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2 years ago

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Can someone help with this please

1 answer:

stiks02 [169]2 years ago

3 0

Answer:

Step-by-step explanation:

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I'll give brainly <3 Please help with these questions <3

Naily [24]

Difference would be 1000x3.0 - 8.8x10000 = -85000
Scientific notation would be 8.5x10^-4

7 0

3 years ago

The following diagram shows parallel lines cut be a transversal. What is the measure of Angle 2?

Crank

2x + 100 = 5x + 55
3x = 45
x = 15

2(15) + 100 = 130

180-130 = 50

so angle 2 is 50 degrees

3 0

3 years ago

Estimate the sum of 23 and 71

Natali [406]

Estimation:

23 --> rounded to 20
71 --> rounded to 70
70 + 20 = 90. I rounded both numbers down so the exact sum must be only a little greater than my estimate.

8 0

4 years ago

Read 2 more answers

A bathtub contains 12 gallons of water. With these 12 gallons, the bathtub is only 30% full. How many gallons of water can the b

Pachacha [2.7K]

Answer:

40 gallons

Step-by-step explanation:

Let the number of gallons that would completely fill the bathtub be represented by N.

Since 12 gallons of water gives 30% of the number of gallons to fill the bathtub, then;

30% of N = 12

$\frac{30}{100}$ x N = 12

30N = 12 x 100

30N = 1200

divide both sides by 30,

N = $\frac{1200}{30}$

= 40

The number of gallons that would completely fill the bathtub is 40 gallons.

6 0

3 years ago

Given the quadratic function f(x) = 4x^2 - 4x + 3, determine all possible solutions for f(x) = 0

solong [7]

Answer:

The solutions to the quadratic function are:

$x=i\sqrt{\frac{1}{2}}+\frac{1}{2},\:x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}$

Step-by-step explanation:

Given the function

$f\left(x\right)\:=\:4x^2\:-\:4x\:+\:3$

Let us determine all possible solutions for f(x) = 0

$0=4x^2-4x+3$

switch both sides

$4x^2-4x+3=0$

subtract 3 from both sides

$4x^2-4x+3-3=0-3$

simplify

$4x^2-4x=-3$

Divide both sides by 4

$\frac{4x^2-4x}{4}=\frac{-3}{4}$

$x^2-x=-\frac{3}{4}$

Add (-1/2)² to both sides

$x^2-x+\left(-\frac{1}{2}\right)^2=-\frac{3}{4}+\left(-\frac{1}{2}\right)^2$

$x^2-x+\left(-\frac{1}{2}\right)^2=-\frac{1}{2}$

$\left(x-\frac{1}{2}\right)^2=-\frac{1}{2}$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

solving

$x-\frac{1}{2}=\sqrt{-\frac{1}{2}}$

$x-\frac{1}{2}=\sqrt{-1}\sqrt{\frac{1}{2}}$ ∵ $\sqrt{-\frac{1}{2}}=\sqrt{-1}\sqrt{\frac{1}{2}}$

as

$\sqrt{-1}=i$

so

$x-\frac{1}{2}=i\sqrt{\frac{1}{2}}$

Add 1/2 to both sides

$x-\frac{1}{2}+\frac{1}{2}=i\sqrt{\frac{1}{2}}+\frac{1}{2}$

$x=i\sqrt{\frac{1}{2}}+\frac{1}{2}$

also solving

$x-\frac{1}{2}=-\sqrt{-\frac{1}{2}}$

$x-\frac{1}{2}=-i\sqrt{\frac{1}{2}}$

Add 1/2 to both sides

$x-\frac{1}{2}+\frac{1}{2}=-i\sqrt{\frac{1}{2}}+\frac{1}{2}$

$x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}$

Therefore, the solutions to the quadratic function are:

$x=i\sqrt{\frac{1}{2}}+\frac{1}{2},\:x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}$

4 0

3 years ago