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2 years ago

CAN SOMEONE HELP ME PLEASE

Mathematics

1 answer:

Tanya [424]2 years ago

6 0

Answer:

$\frac{\sqrt{189x^3y^6} }{\sqrt{3x} } \\= \frac{\sqrt{3^2(7)(3)(x^2)(x)(y^3)^2} }{\sqrt{3x} } \\=\sqrt{\frac{(3xy^3)^2(7)(3x)}{3x} }\\=3xy^3\sqrt{7}$

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4x-12y=-8 what is the answer

umka21 [38]

Answer:

You aren't asked which variable to solve for.

If we're solving for x then

4x = 12y -8

x = 3y - (8/4)

If we're solving for y

-12y = -4x -6

12y = 4x +6

y = (x/3) + (1/2)

Step-by-step explanation:

8 0

2 years ago

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Nancy spent half of her allowance going to the movies. She washed the family car and earned 7 dollars. What is her weekly allowa

Sauron [17]

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as a contractor to make a hole 4 5/8 inches diameter. express the mixed number as an improper fraction.

san4es73 [151]

The mixed number as an improper fraction is 37/8

<h3>How to express the mixed number as an improper fraction?</h3>

The diameter is give as:

Diameter = 4 5/8

Next we convert the above number to an improper fraction

The numerator is calculated as

Numerator = 8 * 4 + 5

This gives

Numerator = 37

Hence, the mixed number as an improper fraction is 37/8

Read more about fractions at:

brainly.com/question/18734292

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7 0

1 year ago

A brick is thrown upward from the top of a building at an angle of 250 above the horizontal, and with an initial speed of 15 m/s

mojhsa [17]

Answer:

(a) 25.08 m

(b) 2.05 m

Step-by-step explanation:

Let the height of the building be 'h'.

Given:

Angle of projection is, $\theta=25$ °

Initial speed is, $u=15\ m/s$

Time of flight is, $t=3.0\ s$

(a)

Consider the vertical motion of the brick.

Vertical component of initial velocity is given as:

$u_y=u\sin\theta\\\\u_y=15\sin(25)=6.34\ m/s$

Vertical displacement of the brick is equal to the height of the building.

So, vertical displacement = $-h$ (Negative sign implies downward motion)

Acceleration is due to gravity in the downward direction. So,

Acceleration is, $g=-9.8\ m/s^2$

Now, using the following equation of motion;

$-h=u_yt+\frac{1}{2}gt^2\\-h=6.34(3)-\frac{9.8}{2}(3)^2\\\\-h=19.02-44.1\\\\-h=-25.08\\\\h=25.08\ m$

Therefore, the building is 25.08 m tall.

(b)

Let the maximum height be 'H'.

At maximum height, the vertical component of velocity is 0 as the brick stops temporarily in the vertical direction.

So, $v_y=0\ m/s$

Now, using the following equation of motion, we have:

$v_y^2=u_y^2+2gH\\\\0=(6.34)^2-2\times 9.8\times H\\\\19.6H=40.2\\\\H=\frac{40.2}{19.6}=2.05\ m$

Therefore, the maximum height of the brick is 2.05 m.

5 0

2 years ago

The value of gold in the year 2010 was approximately $1,800. The value of gold increases by approximately 9.5% a year.

Tju [1.3M]

Hope it helps see below

Answer: 1800e^(.95*x)

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2 years ago