How many cubes with side 2cm are needed to make a cube with side 6cm​

Respuesta:

1

Explicación paso a paso:

Para completar el cuadrado tendremos que sumar la mitad del cuadrado del coeficiente de x a la expresión;

Coeficiente de x = -2

Mitad del coeficiente = (-2/2) = -1

Tomando el cuadrado del resultado = (-1) ^ 2 = 1

Por lo tanto, el número que se debe sumar para obtener un cuadrado perfecto es 1.

Answer:

Approximately 2

Step-by-step explanation:

103 divided by 53 is 1.94

Round 1. 94 up to a whole number is approximately 2 wood pieces.

Answer:

2000

Step-by-step explanation:

set up equation

.7x+x=3400

combine like terms

1.7x=3400

divide by 1.7 on both sides

x=2000

Step-by-step explanation:

since angles in a triangle add up to 180

<vuw=180-(71+23)

=86°

since angles in a straight line add up to 180

<vuf=180-86

=94

Answer:

[tex]theta = \frac{3}{2} \pi + 2k\pi[/tex]

Step-by-step explanation:

+2*k*pi is for the periodicity (if you turn 360° you get back to the same point)

Answer:

Sales tax: 402.80 Final cost: 5,702.80

Step-by-step explanation:

Sales price x sales tax rate = sales tax

5300 x .076 (7.6%) = 402.80

Sales price + tax = final cost

5300 + 402.80 = 5702.80

Answer:

The height of the tree=8.42 m

Step-by-step explanation:

We are given that

Height of Joshua, h=1.45 m

Length of tree's shadow, L=31.65 m

Distance between tree and Joshua=26.2 m

We have to find the height of the tree.

BC=26.2 m

BD=31.65m

CD=BD-BC

CD=31.65-26.2=5.45 m

EC=1.45 m

All right triangles are similar .When two triangles are similar then the ratio of their corresponding sides are equal.

[tex]\triangle ABD\sim \triangle ECD[/tex]

[tex]\frac{AB}{EC}=\frac{BD}{CD}[/tex]

Substitute the values

[tex]\frac{AB}{1.45}=\frac{31.65}{5.45}[/tex]

[tex]AB=\frac{31.65\times 1.45}{5.45}[/tex]

[tex]AB=8.42m[/tex]

Hence, the height of the tree=8.42 m

Answer:

Cylinder H has the greater volume.

Step-by-step explanation:

Recall that the volume of a cylinder is given by:

[tex]\displaystyle V=\pi r^2h[/tex]

Where r is the radius and h is the height.

Cylinder H has a radius of 4.5 meters and a height of 3 meters. Thus, its volume is:

[tex]\displaystyle \begin{aligned} V&=\pi(4.5)^2(3)\\&=60.75\pi \\&\approx190.8518\text{ m}^3\end{aligned}[/tex]

Cylinder J has a diameter of 7 meters and a height of 4.5 meters. The radius is half the diameter, so Cylinder J's radius is 3.5 meters. Thus, its volume is:

[tex]\displaystyle \begin{aligned}V&=\pi(3.5)^2(4.5)\\&=55.125\pi \\&\approx 173.1803\text{ m}^3\end{aligned}[/tex]

Thus, Cylinder H has the greater volume.

GCF=4

4(4x^2/4+-28x/4+-32f/4)

=4(x^2-7x-8f)

Hello,

[tex]4x^2-28x-32\\\\=4(x^2-7x-8)\\\\=4(x^2-8x+x-8)\\\\=4[x(x-8)+1(x-8)]\\\\=4(x-8)(x+1)\\[/tex]

Answer:

The simple answer is we round the square root of the number of data points. For example: 25 data points = 5 bars. 100 data points = 10 bars

Answer:

30 if that is an [tex]x^{2}[/tex] in g(x)

Step-by-step explanation:

f(2) = 2x + 1

      = 2(2) + 1

      = 4 + 1

f(2) = 5      

g(5) = [tex]x^{2}[/tex] + 5

      = [tex]5^{2}[/tex] + 5

      = 25 + 5

g(5) = 30

Answer: 30

Step-by-step explanation:

First find f(2)

f(2) = 2(2) + 1 = 4 + 1 = 5

Therefore g(f(2)) = g(5)

Now solve for g(5)

g(5) = (5)² + 5 = 25 + 5 = 30

Hello!

7(x + 8) = 84 <=>

<=> 7x + 56 = 84 <=>

<=> 7x = 84 - 56 <=>

<=> 7x = 28 <=>

<=> x = 28 : 7 <=>

<=> x = 4

Good luck! :)

Answer:

-6.895

Step-by-step explanation:

11.485

-18.38

______

-6.895

Answer:

Balance = 60x-1500

Step-by-step explanation:

Given data

total number of students= x student

amount paid by each student= Rs 60

Total amount= Rs 60*x

Advance paid for transport= Rs 1500

Hence the balance is = 60x-1500

Answer:

24 m³

Step-by-step explanation:

We need to find the volume of the prism. Given that , rectangular prism with cubic units arranged in two rows of two with six total layers . And the volume of each unit is 1 m³ .

Volume of one unit :-

[tex]\rm\implies Volume_{one\ unit } = 1m^3 [/tex]

By Unitary Method :-

Volume of two cubes , will be :-

[tex]\rm\implies Volume_{two\ unit } = 2m^3 [/tex]

Volume of one layer :-

[tex]\rm\implies Volume_{one\ layer } = 4m^3 [/tex]

Volume of six layers :-

[tex]\rm\implies Volume_{6\ layers } = 4m^3\times 6 \\\\\rm\implies \boxed{\rm Volume_{6\ layers } = 24 \ m^3} [/tex]

Answer:

Step-by-step explanation:

Perimeter of the rectangle = P

Base = b

Height = h

Area = A

P = 2b + 2h

P = 66

STEP 1:

2(2x + 1) + 2(x + 5) = 66

Distribute

4x + 2 + 2x + 10 = 66

STEP 2:

Combine like terms and isolate the variable

6x + 12 = 66

6x = 54

x = 9

STEP 3:

Plug in x

A = (2(9) + 1) * (9 + 5)

STEP 4:

Simplify

A = (18 + 1) * (14)

A = (19)(14)

A = 266

[tex]\displaystyle\bf P=2(2x+1+x+5)=66\\\\6x+12=66\\\\6x=54\\\\\boxed{x=9}\\\\2x+1=19\\\\x+5=14 \\\\S=ab=14\cdot19=266 cm^2[/tex]

Answer:

15 9 3

Step-by-step explanation:

15 9 3

sorry i guessed on this one

but it does work

when they said 3 in the first sentence i helped me guess

Answer:

first term = 15

second term = 9

Third term = 3

Step-by-step explanation:

Let the first term be 3x

Let the second term be = 3x - 6       [ 6 less than first term]

Let the third term be = 1/5 of 3x

                                 [tex]=\frac{1}{5} \times 3x\\\\=\frac{3}{5}x[/tex]

Difference between first and third term = 12

That is ,

           [tex]3x - \frac{3}{5}x = 12\\\\\frac{(3x \times 5) - 3x }{5} = 12\\\\15x - 3x = 12 \times 5\\\\12x = 60\\\\x = \frac{60}{12} = 5\\\\Therefore , \ first \ term = 3x = 3 \times 5 = 15\\\\second \term = 3x - 6 = 3( 5) - 6 = 15 - 6 = 9\\\\Third \ term = \frac{1}{5} \times 3x = \frac{1}{5} \times (3 \times 5) = 3[/tex]

Answer:

Point A, C and D

Point B

Step-by-step explanation:

[tex]{hope it helps[/tex]

thank you, you too! :D

Answer:

x = 140

Step-by-step explanation:

Given the equations

[tex]\frac{2}{5}[/tex] x - [tex]\frac{1}{5}[/tex] y = 98 ( multiply through by 5 to clear the fractions )

2x - y = 490 → (1)

[tex]\frac{2}{7}[/tex] x - [tex]\frac{1}{14}[/tex] y = 55 ( multiply through by 14 to clear the fractions )

4x - y = 770 → (2)

Subtract (1) from (2) term by term to eliminate y

2x + 0 = 280

2x = 280 ( divide both sides by 2 )

x = 140

First graph is the correct one.

when x= 1

y=f(x) = -√1 = -1

when x= 2

y= -√2

The result essentially follows directly from the triple-angle identities,

sin(3A) = 3 sin(A) cos²(A) - sin³(A)

cos(3A) = cos³(A) - 3 sin²(A) cos(A)

Then

sin(3A)/sin(A) = 3 cos²(A) - sin²(A)

cos(3A)/cos(A) = cos²(A) - 3 sin²(A)

and

sin(3A)/sin(A) - cos(3A)/cos(A)

= 3(cos²(A) + sin²(A)) - (sin²(A) + cos²(A))

= 3 - 1

= 2

Step-by-step explanation:

The correct answer is f(x)=3^(x+3)

[tex]\large {\text {$ \sf \cfrac{1}{2x^2} -2 = 0 $}}[/tex]

[tex]\large {\text {$ \sf \cfrac{1}{2x^2}\cdot \:2x^2-2\cdot \:2x^2=0\cdot \:2x^2 $}[/tex]

                 [tex]\searrow[/tex]

                     [tex]\large {\text {$ \sf 1-4x^2=0$}}[/tex]

[tex]\large {\text {$ \sf -4x^2+1 = 0 $}}[/tex]

[tex]\large {\text{$\sf x=\cfrac{-b\pm\sqrt{b^2-4ac}}{2a} \quad\rightarrow\quad x=\cfrac{-0\pm\sqrt{0^2-4\cdot (-4) \cdot1} }{2\cdot (-4) } \:\rightarrow\:\: x=\cfrac{-0^2 \pm4}{2 \cdot(-4)} $}}[/tex]  

                                                             [tex]\huge {\text {$ \sf \downarrow$}}[/tex]

           [tex]\large {\text {$\sf {\bf x_1} = \cfrac{-0+4}{2\left(-4\right)}= \cfrac{-1}{2} $}}[/tex]                        [tex]\large {\text {$\sf {\bf x_2 }=\cfrac{-0-4}{2\left(-4\right)} = \cfrac{1}{2} $}}[/tex]

[tex]\large {\boxed {\boxed { \bf x_1 + x_2= -\cfrac{1}{2}+ \cfrac{1}{2} = 0} }}[/tex]

                                           [tex]\huge {\text {$ \it Alternative \: A $}}[/tex]

Answer:

b

Step-by-step explanation:

12x12

11x11

10x10

9x9

8x8

7x7

6x6

5x5

4x4

3x3

2x2

1x1

add em together equal 650

Answer:

8 m

-----------------Work--------------

2*3 = 6

Ratio

2*4 = 8

Answer:

x = 3, x = 9

Step-by-step explanation:

When solving this problem, keep the general format of a logarithm in mind:

[tex]b^x=y\\log_b(y)=x[/tex]

Where, (b) represents the base, (x) is the exponent, and (y) is the evalutaor. Please note that others might use slightly different terminotoly than what is used in this answer.

One is given the following expression, and is asked to solve for the parameter (x);

[tex]log_5(x-4)=1-log_5(x-8)[/tex]

First, manipulate the exquestion such that all of the logarithmic expressions are on one side. Use inverse operations to do this.

[tex](log_5(x-4))+(log_5(x-8))=1[/tex]

Now use the Logarithmic Base Change rule to simplify. The Logarithmic Base Change rule states the following;

[tex]log_b(x)=\frac{log(x)}{log(b)}[/tex]

Remember, if no base is indicated in a logarithm, then the logarithm's base is (10). Apply the Logarithmic Base Change rule to this problem;

[tex]\frac{log(x-4)}{log(5)}+\frac{log(x-8)}{log(5)}=1[/tex]

Now remove the denominator. Multiply all terms in the equation by the least common denominator; ([tex]log(5)[/tex]) to remove it from the denominator on the left side.

[tex](\frac{log(x-4)}{log(5)}+\frac{log(x-8)}{log(5)}=1)*(log(5))[/tex]

[tex]log(x-4)+log(x-8)=log(5)[/tex]

All logarithms have the same base, the left side of the equation has the addition of logarithms. This means that one can apply the Logarithm product rule. The logarithm product rules the following;

[tex]log_b(x*y)=(log_b(x))+(log_b(y))[/tex]

This rule can be applied in reverse to simplify the left side of the equation. Rather than rewriting the product of logarithms as two separate logarithms being added, one can rewrite it as one logarithm getting multiplied.

[tex]log(x-4)+log(x-8)=log(5)[/tex]

[tex]log((x-4)(x-8))=log(5)[/tex]

Now used inverse operations to bring all of the terms onto one side of the equation:

[tex]log((x-4)(x-8))=log(5)[/tex]

[tex]log((x-4)(x-8))-log(5)=0[/tex]

Similar to the Logarithm product rule, the Logarithm quotient rule states the following;

[tex]log_b(x/y)=(log_b(x))-(log_b(y))[/tex]

One can apply this rule in reverse here to simplify the logarithms on the left side:

[tex]log((x-4)(x-8))-log(5)=0[/tex]

[tex]log(\frac{(x-4)(x-8)}{5})=0[/tex]

The final step in solving this equation is to use the Logarithm of (1) property. This property states the following:

[tex]log_b(1)=0[/tex]

When applying this property here, one can conclude that the evaluator must be equal to (1), therefore, the following statements can be made.

[tex]log(\frac{(x-4)(x-8)}{5})=0[/tex]

[tex]\frac{(x-4)(x-8)}{5}=1[/tex]

Inverse operations,

[tex]\frac{(x-4)(x-8)}{5}=1[/tex]

[tex](x-4)(x-8)=5[/tex]

[tex](x-4)(x-8)-5=0[/tex]

Simplify,

[tex](x-4)(x-8)-5=0[/tex]

[tex]x^2-12x+32-5=0[/tex]

[tex]x^2-12x+27=0[/tex]

Factor, rewrite the quadratic expression as the product of two linear expressions, such that when the linear expressions are multiplied, the result is the quadratic expression:

[tex]x^2-12x+27=0[/tex]

[tex](x-3)(x-9)=0[/tex]

Now use the zero product property to solve. The zero product property states that any number times (0) equals (0).

[tex]x=3,x=9[/tex]

Hello,

I suppose the question is solve for x.

[tex]\displaystyle log_5\ (a)=\dfrac{ln (a)}{ln (5)} \\\\log_5(x-4)=1-log_5(x-8)\\\\\dfrac{ln(x-4)}{ln(5)} =1- \dfrac{ln(x-8)}{ln(5)}\\\\ln(x-4)=ln(5)-ln(x-8)\\\\ln(x-4)+ln(x+8)=ln(5)\\\\ln((x-4)*(x-8))=ln(5)\\\\(x-4)*(x-8)=5\\\\x^2-12x+27=0\\\\\Delta=12^2-4*27=36=6^2\\\\x=9\ or\ x=3\\\\Sol=\{3,9\}\\[/tex]

For Moderators,

this is a mathematical resolution without any bla-bla sentences that you will easily find. (I can not do it sorry)

Answer:

135g

Step-by-step explanation:

[tex]\boxed{mass = density \times volume}[/tex]

Given: density= 5g/cm³, volume= 27cm³

Mass

= 5 ×27

= 135g

Answer:

Answer D is TRUE because on the graph if you were to keep going left even after the picture of the graph ends, the graph will still keep going up infinitely.

Step-by-step explanation: