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3 years ago

PLEASE HELP ASAP !! (Picture included)

Mathematics

1 answer:

slega [8]3 years ago

8 0

This is the question that's gonna put me over the top and make me a Brainly Genius. Of course working for free doing middle school math homework doesn't really sound like a genius, but I'm doing it anyway.

This one is the Pythagorean Theorem of course:

$(x+3)^2 + (x+4)^2 = (x+10)^2$

$x^2 + 6x + 9 + x^2 + 8x + 16 = x^2 + 20 x + 100$

$x^2 - 6x - 75 = 0$

The Shakespeare Quadratic Formula (2b or -2b) is a little shortcut for the quadratic formula when the middle term is even.

$x^2-2bx+c \textrm{ has zeros } x=b\pm \sqrt{b^2-c}$

So we get

$x = 3 \pm \sqrt{3^2 -(-75)} = 3 \pm \sqrt{84} = 3 \pm 2 \sqrt{21}$

The minus root is less than -6 giving negative sides, so we reject it and conclude

$x = 3 + 2 \sqrt{21}$

Choice C

When I hit Add I officially become a Genius. Maybe I should retire.

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What are the amplitude, period, and phase shift of the given function? f(t)=-1/2sin(3t-2pi)

Romashka [77]

Answer:

The $amplitude=|A|=\frac{1}{2}$ , $period=3$ and $\text{phase shift}=\frac{2\pi}{3}$ .

Step-by-step explanation:

The given function is

$f(t)=-\frac{1}{2}\sin (3t-2\pi)$ .... (1)

The general form of sine function is

$F(t)=A\sin (Bt-C)+D$ ....(2)

where, |A| is the amplitude, B is period, D is the vertical shift (up or down), and C/B is used to find the phase shift.

On comparing (1) and (2), we get

$A=-\frac{1}{2}$

$B=3$

$C=2\pi$

$D=0$

So,

$amplitude=|A|=\frac{1}{2}$

$period=3$

$\text{phase shift}=\frac{2\pi}{3}$

Therefore $amplitude=|A|=\frac{1}{2}$ , $period=3$ and $\text{phase shift}=\frac{2\pi}{3}$ .

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3 years ago

give the slope of the line (if it exists) and the y intercept (if it exists) and graph the line 5y=3. what is the slope?

OLga [1]

Answer:

Step-by-step explanation:

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3 years ago

The probability all of the students in this class will be present and on time is 3/10. The probability all of the students are p

cestrela7 [59]

Answer:

3/4

Step-by-step explanation:

A: students present
B: students on time

P(all students present and on time) = P(A and B) = 3/10
P(all students present) = P(A) = 2/5

P(A and B) = P(A).P(B|A) where P(B|A) is the probability of everyone being on time given that everyone is present

So P(B|A) = P(A and B) /P(A) = 3/10 ÷ 2/5 = 3/10 * 5/2 = 15/20 which can be reduced to 3/4 by dividing numerator and denominator by 5

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2 years ago

JJill begins the summer with $500 in her savings account. Each week, she withdraws $20. Her brother, Sam,starts the summer with

meriva

Answer:

7th week

Step-by-step explanation:

Let us represent the number of weeks as x

Jill begins the summer with $500 in her savings account. Each week, she withdraws $20.

$500 - $20× x

500 - 20x

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$150 + $30 × x

150 + 30x

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We solve this be Equating both Equations together

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