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2 years ago

Explain the steps needed to solve the inequality. -3x + 4 > 25

Mathematics

1 answer:

r-ruslan [8.4K]2 years ago

6 0

Step-by-step explanation:

-3x + 4 > 25

-3x > 21 (subtract 4 on both sides)

x < -7. (divide -3 on both sides, flip the sign)

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Graph the system below amd write its solution 2x+y=6

Anastasy [175]

The answer to the question

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3 years ago

What are the solutions of x^2+6x=10

posledela

Answer:

x=-3+√19, x=-3-√19

Step-by-step explanation:

x^2+6x=10

x^2+6x-10=10-10

x^2+6x-10=0

a=1 b=6 c=-10 - use the quadratic formula

x=-3+√19, x=-3-√19

I hope this helps!

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3 years ago

Please help! Consider the given circle with the shaded sector xy and central angle 225. The circumference is 30 units

OleMash [197]

$\textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ C=30\pi \end{cases}\implies 30\pi =2\pi r\implies \cfrac{30\pi }{2\pi }=r\implies \boxed{15=r} \\\\[-0.35em] ~\dotfill$

$\textit{arc's length}\\\\ s = \cfrac{\theta \pi r}{180}~~ \begin{cases} r=radius\\ \theta =\stackrel{degrees}{angle}\\[-0.5em] \hrulefill\\ r=15\\ \theta =225 \end{cases}\implies \widehat{XY}=\cfrac{(225)\pi (15)}{180}\implies \boxed{\widehat{XY}\approx 58.90} \\\\[-0.35em] ~\dotfill\\\\ \textit{area of a circle}\\\\ A=\pi r^2\qquad \qquad A=\pi (15)^2\implies A=225\pi \implies \boxed{A\approx 706.86}$

6 0

2 years ago

If f(x) varies directly with x, and f(x) = 12 when x = 8, write the direct linear variation equation.

Feliz [49]

<span>f(x) varies directly with x.
f(x) = kx (k is the constant of proportionality)
if x = 8,
12 = 8 k
k = 12/8
k = 3/2
So the equation would be:
f(x) = (3/2)x
</span><span>
</span>

4 0

3 years ago

Show that if the vector field F = Pi + Qj + Rk is conservative and P, Q, R have continuous first-order partial derivatives, then

olchik [2.2K]

Answer:

It is proved that $\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}, \frac{\partial P}{\partial z}=\frac{\partial R}{\partial x}, \frac{\partial Q}{\partial z}=\frac{\partial R}{\partial y}$

Step-by-step explanation:

Given vector field,

$F=P\uvec{i}+Q\uvec{j}+R\uvec{k}$

Where,

$P=f_x=\frac{\partial f}{\partial x}, Q=f_y=\frac{\partial f}{\partial y}, R=f_z=\frac{\partial f}{\partial z}$

To show,

$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}, \frac{\partial P}{\partial z}=\frac{\partial R}{\partial x}, \frac{\partial Q}{\partial z}=\frac{\partial R}{\partial y}$

Consider,

$\frac{\partial P}{\partial y}=\frac{\partial}{\partial y}(\frac{\partial f}{\partial x})=\frac{\partial^2 f}{\partial y\partial x}=\frac{\partial^2 f}{\partial x\partial y}=\frac{\partial }{\partial x}(\frac{\partial f}{\partial y})=\frac{\partial Q}{\partial x}$

$\frac{\partial P}{\partial z}=\frac{\partial}{\partial z}(\frac{\partial f}{\partial x})=\frac{\partial^2 f}{\partial z\partial x}=\frac{\partial^2 f}{\partial x\partial z}=\frac{\partial }{\partial x}(\frac{\partial f}{\partial z})=\frac{\partial R}{\partial x}$

$\frac{\partial Q}{\partial z}=\frac{\partial}{\partial z}(\frac{\partial f}{\partial y})=\frac{\partial^2 f}{\partial z\partial y}=\frac{\partial^2 f}{\partial y\partial z}=\frac{\partial}{\partial y}(\frac{\partial f}{\partial z})=\frac{\partial R}{\partial y}$

Hence proved.

4 0

2 years ago