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MakcuM [25]
3 years ago
7

I need help on this pleaseee

Mathematics
2 answers:
Nadya [2.5K]3 years ago
6 0

Answer:

2 pairs are 18

4 pairs are 36

6 pairs are 48

1 pair is 6

Step-by-step explanation:

Leona [35]3 years ago
3 0

Answer:

QUESTION:

I need help on this pleaseee...

ANSWER:

2 pairs of pants = $18

4 pairs of pants = $36

6 pairs of pants = $54

1 pair of pants = $6

Step-by-step explanation:

Hope that this helps you out! :)          

If you have any questions please put them in the comment section below this answer.          

Have a great rest of your day/night!          

Please thank me on my profile if this answer has helped you.

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An acute, isosceles triangle has a vertex angle measuring 78 degrees. What are the measures of the other two angles?
mixas84 [53]

Answer:

Step-by-step explanation:

180-78=102

102/2=51

6 0
3 years ago
Read 2 more answers
Solve the system by substitution.<br> y = 5x<br> y = 9x + 4
SVEN [57.7K]

Answer:

(-1, -5)

Step-by-step explanation:

\left \{ {{y=5x} \atop {y=9x+4}} \right.

1. We must calculate x, for that we must substitute y, so we have:

5x=9x+4

2. Now we solve the equation.

5x-9x=4

-4x=4

x=-1

3. Now we must calculate y

y=5x

y=5(-1)

y=-5

And the answer is (-1, -5)

5 0
3 years ago
4x(7x+5) what is the answer
guapka [62]
The answer is 28x^2+20x.

Hope this helps.
8 0
3 years ago
Read 2 more answers
Find the equation of the plane passing through the points
erica [24]

Extract the normal vectors from the given planes:

-2x+y+z+2=0\implies\vec n_1=(-2,1,1)

x+y-3z+1=0\implies\vec n_2=(1,1,-3)

(which are unique up to their signs, meaning either \vec n_1 or -\vec n_1 are valid choices for the normal vector)

The third plane must be perpendicular to both these given planes, which means it would be parallel to both \vec n_1 and \vec n_2, which in turn means its own normal vector \vec n_3 should be perpendicular to both \vec n_1 and \vec n_2.

Enter the cross product:

\vec n_3=\vec n_1\times\vec n_2=(-4,-5,-3)

or (4, 5, 3), which also works.

The given plane passes through (-1, 1, 4), so its equation is

(x+1,y-1,z-4)\cdot\vec n_3=0

Simplify:

(x+1,y-1,z-4)\cdot(4,5,3)=0

4(x+1)+5(y-1)+3(z-4)=0

\boxed{4x+5y+3z=13}

3 0
3 years ago
I need help with 1 and 2 please :)
KIM [24]
The first one should be 7/26
4 0
3 years ago
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