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katovenus [111]
2 years ago
11

Write the quadratic equation in standard form and then choose the value of "b." (2x - 1)(x + 5) = 0

Mathematics
1 answer:
Sunny_sXe [5.5K]2 years ago
3 0
If we were to foil
after experieence

we know
ax²+bx+c=0

and
in form
(ax+b)(cx+d)=0
if we expand it, we get
acx²+bcx+adx+bd=0
or
(ac)x²+(bc+ad)x+(bd)=0
compare to
ax²+bx+c=0
we notice that the middle terms (x terms) are

b=(bc+ad)
so

in form
(2x-1)(1x+5)
b=bc+ad=(-1*1+2*5)=-1+10=9

b=9
or you could just expand it
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Well... if 450 is 90% of the crates then 500 is 100%. 450 crates is 0.90% of 500 crates.
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Sapphire and Trinity are shopping for school clothes. Sapphire has $50 and a coupon for a $10 discount at a boutique where each
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Answer:

susan's: 10/3

Bonnie's:5

Step-by-step explanation:

6 0
2 years ago
Find the value(s) of x where the tangent to the graph of y=e^5x is parallel to the
Alik [6]

Answer:

There are none.

Step-by-step explanation:

<u>No calculus involved:</u>

The line, in slope-intercept form, has equation y=-10x+17, ie is always decreasing (easy to spot applying the definition)

Meanwhile, y=e^{5x} is always increasing over its domain.

At no point the tangent will be decreasing.

<u>Let's use calculus</u>

We are to solve the equation y'(x) = -10 \rightarrow 5e^{5x} = -10 \rightarrow e^{5x}=-2 which has no real solutions.

8 0
2 years ago
Can someone solve this?
ladessa [460]
  • First question:

Recall that \cos^2x+\sin^2x=1 and \sqrt{x^2}=|x| for all x. So

\sqrt{1-\cos^2x}=\sqrt{\sin^2x}=|\sin x|

\sqrt{1-\sin^2x}=\sqrt{\cos^2x}=|\cos x|

For 0, we expect both \cos x>0 and \sin x>0 (i.e. the sine and cosine of any angle that lies in the first quadrant must be positive). By definition of absolute value, |x|=x if x>0.

So we have

\dfrac{\sqrt{1-\cos^2x}}{\sin x}+\dfrac{\sqrt{1-\sin^2x}}{\cos x}=\dfrac{\sin x}{\sin x}+\dfrac{\cos x}{\cos x}=1+1=\boxed{2}

making H the answer.

  • Second question:

C is always true, because the inequality reduces to x > y.

6 0
3 years ago
Find the nth term of the arithmetic sequences<br> a1=5,d=6,n=11
ki77a [65]

here's the solution,

  • n = 11
  • a = 5 ( a = first term )
  • d = 6 ( d = common difference )

we know,

=》

nth  \: \: term \:  = a   \: + (n - 1) \times d

=》

11th \:  \: term   = 5 + (11 - 1) \times 6

=》

11th \:  \: term = 5 + (10 \times 6)

=》

11th \:  \: term  = 5 + 60

=》

11th \:  \: term = 65

nth term ( 11th term ) = 65

3 0
3 years ago
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