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2 years ago

BELL RINGER - 12/16/2020

Mathematics

2 answers:

Trava [24]2 years ago

5 0

Answer:

There is a 6% chance

Step-by-step explanation:

user100 [1]2 years ago

4 0

Answer:

6% chance

Step-by-step explanation:

12+19+14+15 = 60 60 divided by 100 = 0.6 = 6%

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What are the zeros of the function f(x) = x2 + 8x + 4, expressed in simplest radical form?

klio [65]

Answer:

x = -4 ± 2√3

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Expand this expression (x-y)²<br>

pochemuha

Answer:

x^2 - 2xy + y^2.

Step-by-step explanation:

(x - y)^2

= (x - y)(x - y)

= x(x -y) - y(x - y)

= x^2 - xy - xy + y^2

= x^2 - 2xy + y^2.

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2 years ago

Given the function f(x) = -x + 4, then what is f(-3x) as a simplified<br> polynomial?

quester [9]

Answer:

3x+4

Step-by-step explanation:

f(x) = - x+4

Substitute in x=-3x

f(-3x)= - (-3x) + 4 = 3x+4

5 0

2 years ago

Is the answer a, b, c, or d.

ale4655 [162]

The correct answer is d.

We have the following system of linear equations:

(I) $4x+7y=-14$
(II) $8x+5y=8$

Let's use the elimination method, then let's multiply the equation (1) $\times(-2)$ and subtracting (I) and (II):

(I) $-2(4x+7y)=-2(-14)$
∴ $-8x-14y=28$

(I) $-8x-14y=28$
(II) $8x+5y=8$
____________________
(III) $-9y=36$

∴ $y=-4$

We can find the value of x by substituting y either in (I) or (II). Thus, from (I):

$4x+7(-4)=-14$
∴ $4x-28=-14$
∴ $4x=14$
∴ $x=\frac{14}{4}$
∴ $\boxed{x=\frac{7}{2}}$

Let's substitute the values of x and y into (I) and (2)

(I) $4(\frac{7}{2})+7(-4)=14-28=-14$
(II) $8(\frac{7}{2})+5(-4)=28-20=8$

Finally the answer is:

$\boxed{x=\frac{7}{2} \rightarrow x= 3 \frac{1}{2}}$

4 0

2 years ago

Solve the given differential equation by using an appropriate substitution. the de is homogeneous. (y2 yx) dx − x2 dy = 0

Schach [20]

The solution for the given differential equation is lnlxl = $\frac{-x}{y} +C$

Given,

$(y^{2} +y^{x} )dx - x^{2} dy =0$

Here,

x = vy, dx = vdy + ydv

y = ux, dy = udx + xdu

Then,

$((ux)^{2} +(ux)x)dx-x^{2} (udx+xdu)=0\\=u^{2} x^{2} dx+ux^{2} dx-x^{2} udx-x^{3} du=0\\=u^{2} x^{2} dx=x^{3} du\\=\frac{x^{2} }{x^{3} } dx=\frac{1}{u^{2} } du$