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Viktor [21]
3 years ago
9

A total that is divided into equal groups is called the

Mathematics
1 answer:
brilliants [131]3 years ago
8 0

<em>Greetings!</em>

In division, the number being divided is considered the dividend, and the number dividing is called the<em> divisor</em>. The answer is the <em>quotient</em>.

<em>A total that is divided into equal groups is called the </em><u><em>dividend</em></u><em>.</em>

Hope this helps!

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The line through (-2, 3) with slope 2/3 in point slope form​
Debora [2.8K]

Answer:

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

Step-by-step explanation:

slope = 2/3

point-slope form --> y - y1 = m(x - x1)

y - 3 = 2/3(x - -2)

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

point slope form of the line is y - 3 = 2/3(x + 2)

3 0
2 years ago
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Write the equation of the line that passes through the points (1, 1) and (2, 4). Write your answer in slope-intercept form
navik [9.2K]

Answer:

The answer in slope-intercept form is y = 3x - 2.

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3 years ago
Given a triangle with perimeter 63 cm, one side of which is 21 cm, and one of the medians is perpendicular to one of its angle b
kodGreya [7K]

Answer:

The lengths of all sides of the triangle will be 21 cm, 21 cm, and 21 cm

Step-by-step explanation:

Given that the perimeter of the triangle = 63 cm

And the length of one side = 21 cm

and one of the medians is perpendicular to one of its angle bisectors

Since triangle has 3 sides,

Median = 63 ÷ 3 = 21

Therefore each side length = 21 cm

3 0
2 years ago
An object is launched from a launching pad 144 ft. above the ground at a velocity of 128ft/sec. what is the maximum height reach
ollegr [7]

Answer:

18) a. h(x) = -16x² + vx + h(0) ⇒ h(x) = -16x² + 128x + 144

b. The maximum height = 400 feet

c. Attached graph

19) The rocket will reach the maximum height after 4 seconds

20) The rocket hits the ground after 9 seconds

Step-by-step explanation:

* Lets study the rule of motion for an object with constant acceleration

# The distance S = ut ± 1/2 at², where u is the initial velocity, t is the time

  and a is the acceleration of gravity

# The vertical distances h in x second is h(x) - h(0), where h(0)

   is the initial height of the object above the ground

∵ h(x) = vx + 1/2 ax², where h is the vrtical distance, v is the initial

  velocity, a is the acceleration of gravity (32 feet/second²) and x

  is the time

18)a.

∵ The value of a = -32 ft/sec² ⇒ negative because the direction

   of the motion

  is upward

∴ h(x) - h(0) = vx - (1/2)(32)(x²) ⇒ (1/2)(32) = 16

∴ h(x) = vx - 16x² + h(0)

∴ h(x) = -16x² + vx + h(0) ⇒ proved

* Find the height of the object after x seconds from the ground

∵ h(0) = 144 and v = 128 ft/sec

∴ h(x) = -16x² + 128x + 144

b.

* At the maximum height h'(x) = 0

∵ h'(x) = -32x + 128

∴ -32x + 128 = 0 ⇒ subtract 128 from both sides

∴ -32x = -128 ⇒ ÷ -32

∴ x = 4 seconds

- The time for the maximum height = 4 seconds

- Substitute this value of x in the equation of h(x)

∴ The maximum height = -16(4)² + 128(4) + 144 = 400 feet

c. Attached graph

19)

- The object will reach the maximum height after 4 seconds

20)

- When the rocket hits the ground h(x) = 0

∵ h(x) = -16x² + 128x + 144

∴ 0 = -16x² + 128x + 144 ⇒ divide the two sides by -16

∴ x² - 8x - 9 = 0 ⇒ use the factorization to find the value of x

∵ x² - 8x - 9 = 0

∴ (x - 9)( x + 1) = 0

∴ x - 9 = 0 OR x + 1 = 0

∴ x = 9 OR x = -1

- We will rejected -1 because there is no -ve value for the time

* The time for the object to hit the ground is 9 seconds

8 0
3 years ago
Helppppppp! 15 points!
Vika [28.1K]

{\boxed{\mathcal{\purple{D.\:81}}}} ✅

\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\orange{:}}}}}

( {3})^{ \frac{11}{5} }  \div  {3}^{ \frac{ - 9}{5} }  \\  \\=(  {3})^{ \frac{11}{5} - ( \frac{ - 9}{5} ) }  \\ \\ = ( {3})^{ \frac{11}{5}   +  \frac{9}{5} }  \\ \\ = ( {3})^{ \frac{11 + 9}{5} } \\ \\  = ( {3})^{ \frac{20}{5} }  \\ \\  = ( {3})^{4}  \\ \\ = ( \: 3 \times 3 \times 3 \times 3 \: ) \\ \\ = 81

<u>Note</u>:-

{a}^{m}  \div  {a}^{n}  =  {a}^{m - n}

\sf\red{(-)\:x\:(-)\:=\:+}

\large\mathfrak{{\pmb{\underline{\orange{Happy\:learning }}{\orange{!}}}}}

4 0
2 years ago
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