An unknown metal having a mass of 287.8 g was added to a graduated cylinder that contained
31.47 mL of water. After the addition of the metal, the water level rose to 58.85 mL. Determine
the volume of the metal. Calculate the density of the metal using dimensional analysis.

Answer:

less then 90º

Step-by-step explanation:

an acute angle is an angle which is less then 90º

Answer:

$14.50

Step-by-step explanation:

Is the answer because we don't put a decimmel pint after the number

The property of real numbers demonstrated in the given statement is the "Reflexive Property of Equality."

The Reflexive Property of Equality is one of the fundamental properties of real numbers, stating that any real number is always equal to itself. In mathematical terms, for any real number 'a,' the Reflexive Property is expressed as:

a = a

This property is self-evident, as it simply means that any quantity or value is identical to itself. In the provided statement, the cost of one item sold for $14.50 is precisely $14.50, showing that the item's cost is equal to itself.

To further illustrate this concept, consider any real number 'x.' According to the Reflexive Property, we have:

x = x

This applies to all real numbers, whether they are whole numbers, decimals, fractions, negative numbers, or irrational numbers. For example:

1 = 1

5.3 = 5.3

-2 = -2

π = π (pi)

No matter what value 'x' represents, the Reflexive Property holds true for all real numbers. It is a fundamental principle that underpins many mathematical operations and proofs.

In summary, the Reflexive Property of Equality states that any real number is equal to itself. The given statement, "The cost of one item sold for $14.50 is $14.50," exemplifies this property, showing that $14.50 is indeed equal to $14.50, thus reaffirming the reflexive nature of real numbers.

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Answer:n =21

Step-by-step explanation:

Step-by-step explanation:

[tex]thank \: you[/tex]

Therefore, In the similarity transformation of ΔACB to ΔEFD, ΔACB was dilated by a scale factor of 2, reflected across the y-axis, and moved through the translation 2 units

What is graph transformation?

The technique of altering an existing graph or graphed equation to create a different version of the following graph is known as graph transformation. In particular, the change of algebraic equations, it is a typical kind of algebraic problem.

Graphs can be extended, rotated, inverted, or any combination of these transformations; they can also be translated or moved along the xyxy plane. The formulas "move the function f(x)f(x) by cc units in some direction," "stretch the function f(x)f(x) by cc units," and "rotate the function f(x)f(x) by xx units about the xx-, yy-, or zz-axis" are frequently used in mathematical issues. In every instance, the transformation has specific, calculable effects on the underlying function.

Since the triangle ABC and DEF is similar

So we can write,

DE/BC=EF/AB=FD/CA

scale factor =Length of CA/length of DF

                     =6/3

                      =2:1

Reflected at y axis i.e x=0

and raised by 2 units.

Therefore, In the similarity transformation of ΔACB to ΔEFD, ΔACB was dilated by a scale factor of 2, reflected across the y-axis, and moved through the translation 2 units

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Step-by-step explanation:

for a=

[tex] \frac{s - n}{n} [/tex]

Answer:

False, assuming "combined" means "added or subtracted and would simplify."

[tex]x^2[/tex] and [tex]x[/tex] cannot be combined with addition or subtraction.  They are not like terms.

Step-by-step explanation:

Combining like terms only works if the variables and the powers on the variables match perfectly.  

Examples:

[tex]x^2[/tex] and [tex]x[/tex] don't have matching powers, so they're not like terms.

[tex]x^2[/tex] and [tex]y^2[/tex] have the same powers, but the variables don't match, so they're not like terms.

Anything can be multiplied together though.  So that's a different situation.

Answer:

[tex]\tt{ \green{P} \orange{s} \red{y} \blue{x} \pink{c} \purple{h} \green{i} e}[/tex]

Answer:

The selling price of 21 articles is equal to the cost price of 18 articles. Find the loss percent. Answer: answer is 100/7 %

Answer:

ball$

Step-by-step explanation:

<3

Answer:

False.

Step-by-step explanation:

.

17+1= 18

12= -6

12-17= -5 + 1 =-4.

Answer:

False

Step-by-step explanation:

12- (17 + 1) = (12 - 17) + 1

PEMDAS says parentheses first

12- (18) = (-5) + 1

Then add and subtract

-6 = -4

They are not equal

interval notation of x<2

x=1

Answer:

Step-by-step explanation:

[tex]U=\{(x,y)\ \zeta\ y=|x+1| \ and\ x \in \{-2,-1,0,1,2\}\ \}\\\\\begin{array}{|c|c|}x&y\\---&---\\-2&2\\-1&0\\0&2\\1 & 4\\2&6\\---&---\\\end{array}\\\\U=\{(-2,2),(-1,0),(0,2),(1,4),(2,6)\}\\[/tex]

Answer:

2 / (-2,2)

Step-by-step explanation:

Answer:

- 6[tex]x^{4}[/tex] + 14x³ + 8x² - 18x - 10

Step-by-step explanation:

Given

(3x² - 4x - 5)(- 2x² + 2x + 2)

Each term in the second factor is multiplied by each term in the first factor, that is

3x²(- 2x² + 2x + 2) - 4x(- 2x² + 2x + 2) - 5(- 2x² + 2x + 2) ← distribute parenthesis

= - 6[tex]x^{4}[/tex] + 6x³ + 6x² + 8x³ - 8x² - 8x + 10x² - 10x - 10 ← collect like terms

= - 6[tex]x^{4}[/tex] + 14x³ + 8x² - 18x - 10

Answer:

Step-by-step explanation:

Given,

[tex] \frac{ \sqrt[4]{9} \sqrt{9} }{ \sqrt[4]{ {9}^{5} } } [/tex]

[tex] = \frac{ \sqrt[4]{9} \sqrt{9} }{ \sqrt[4]{9 \times 9 \times 9 \times 9 \times 9} } [/tex]

[tex] = \frac{ \sqrt[4]{9} \sqrt{9} }{9 \sqrt[4]{9} } [/tex]

[tex] = \frac{ \sqrt{9} }{9} [/tex]

[tex] = \frac{ {9}^{ \frac{1}{2} } }{ {9}^{1} } [/tex]

[tex] = {9}^{ \frac{1}{2} - 1} [/tex]

[tex] = {9}^{ - \frac{1}{2} } (ans)[/tex]

Answer:

<ALS - <2 = <1

< LDA + <ADE = 180  degrees

Step-by-step explanation:

<ALS = <1+<2

<ALS - <2 = <1

< LDA + <ADE = 180  degrees since it forms a straight line

Answer:

97

Step-by-step explanation:

388 (number of tickets) / 4 (number of groups (adults, and the 3 times as many kids))

If the number of student tickets sold was three times the number of adult tickets sold, 97 adult tickets were sold.

Let's assume the number of adult tickets sold is 'x.'

According to the information given, the number of student tickets sold was three times the number of adult tickets sold. So, the number of student tickets sold is 3x.

The total number of tickets sold is the sum of the adult and student tickets, which is 388:

x + 3x = 388

Now, combine like terms:

4x = 388

To find the value of 'x,' divide both sides of the equation by 4:

x = 388 / 4

x = 97

To verify the result, let's find the number of student tickets:

Number of student tickets = 3 * 97 = 291

Now, let's add the adult and student tickets:

97 (adult) + 291 (student) = 388

The result checks out, and we have 97 adult tickets and 291 student tickets, which together sum up to the total of 388 tickets sold for the school play.

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Answer:

I have ans in the picture form.

Step-by-step explanation:

Take it

Answer:

33

Step-by-step explanation:

608pages/16h = 33pages/h

Step-by-step explanation:

[tex]thank \: you[/tex]

Answer:

[tex] \frac{1}{3} (2x - y) = z \\ \frac{2}{3} x - \frac{y}{3} = z \\ \frac{2}{3} x = z + \frac{y}{3} \\ \frac{2}{3} x = \frac{3z + y}{3} \\ x = \frac{3z + y}{3} \div \frac{2}{3} \\ x = \frac{3z + y}{3} \times \frac{3}{2} = \frac{3(3z + y)}{6} \\ \\ x = \frac{3z + y}{2} \\ \\ x = \frac{3z}{2} + \frac{y}{2} [/tex]

I hope I helped you^_^

The median of that data set is 9.

Sort the numbers from least to greatest and find the middle number (that should be your median)

If there are 2 numbers in the middle of the set, then you take the average and that would be your median.

We have that the distance around the park to the nearest yard,QM to the nearest yard and the Time of travel for man's jog is given respectively as

From the Question we are told that

[tex]PQ=(50-10)=40\\\\QR=(80-10)70[/tex]

Generally the equation for the Resultant PR is mathematically given as

[tex]PR=\sqrt{PQ^2+QR^2}\\\\PR=\sqrt{70^2+40^2}\\\\PR=80yd[/tex]

a)

Generally the equation for the Distance around they park is mathematically given as

[tex]X=PQ+QR+PR\\\\X=80+70+40\\\\[/tex]

[tex]X=190yd[/tex]

b)

Generally the equation for the Path QM is mathematically given as

From the diagram we see that

[tex]QM=PM=MR\\\\QM=\frac{80}{2}\\\\[/tex]

[tex]QM=40yd[/tex]

c)

From the question we are told

Average speed of [tex]150yd/m[/tex]

Total distance traveled

[tex]D_t=40+40+40+70+40\\\\D_t=230yd[/tex]

Therefore

Time

[tex]T=\frac{D_t}{V}\\\\T=\frac{230}{150}[/tex]

[tex]T=1.5minutes.[/tex]

in conclusion

The distance around the park to the nearest yard,QM to the nearest yard and the Time of travel for man's jog is given respectively as

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The cube root of 27 is 3!!

Answer:

a= 1 , b= 3 , c= –3

[tex]x = \frac{ - b \frac{ + }{} \sqrt{ {b}^{2} - 4ac} }{2a} \\ x = \frac{ - 3 \frac{ + }{} {} \sqrt{ {3}^{2} - 4(1)( - 3)} }{2(1)} \\ = \frac{ - 3 \frac{ + }{} \sqrt{9 + 12} }{2} = \frac{ - 3 \frac{ + }{} \sqrt{21} }{2} \\ \\ x = \frac{ - 3 \frac{ + }{} \sqrt{21} }{2} \\ \\ x = 0.79 \\ x = - 3.79[/tex]

I hope I helped you^_^

Answer:

m = -18

Step-by-step explanation:

6 + m/2 = -3

Subtract 6 from each side

6 + m/2 -6 = -3-6

m/2 = -9

Multiply each side by 2

m/2 * 2 = -9*2

m = -18

[tex]\huge\text{Hey there!}[/tex]

[tex]\huge\boxed{\mathsf{6+ \dfrac{m}{2}=-3}}}\\\huge\boxed{\rightarrow\mathsf{6 + \dfrac{1}{2}m = -3}}[/tex]

[tex]\huge\boxed{\rightarrow \mathsf{\dfrac{1}{2}m+ 6 = -3}}\\\\\huge\boxed{\text{SUBTRACT 6 to BOTH SIDES}}\\\huge\boxed{\mathsf{\dfrac{1}{2}m + 6 - 6 = -3 + 3}}[/tex]

[tex]\huge\boxed{\mathsf{\dfrac{1}{2}m = -9}}[/tex]

[tex]\huge\boxed{\text{MULTIPLY 2 to BOTH SIDES}}[/tex]

[tex]\huge\boxed{\mathsf{\dfrac{1}{2}m \times 2 = 2 \times -9}}[/tex]

[tex]\huge\boxed{\mathsf{m = 2\times-9}}\\\huge\boxed{\mathsf{2\times-9 = m}}\\\huge\boxed{\mathsf{2\times-9 = \bf -18}}\\\\\huge\boxed{\textsf{Therefore, your answer: \bf -18}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]

~[tex]\huge\boxed{\frak{Amphitrite1040:)}}[/tex]