathematics

2 answers:

kaheart [24]4 years ago

6 0

Answer:

First option

A. y = 3.15 + 2.75(x - 1)

Step-by-step explanation:

d = 5.9 - 3.15 = 2.75

f(1) = 3.15

so the rule:

y = f(1) + d(n - 1)

y = 3.15 + 2.75(x - 1)

Mashutka [201]4 years ago

4 0

A.y=3.15+2.75(x-1) because if you substitute the x values in the equation you will get all the y values

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Combine the radicals 9✔x -3✔x

Anit [1.1K]

Answer:

6√x

Step-by-step explanation:

Given :-

9√x - 3√x .

For combining , take √x as common ,

=> √x ( 9 - 3 )

=> √x * 6

=> 6√x

7 0

3 years ago

Half a number is thirty-two

alukav5142 [94]

Answer:

64?

Step-by-step explanation:

32+32=64

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3 years ago

Let h(x) = 8x-3 What is h(5)? A. 40 B. 37 C. 43 D. 45

Alex

I believe your answer is B.37 because if u make the x a 5 like shown in the equation it now looks like h(5)=8(5)-3 so now you solve it and i got 37 i hope this helps you!

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3 years ago

Show that a = −1 + √3i and b = 2 satisfy 1/a+b=1/a + 1/b

Zarrin [17]

Answer:

LHS = $\frac{1 - \sqrt3i}{4}$ = RHS = $\frac{1 - \sqrt3i}{4}$

Step-by-step explanation:

Data provided in the question:

a = −1 + √3i and b = 2

to prove:

$\frac{1}{a+b}=\frac{1}{a} + \frac{1}{b}$

Considering the LHS

⇒ $\frac{1}{a+b}$

substituting the value of a and b, we get

⇒ $\frac{1}{−1 + \sqrt3i+2}$

⇒ $\frac{1}{1 + \sqrt3i}$

on multiplying and dividing by conjugate ( 1 - √3i )

we get

$\frac{1}{1 + \sqrt3i}\times\frac{1 - \sqrt3i}{1 - \sqrt3i}$

$\frac{1 - \sqrt3i}{(1^2 - (\sqrt3i)^2}$

$\frac{1 - \sqrt3i}{1 + 3}$ (as (√i)² = -1 )

$\frac{1 - \sqrt3i}{4}$

Now,

considering the RHS

$\frac{1}{a} + \frac{1}{b}$

substituting the value of a and b, we get

⇒ $\frac{1}{-1 + \sqrt3i} + \frac{1}{2}$

⇒ $\frac{2\times1 + ( -1 + \sqrt3i)\times1}{(-1 + \sqrt3i)\times2}$

⇒ $\frac{2 + ( -1 + \sqrt3i)}{(-1 + \sqrt3i)\times2}$

⇒ $\frac{1 + \sqrt3i}{(-1 + \sqrt3i)\times2}$

now,

on multiplying and dividing by conjugate ( -1 - √3i )

we get

$\frac{1 + \sqrt3i}{(−1 + \sqrt3i)\times2}\times\frac{-1 - \sqrt3i}{-1 - \sqrt3i}$

$\frac{1 + \sqrt3i}{(−1 + \sqrt3i)\times2}\times\frac{-1( 1 + \sqrt3i)}{-1 - \sqrt3i}$

$\frac{(1 + \sqrt3i}^2\times(-1){((-1)^2 - (\sqrt3i)^2)\times2}$

$\frac{(1^2 + (\sqrt3i)^2+2(1)(\sqrt3i)\times(-1)}{(1 + 3)\times2}$

$\frac{(1 - 3 + 2\sqrt3i)\times(-1)}{(4)\times2}$

$\frac{(-2 + 2\sqrt3i)\times(-1)}{(4)\times2}$

$\frac{-2( 1 - 2\sqrt3i)\times(-1)}{(4)\times2}$

$\frac{( 1 - 2\sqrt3i)}{(4)}$

Since, LHS = RHS

hence satisfied

3 0

3 years ago

Round 7.2352297363 to nearest millionth

Elodia [21]

Answer:

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Step-by-step explanation

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6 0

3 years ago

Read 2 more answers