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3 years ago

ALMOST DONE! TRY TO ANWSER CORRECTLY‼️‼️‼️‼️‼️‼️‼️‼️‼️‼️‼️‼️‼️

Mathematics

2 answers:

Bas_tet [7]3 years ago

8 0

The answer is true lol

Nat2105 [25]3 years ago

3 0

It’s answer would be true

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If f(x) = 3^x + 10x and g(x) = 2x - 4, find (f + g)(x)

Karo-lina-s [1.5K]

Answer:

3^x + 10x+2x-4

Step-by-step explanation:

f(x) = 3^x + 10x and g(x) = 2x - 4

(f + g)(x)= 3^x + 10x+2x-4

3 0

2 years ago

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Simplify (2x²-6x-8)÷(x-4)

Norma-Jean [14]

Answer: Heyaa!

The Answer is 2x+2

Step-by-step explanation:

Simplify the expression.

hopefully this helps you !

- Matthew ~

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1 year ago

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What is the gcf for 28 and 70

Lena [83]

28 factors: 1,2,4,7,14,28
70 factors: 1,2,5,7,10,14,35,70
GCF : 14

8 0

3 years ago

Short Response

garik1379 [7]

Answer:

(a)2(89x+84)

(b) $2(89x+84-x^2\pi)$

Step-by-step explanation:

The dimensions of the larger rectangular field are:

Length=5x + 12; Width = 9x + 14.

The dimensions of the smaller rectangular soccer field are:

Length=5x; Width = 9x.

(a)Area of the part of the field that is outside the soccer field

=Area of the larger rectangular field - Area of the Soccer Field

=(5x+12)(9x+14)-5x(9x)

=(5x)(9x)+70x+108x+168-5x(9x)

=178x+168

=2(89x+84)

(b)Radius of the Semicircular Fountain =2x

From Part (a),

Area of the larger rectangular field - Area of the Soccer Field=178x+168

Area of the Semicircular Fountain = $\dfrac{\pi r^2}{2} =\dfrac{\pi (2x)^2}{2} =\dfrac{4x^2\pi}{2} =2x^2\pi$

Area of the Field that does not include the soccer field or the fountain.

=Area of the larger rectangular field - Area of the Soccer Field-Area of the Semicircular Fountain

$=178x+168-2x^2\pi\\=2(89x+84-x^2\pi)$

8 0

3 years ago

A miniature quadcopter is located at xi = −1.75 m and yi = −5.70 m at t = 0 and moves with an average velocity having components

Black_prince [1.1K]

The quadcopter's average velocity is given in terms of change in position by

$\vec v_{\rm av}=\dfrac{\Delta x}{\Delta t}\,\vec\imath+\dfrac{\Delta y}{\Delta t}\,\vec\jmath$

where $\Delta x=x_f-x_i$ , the difference in the quadcopter's final and initial positions and $\Delta t=(1.60-0)\,\mathrm s=1.60\,\mathrm s$ .

The $x$ -component of the average velocity is 2.70 m/s, so

$2.70\dfrac{\rm m}{\rm s}=\dfrac{x_f-(-1.75\,\mathrm m)}{1.60\,\rm s}\implies \boxed{x_f=2.57\,\mathrm m}$

and the $y$ -component is -2.50 m/s, so

$-2.50\dfrac{\rm m}{\rm s}=\dfrac{y_f-(-5.70\,\mathrm m)}{1.60\,\rm s}\implies\boxed{y_f=-9.70\,\mathrm m}$

4 0

2 years ago