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3 years ago

the quadratic of a certain quadratic function has no x-interecepts which of the following are possible values for discriminants

Mathematics

1 answer:

Snowcat [4.5K]3 years ago

4 0

Answer:

Some possible values are -5, -3, -2.

Step-by-step explanation:

Basically if d(discriminant)

d > 0 = 2 x intercepts,

d = 0 means one x intercept

d < 0 = 0 x intercepts

:) Hoped this helped :)

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Which of the following equations represents a nonlinear function?

Ainat [17]

Y=square root of x

All the others would be a straight line with an unchanging slope (meaning they're linear). Y=rootx is not linear.
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A stray dog ate 40 of your muffins. That was 5/6 of all them! How many are left?

lilavasa [31]

I think that there are 33 left.

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3 years ago

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F⃗ (x,y)=−yi⃗ +xj⃗ f→(x,y)=−yi→+xj→ and cc is the line segment from point p=(5,0)p=(5,0) to q=(0,2)q=(0,2). (a) find a vector pa

DerKrebs [107]

a. Parameterize $C$ by

$\vec r(t)=(1-t)(5\,\vec\imath)+t(2\,\vec\jmath)=(5-5t)\,\vec\imath+2t\,\vec\jmath$

with $0\le t\le1$ .

b/c. The line integral of $\vec F(x,y)=-y\,\vec\imath+x\,\vec\jmath$ over $C$ is

$\displaystyle\int_C\vec F(x,y)\cdot\mathrm d\vec r=\int_0^1\vec F(x(t),y(t))\cdot\frac{\mathrm d\vec r(t)}{\mathrm dt}\,\mathrm dt$

$=\displaystyle\int_0^1(-2t\,\vec\imath+(5-5t)\,\vec\jmath)\cdot(-5\,\vec\imath+2\,\vec\jmath)\,\mathrm dt$

$=\displaystyle\int_0^1(10t+(10-10t))\,\mathrm dt$

$=\displaystyle10\int_0^1\mathrm dt=\boxed{10}$

d. Notice that we can write the line integral as

$\displaystyle\int_C\vecF\cdot\mathrm d\vec r=\int_C(-y\,\mathrm dx+x\,\mathrm dy)$

By Green's theorem, the line integral is equivalent to

$\displaystyle\iint_D\left(\frac{\partial x}{\partial x}-\frac{\partial(-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=2\iint_D\mathrm dx\,\mathrm dy$

where $D$ is the triangle bounded by $C$ , and this integral is simply twice the area of $D$ . $D$ is a right triangle with legs 2 and 5, so its area is 5 and the integral's value is 10.

4 0

3 years ago

I need help please help me

iogann1982 [59]

Method 1 :

replace x and y by their value 1 and 3

3 = 4(1) - 1 = 4-1 = 3

2(1) + 3 = 2 + 3 = 5

correct

Method 2 :

y = 4x - 1

2x + y = 5

y = 4x - 1

y = -2x + 5

y - y = 4x - 1 - ( -2x + 5 )

0 = 4x - 1 + 2x - 5

6x - 6 = 0

6x = 6

x = 6/6 = 1

y = 4x - 1

y = 4(1) - 1

y = 3

correct

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2 years ago

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How do you solve interquartile range?

77julia77 [94]

By Hand
Step 1:
Put the numbers in order.
1, 2, 5, 6, 7, 9, 12, 15, 18, 19, 27.

Step 2:
Find the median.
1, 2, 5, 6, 7, 9, 12, 15, 18, 19, 27.

Step 3:
Place parentheses around the numbers above and below the median.
Not necessary statistically, but it makes Q1 and Q3 easier to spot.
(1, 2, 5, 6, 7), 9, (12, 15, 18, 19, 27).

Step 4:
Find Q1 and Q3
Think of Q1 as a median in the lower half of the data and think of Q3 as a median for the upper half of data.
(1, 2, 5, 6, 7), 9, ( 12, 15, 18, 19, 27). Q1 = 5 and Q3 = 18.

Step 5:
Subtract Q1 from Q3 to find the interquartile range.
18 – 5 = 13.

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3 years ago

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