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3 years ago

Use substitution to solve y= 4x - 6 and 5x +3y = -1HELP ASAP PLEASE!!!

Mathematics

1 answer:

ella [17]3 years ago

7 0

So the answer would be x=1 and y=-2. Hope this helps

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Solve the compound inequality.

aivan3 [116]

Option D:

$x$

Solution:

Given compound inequality:

$6-x>15 \text { or } 2 x-9 \geq 3$

Let us first solve $6-x>15$ .

Subtract 6 from both sides.

$6-x-6>15-6$

$-x>9$

To reverse the inequality multiply by -1 on both sides.

$(-x)(-1)$

$x$

Now solve the next inequality $2 x-9 \geq 3$ .

Add 9 on both sides.

$2 x-9+9 \geq 3+9$

$2 x \geq 12$

Divide by 2 on both sides.

$$\frac{2 x}{2} \geq \frac{12}{2}$

$x \geq 6$

Combine the inequality.

$x$

Option D is the correct answer.

5 0

3 years ago

What is the interquarile range of this box plot

wariber [46]

IQR = Q3 - Q1

= 67 - 45

= 22

Answer is B. 22

4 0

3 years ago

Read 2 more answers

The measure of an angle is 8 degrees less than three times the measure of another angle. If the two angles are supplementary (tw

Kitty [74]

Answer:

The measure of the larger is $133^{\circ}$ .

Step-by-step explanation:

Given:

The measure of an angle is $8$ degrees less than three times the measure of another angle.

The two angles are supplementary.

To find: The measure of the larger angle.

Solution:

Let the measure of the smaller angle be $x^{\circ}$ .

Then the measure of the larger angle be $3x^{\circ}-8^{\circ}$ .

The two angles are supplementary, so their sum is $180^{\circ}$ .

So, $x^{\circ}+3x^{\circ}-8^{\circ}=180^{\circ}$

$\Rightarrow 4x^{\circ}-8^{\circ}=180^{\circ}$

$\Rightarrow 4x^{\circ}=180^{\circ}+8^{\circ}$

$\Rightarrow 4x^{\circ}=188^{\circ}$

$\Rightarrow x^{\circ}=\frac{188^{\circ}}{4}$

$\Rightarrow x^{\circ}=47^{\circ}$

So, the measure of the smaller angle is $47^{\circ}$ .

And, the measure of the larger angle is $3\times47^{\circ}-8^{\circ}=133^{\circ}$ .

Hence, the measure of the larger is $133^{\circ}$ .

3 0

2 years ago

If (x-y)^2=71 and x^2+y^2=59 what is the value of xy?

Lena [83]

Solve the equations
first one
take the sqrt of both sides
x-y=√71
add y to both sides
x=y+√71
sub y+√71 for x in other part

(y+√71)²+y²=59
y²+2y√71+71+y²=59
2y²+2y√71+71=59
minus 59 both sides
2y²+2y√71+12=0
factor out 2
2(y²+y√71+6)=0
use quadratic equation or something
y= $\frac{- \sqrt{71}- \sqrt{47} }{2}$ and $\frac{- \sqrt{71}+ \sqrt{47} }{2}$

sub those for x

x=y+√71
note: √71=(2√71)/2

x= $\frac{- \sqrt{71}- \sqrt{47} +2\sqrt{71}}{2}$ , $\frac{\sqrt{71}- \sqrt{47}}{2}$
or
x= $\frac{- \sqrt{71}+ \sqrt{47} +2\sqrt{71}}{2}$ , $\frac{\sqrt{71}+ \sqrt{47}}{2}$

xy= $\frac{\sqrt{71}- \sqrt{47}}{2}$ times $\frac{-\sqrt{71}- \sqrt{47}}{2}$ or $\frac{\sqrt{71}+ \sqrt{47}}{2}$ times $\frac{-\sqrt{71}+ \sqrt{47}}{2}$

the result is -6 both times

xy=-6

5 0

3 years ago

Compute the differential of surface area for the surface S described by the given parametrization.

AysviL [449]

With $S$ parameterized by

$\vec r(u,v)=\langle e^u\cos v,e^u\sin v,uv\rangle$

the surface element $\mathrm dS$ is

$\mathrm dS=\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

We have

$\dfrac{\partial\vec r}{\partial u}=\langle e^u\cos v,e^u\sin v,v\rangle$

$\dfrac{\partial\vec r}{\partial v}=\langle -e^u\sin v,e^u\cos v,u\rangle$

with cross product

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\langle ue^u\sin v-ve^u\cos v,-ve^u\sin v-ue^u\cos v,e^{2u}\cos^2v+e^{2u}\sin^2v\rangle$

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\langle e^u(u\sin v-v\cos v),-e^u(v\sin v+u\cos v),e^{2u}\rangle$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=\sqrt{e^{2u}(u\sin v-v\cos v)^2+e^{2u}(v\sin v+u\cos v)^2+e^{4u}}$

$\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=e^u\sqrt{u^2+v^2+e^{2u}}$

So we have

$\mathrm dS=\boxed{e^u\sqrt{u^2+v^2+e^{2u}}\,\mathrm du\,\mathrm dv}$

8 0

3 years ago