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GaryK [48]
2 years ago
7

After being rearranged and simplified which equation cannot be solved using the quadratic formula?

Mathematics
2 answers:
sleet_krkn [62]2 years ago
8 0

Answer:

B

Step-by-step explanation:

5x^2 -3x+10 = 5x^2

Subtracting 5x^2 from each side

5x^2-5x^2 -3x+10 = 5x^2-5x^2

-3x+10 =0

We cannot use the quadratic formula since a =0

Marina86 [1]2 years ago
7 0

Answer:

B

Step-by-step explanation:

The answer is B because when you simplify it you get -3x + 10 = 0. Since this is no longer a quadratic equation, you can't solve it using the quadratic formula.

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vredina [299]

You're looking for the extreme values of x^2+3y^2+13x subject to the constraint x^2+y^2\le1.

The target function has partial derivatives (set equal to 0)

\dfrac{\partial(x^2+3y^2+13x)}{\partial x}=2x+13=0\implies x=-\dfrac{13}2

\dfrac{\partial(x^2+3y^2+13x)}{\partial y}=6y=0\implies y=0

so there is only one critical point at \left(-\dfrac{13}2,0\right). But this point does not fall in the region x^2+y^2\le1. There are no extreme values in the region of interest, so we check the boundary.

Parameterize the boundary of x^2+y^2\le1 by

x=\cos u

y=\sin u

with 0\le u. Then t(x,y) can be considered a function of u alone:

t(x,y)=t(\cos u,\sin u)=T(u)

T(u)=\cos^2u+3\sin^2u+13\cos u

T(u)=3+13\cos u-2\cos^2u

T(u) has critical points where T'(u)=0:

T'(u)=-13\sin u+4\sin u\cos u=\sin u(4\cos u-13)=0

(1)\quad\sin u=0\implies u=0,u=\pi

(2)\quad4\cos u-13=0\implies\cos u=\dfrac{13}4

but |\cos u|\le1 for all u, so this case yields nothing important.

At these critical points, we have temperatures of

T(0)=14

T(\pi)=-12

so the plate is hottest at (1, 0) with a temperature of 14 (degrees?) and coldest at (-1, 0) with a temp of -12.

4 0
1 year ago
A lab technician is tested for her consistency by making multiple measurements of the cholesterol level in one blood sample. The
Zanzabum

Step-by-step explanation:

Given precision is a standard deviation of s=1.8, n=12,  target precision is a standard deviation of σ=1.2

The test hypothesis is

H_o:σ <=1.2

Ha:σ > 1.2

The test statistic is

chi square = \frac{(n-1)s^2}{\sigma^2}

=\frac{(12-1)1.8^2}{1.2^2}

=24.75

Given a=0.01, the critical value is chi square(with a=0.01, d_f=n-1=11)= 3.05 (check chi square table)

Since 24.75 > 3.05, we reject H_o.

So, we can conclude that her standard deviation is greater than the target.

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2 years ago
In the figure, the slope of mid-segment DE is -0.4. The slope of segment AC is . i need help fast plz
AlexFokin [52]
It will have the same value. The slope of the mid-segment of DE is just equal to the slope of the segment of AC. This is because of the mid-segment theorem stating that both slopes of a triangle have same value. So the answer is -0.4 for both segments.
8 0
1 year ago
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2. Starting at a fixed time, each car entering an intersection is observed to see whether it turns left (L), right (R), or goes
jasenka [17]

Answer:

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The possible values of X are integers greater than zero, which are known as the Natural numbers.

If X = number of cars observed, simply count the number of letters in each outcome for the value of X:

Outcome = RRL, AARRL, AARL, RRAL, ARL

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natita [175]

Answer:A

Step-by-step explanation:

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