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2 years ago

Pls guys need help...i have test tomorrow

Mathematics

2 answers:

Sergeu [11.5K]2 years ago

4 0

Answer:

18x

Step-by-step explanation:

3x^2 - 27 = -18x

x^2 - 3x = -1x

-18x / -1x = 18x

3x^2 = 9

x^2 = 2x

zalisa [80]2 years ago

4 0

Answer: 18x

Step-by -step: first you started with 3x^2 which is 9x then go to 9x-27 which is 18x
Next you go to x^2 with is also know as 1x^2 which can be 2. Then you subtract 2x -3x which can be 1x. Finally you divide 18x to 1x giving you the answer 18x

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dana is flying a kite whose string is 65 meters long and makes a 70 degree angle with the ground. How far is the kite above the

SSSSS [86.1K]

Here you would want to use the sine function, since it’s making a triangle with the ground, and length of the string, the height to the kite. This can be found by 65sin(70°) which is ≈ 61.08 meters

3 0

3 years ago

Write the equation in standard form for the circle that has a diameter with endpoints (22,0) and (2,0)

vovikov84 [41]

Answer:

(x - 12)² + y² = 100

Step-by-step explanation:

The standard form of the equation of a circle is;

(x - a)² + (y - b)² = r²

where:

a and b are the coordinates of the centre of the circle

r is the radius

We are given the coordinates of the endpoints of the diameter as; (22,0) and (2,0)

Thus, the centre of the circle would be at the mid point of the endpoints of the diameter.

Coordinates of the centre is;

((22 + 2)/2), (0 +0)/2))

This is;

(12, 0)

So, a = 12 and b = 0

Now,to get the radius r, we will use the formula;

r = √[(x2 - x1)² + (y2 - y1)²]

Where;

(x1, y1) and (x2, y2) are 2 points namely (12,0) and (22, 0)

r = √[(12 - 22)² + (0 - 0)²]

r = √(-10)²

r = √100

r = 10

Thus,equation of the circle is;

(x - 12)² + (y - 0)² = 10²

(x - 12)² + y² = 100

7 0

3 years ago

A girl's feet are negative 2 over 4 yards from the surface of a pool. A boy's feet are negative 3 over 4 yards from the surface

frozen [14]

The girls are closer. Her feet are 1/2 the distance away from the surface pool, the boys are 3/4 away.

6 0

2 years ago

Plz help I will mark you brainlist plz

joja [24]

Answer:

m = 12.9g

v = 8cm3

d = ?

d = m/v

d = 12.9g/8cm3

D = 1.6125g/cm3

5 0

2 years ago

Cylinders A and B are similar solids. The base of cylinder A has a circumference of 47 units. The base of cylinder B has an

fiasKO [112]

<h3><u>Question:</u></h3>

Cylinders A and B are similar solids. The base of cylinder A has a circumference of 4π units. The base of cylinder B has an area of 9π units.

The dimensions of cylinder A are multiplied by what factor to produce the corresponding dimensions of cylinder B?

<h3><u>Answer:</u></h3>

Dimensions of cylinder A are multiplied by $\frac{3}{2}$ to produce the corresponding dimensions of cylinder B

<h3><u>Solution:</u></h3>

Cylinders A and B are similar solids.

The base of cylinder A has a circumference of $4 \pi$ units

The base of cylinder B has an area of $9 \pi$ units

Let "x" be the required factor

From given question,

Dimensions of cylinder A are multiplied by what factor to produce the corresponding dimensions of cylinder B

Therefore, we can say,

$\text{Dimensions of cylinder A} \times x = \text{Dimensions of cylinder B }$

<h3><u>Cylinder A:</u></h3>

The circumference of base of cylinder (circle ) is given as:

$C = 2 \pi r$

Where "r" is the radius of circle

Given that base of cylinder A has a circumference of $4 \pi$ units

Therefore,

$4 \pi = 2 \pi r\\\\r = 2$

Thus the dimension of cylinder A is radius = 2 units

<h3><u>Cylinder B:</u></h3>

The area of base of cylinder (circle) is given as:

$A = \pi r^2$

Given that, the base of cylinder B has an area of $9 \pi$ units

Therefore,

$\pi r^2 = 9 \pi\\\\r^2 = 9\\\\r = 3$

Thus the dimension of cylinder B is radius = 3 units

$\text{Dimensions of cylinder A} \times x = \text{Dimensions of cylinder B }\\\\2 \times x = 3\\\\x = \frac{3}{2}$

Thus dimensions of cylinder A are multiplied by $\frac{3}{2}$ to produce the corresponding dimensions of cylinder B

5 0

2 years ago

Pls guys need help...i have test tomorrow​

Pls guys need help...i have test tomorrow