MATHEMATICS HIGH SCHOOL

Find the following measure for this figure. Volume =
18 cubic units
24 cubic units
36 cubic units
Find the following measure for this figure. Volume = 18 - 1

Answers

Answer 1
Answer: The figure on the picture is parallelepiped (or right rectangular prism), whose volume is icalculated as V=lwh, where l measures the length, w measures the width and measures the altitude of parallelepiped.

In your case, l=4 units, w=2 units, h=3 units and V=4×2×3=24 cubic units.
Answer 2
Answer:

Answer:

So the answer is 24 cubic units

Step-by-step explanation:

l=4 units,

w=2 units,

h=3 units and

V=4×2×3=24 cubic units.

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Write an algebraic expression for the following word phrase: the quotient of r and 12?

Answers

quotient= division so 12/r would be your answer 

Final answer:

The algebraic expression for the word phrase 'the quotient of r and 12' is r/12. This represents a division operation where r is divided by 12.

Explanation:

The phrase 'the quotient of r and 12' can be translated into an algebraic expression as r/12. In this expression, 'quotient' refers to the mathematical operation of division. Hence, the expression means that the variable 'r' is divided by '12'.

Learn more about Algebra here:

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0.5/10 5/100 are these ratios equal?

Answers

Answer:

No

Step-by-step explanation:

5/100 would be 0.05 as a decimal

The width,w, of a rectangular garden is X -2 the area of the garden is X^3-2X-4 what is an expression for the length of the garden? A. X^2-2x-2
B. X^2+2x-2
C. X^2-2x+2
D. X^2+2x+2

Answers

Answer:

D.x^2+2x+2

Step-by-step explanation:

We know that,

The area of a rectangle is,

A = l × b,

Where, l is the length of the rectangle,

w is the width of the rectangle,

Given,

A = x^3-2x-4

w=(x-2)

By substituting values,

x^3-2x-4=(x-2)l

\implies l = (x^3-2x-4)/(x-2)=x^2+2x+2 ( By long division shown below )

Hence, the length of the rectangular garden isx^2+2x+2

Option D is correct.
A=lw \Rightarrow l=(A)/(w)
A - area, l - length, w - width

w=x-2 \nA=x^3-2x-4 \n \nl=(x^3-2x-4)/(x-2)=(x^3-2x^2+2x^2-4x+2x-4)/(x-2)=(x^2(x-2)+2x(x-2)+2(x-2))/(x-2)= \n=((x^2+2x+2)(x-2))/(x-2)=x^2+2x+2

The answer is D. x²+2x+2.

In PQR, the measure of

Answers

Answer:

17.9 degrees

Step-by-step explanation:

tangent x= 3.2/9.9

tangent x= 0.323

x=tan^-1(0.323)

x=17.9 degrees

Which of the following is not a name of a valid chemical formula? Select one:
a. carbon chloride
b. nitrogen monoxide
c. copper (II) carbonate
d. chloric acid

Answers

Among the compounds given, the one that is not valid is a. carbon chloride. a compound composed of non-metals should have prefixes that is related to the number of atoms of the elements present. In this case, the other cmopounds are considered valid but not A 
The correct answer is: Option (a) Carbon Chloride

Explanation:
Carbon has four valence electrons; therefore, it would require 4 more electrons to become stable. Chlorine, however, has seven (7) valence electrons. So it would require only one electron to become stable. In this case, as in the name, carbon chloride, the word chloride refers to the compound that is stable, which in this case cannot be possible (as carbon requires 4 electrons, and chlorine only has 7). If chlorine had 4 valence electrons, then it would be possible. Hence, the correct option is (a) Carbon Chloride. 

How to solve number 21 :/

Answers

\left(\begin{array}{cc}1&1\n-2&1\end{array}\right) \bold{M}= 2\left(\begin{array}{cc}1&0\n0&1\end{array}\right)\n \bold{M}=2\frac{ \left(\begin{array}{cc}1&0\n0&1\end{array}\right) }{ \left(\begin{array}{cc}1&1\n-2&1\end{array}\right)}\n \bold{M}=2\left(\begin{array}{cc}1&0\n0&1\end{array}\right) \left(\begin{array}{cc}1&1\n-2&1\end{array}\right)^(-1)

\left(\begin{array}{cc}1&1\n-2&1\end{array}\right)^(-1)=(1)/(1-(-2))\left(\begin{array}{cc}1&-1\n2&1\end{array}\right)\n\left(\begin{array}{cc}1&1\n-2&1\end{array}\right)^(-1)=(1)/(3)\left(\begin{array}{cc}1&-1\n2&1\end{array}\right)\n\left(\begin{array}{cc}1&1\n-2&1\end{array}\right)^(-1)=\left(\begin{array}{cc}(1)/(3)&-(1)/(3)\n(2)/(3)&(1)/(3)\end{array}\right)\n\n
\bold{M}= 2\left(\begin{array}{cc}1&0\n0&1\end{array}\right)\left(\begin{array}{cc}(1)/(3)&-(1)/(3)\n(2)/(3)&(1)/(3)\end{array}\right)\n\bold{M}= 2\left(\begin{array}{cc}(1)/(3)&-(1)/(3)\n(2)/(3)&(1)/(3)\end{array}\right)\n\bold{M}= \left(\begin{array}{cc}(2)/(3)&-(2)/(3)\n(4)/(3)&(2)/(3)\end{array}\right)
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