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2 years ago

Find the sum or difference. 10/2x - 3/2x

Mathematics

1 answer:

mr Goodwill [35]2 years ago

4 0

Answer:

7/2x

Step-by-step explanation:

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HOWDY YALL 333 111 222 444 POOP

Savatey [412]

Answer:

333 111 poop to you to

Step-by-step explanation:

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2 years ago

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Abbey is adopting a kitten. Abbey has a rectangular container to bring the kitten home in. The container has a volume of 3,100

solniwko [45]

L=10.6in
H=15.6in
B=?

$\\ \rm\Rrightarrow V=LBH$

$\\ \rm\Rrightarrow 3100.5=10.6(15)B$

$\\ \rm\Rrightarrow 3100.5=159B$

$\\ \rm\Rrightarrow B=\dfrac{3100.5}{159}+19.5$

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2 years ago

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Find the point on the plane 2x+5y+z=8 that is nearest to (2,0,1).

Svetach [21]

Answer:

(2.2, 0.5, 1.1)

Step-by-step explanation:

The parametric equation of the line normal to the plane and through point (2, 0, 1) can be written ...

... L = (2, 0, 1) + t(2, 5, 1)

We want to find the value of t (and the corresponding point) that makes L satisfy the equation of the plane.

... 2(2+2t) +5(0 +5t) +1(1+t) = 8 . . . . . put values from L in for x, y, z in plane

... 5 + 30t = 8 . . . . . simplify

... t = (8 -5)/30 = 0.1 . . . . solve for t (subtract 5, divide by 30)

For this value of t, L is ...

... (2, 0, 1) + 0.1(2, 5, 1) = (2.2, 0.5, 1.1)

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3 years ago

There is a bag filled with 6 blue and 5 red marbles.

sergij07 [2.7K]

Answer:

0 percent

Step-by-step explanation:

it is a good question

in the question only there is written that he took *3 marbles*

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3 years ago

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From the side view, a gymnastics mat forms a right triangle with other angles measuring 60° and 30°. The gymnastics mat extends

valkas [14]

Answer: The mat is 4.33 ft high off the ground.

Step-by-step explanation:

Since we have given that

Angle of elevation with the first triangle = 30°

Angle of elevation with the second triangle = 60°

Length at which gymnastics mat extends across the floor = 5 feet

so, As shown in the figure:

We need to find the height of the mat off the ground.

If CD = 5 ft,

Let, AB = y, DC = x.

In Δ ABC,

$\tan 60^\circ=\frac{AB}{BC}\\\\\frac{\sqrt{3}}{2}=\frac{y}{x}\\\\x=\frac{y}{\sqrt{3}}$

Similarly, in Δ ACD,

$\tan 30^\circ=\frac{AB}{BD}\\\\\frac{1}{\sqrt{3}}=\frac{y}{x+5}\\\\\frac{1}{\sqrt{3}}=\frac{y}{\frac{y}{\sqrt{3}}+5}\\\\\frac{1}{\sqrt{3}}=\frac{y\sqrt{3}}{y+5\sqrt{3}}\\\\3y=y+5\sqrt{3}\\\\2y=5\sqrt{3}\\\\y=\frac{5\sqrt{3}}{2}\\\\y=4.33\ ft$

Hence, the mat is 4.33 ft high off the ground.

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3 years ago

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