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3 years ago

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3x+ 32 +7x - 22 = 180

1 answer:

Sedaia [141]3 years ago

6 0

Add 3x and 7x because they’re on the same and when you added it you’ll get 10x and than do 32+-22 and you’ll get 10 because 32-22 is 10 and you subtract the 10 to 180 and you’ll get 10x=170 and divide 170 by 10x to get x by itself when you divided you’ll get x=17 and that’s the answer.

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Find equations of the spheres with center(1, −1, 6)that touch the following planes.a) xy-planeb) yz-planec) xz-plane

Misha Larkins [42]

Given :

Center of sphere , C( 1 , -1 , 6 ) .

To Find :

Find equations of the spheres with center (1, −1, 6) that touch the following planes.a) xy-plane b) yz-plane c) xz-plane .

Solution :

a)

Distance of the point from xy-plane is :

d = 6 units .

So , equation of circle with center C and radius 6 units is :

$(x-1)^2+(y-(-1))^2+(z-6)^2=6^2\\\\(x-1)^2+(y+1)^2+(z-6)^2=36$

b)

Distance of point from yz-plane is :

d = 1 unit .

So , equation of circle with center C and radius 1 units is :

$(x-1)^2+(x+1)^2+(z-6)^2=1^2\\\\(x-1)^2+(x+1)^2+(z-6)^2=1$

c)

Distance of point from xz-plane is :

d = 1 unit .

So , equation of circle with center C and radius 1 units is :

$(x-1)^2+(x+1)^2+(z-6)^2=1^2\\\\(x-1)^2+(x+1)^2+(z-6)^2=1$

Hence , this is the required solution .

4 0

3 years ago

What do I do on this problem please help

insens350 [35]

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4 years ago

Melissa said that each member of her group completed 5/12 foot of a tower for a garden display. There were 3 members in her grou

Setler [38]

A) 5/12×3=1 1/4 I know this for sure

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3 years ago

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Let V=ℝ2 and let H be the subset of V of all points on the line 4x+3y=12. Is H a subspace of the vector space V?

sergejj [24]

Answer:

No, it isn't.

Step-by-step explanation:

We have $V=IR^{2}$ and let $H$ be the subset of $V$ of all points on the line

$4x+3y=12$

We need to find if $H$ is a subspace of the vector space $V$ .

In $IR^{2}$ all the possibilities for own subspace of the vector space $IR^{2}$ are :

$IR^{2}$ itself.

The vector $0_{IR^{2}}=\left[\begin{array}{c}0&0\end{array}\right]$

All lines in $IR^{2}$ that passes through the origin ( $0_{IR^{2}}=\left[\begin{array}{c}0&0\end{array}\right]$ )

We know that $H$ is the subset of $IR^{2}$ of all points on the line $4x+3y=12$

If we look at the equation, the point $\left[\begin{array}{c}0&0\end{array}\right]$ doesn't verify it because :

$4x+3y=12\\4(0)+3(0)=12\\0=12$

Which is an absurd. Therefore, $H$ doesn't contain the origin (and $H$ is a line in $IR^{2}$ ). Finally, it can't be a vector space of $V=IR^{2}$

8 0

3 years ago

Please help me ASAP?!!

Blababa [14]

Answer:

(0,-5) and (-4,3)

Step-by-step explanation:

The given equations are

$\left \{ {{x^2+y^2=25} \atop {2x+y=-5}} \right.$

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$y=-2x-5$

Substitute into the first equation;

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Expand:

$x^2+4x^2+20x+25=25$

$x^2+4x^2+20x+25-25=0$

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$\Rightarrow x=0,x=-4$

When x=0,

$y=-2(0)-5=-5$

This gives; (0,-5)

When x=-4

$y=-2(-4)-5=3$

This gives (-4,3)

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3 years ago

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