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3 years ago

13

I need help with a math problem

1 answer:

Neko [114]3 years ago

7 0

Which one I can help with anyone

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Let sin(47o)=0.7314 . Enter an angle measure (β ), in degrees, for cos(β)=0.7314

amid [387]

Sinx = cosb => x + b = 90 <=> b = 90 - 47 = 43o

8 0

3 years ago

Can you help me answer these two questions please

vredina [299]

The answer foe 21 is d

5 0

3 years ago

Read 2 more answers

There are 4 jacks in a standard deck of 52 playing cards. If patricia selects a card at random, what is the probability that it

sammy [17]

Answer:

(C) 1/13

Step-by-step explanation:

Since, In a pack of 52 playing cards, there are 4 jacks in total.

If patricia selects a card at random, then the probability that it will be a jack will be:

Probability it is a jack= $\frac{Total number of jacks}{Total number of cards}$

= $\frac{4}{52}$

= $\frac{1}{13}$

Which is the required probability.

5 0

3 years ago

Read 2 more answers

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

2 years ago

Give an example of a quadratic equation with non-real solutions.

alexgriva [62]

Answer:

x² + 1 = 0

Step-by-step explanation:

You can form any with discriminant negative (B²-4AC < 0)

An example could be:

x² + 1 = 0

x² = -1 which has no real roots/solutions

3 0

3 years ago