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VLD [36.1K]
3 years ago
7

Please help I need this done by tomorrow Thank you

Mathematics
2 answers:
PtichkaEL [24]3 years ago
5 0

19. C

20. B

21. 200

22. im not sure if im right but i think p+6

23. C

24. Can't see

Pavlova-9 [17]3 years ago
3 0

Question 1:

For this case we must find the value of the following expression:

4 ^ 3-3 ^ 3

So we have to:

4 ^ 3 = 4 * 4 * 4 = 4 ^ 2 * 4 = 16 * 4 = 64\\3 ^ 3 = 3 * 3 * 3 = 3 ^ 2 * 3 = 9 * 3 = 27

Substituting we have:

64-27 =\\37

Thus, the value of the expression is 37.

Answer:

4 ^ 3-3 ^ 3 = 37

Option C

Question 2:

For this case we have the following expression:

2 * 3 ^ 2

We must simplify it!

We have to:

3 ^ 2 = 3 * 3 = 9

Rewriting the expression we have

2 * 9 =

2 times 9 is 18.

Finally we have to:

2 * 3 ^ 2 = 18

Answer:

2 * 3 ^ 2 = 18

Option B

Question 3:

For this case we have that by definition, the perimeter of a square is given by:

P = 4l

Where:

l: It's the side of the square

In this case they tell us that:

l = 50 \ in

Thus:

P = 4 * 50\\P = 200 \ in

Now we must do a conversion. By definition we have to:

1 foot equals 12 inches. So:

200 \ in * \frac {1} {12} * \frac {ft} {in} = \frac {50} {3} ft = 16.67ft\\

Answer:

The perimeter of the garden is 16.67 \ ft

Question 4:

For this case we have to:

Let "p" be the variable that represents the number of pages Nora read yesterday.

If today I read 6 more than yesterday, then we can write the following expression:

p + 6

So, the number of pages that Nora read today is p + 6

ANswer:

The number of pages that Nora read today isp + 6

Question 5:

For this case we have that by definition, the area of ​​a triangle is given by:

A = \frac {b * h} {2}

Where:

b: It is the base of the triangle

h: It is the height

We have as data that:

b = 5 \ cm\\h = 8 \ cm

Substituting we have:A = \frac {5 * 8} {2} = \frac {40} {2} = 20 \ cm ^ 2

The area of ​​the triangle is20 \ cm ^ 2

Answer:

Option B

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Solve for x in the equation 2x^2+3x-7=x^2+5x+39
Shalnov [3]
Hey there, hope I can help!

\mathrm{Subtract\:}x^2+5x+39\mathrm{\:from\:both\:sides}
2x^2+3x-7-\left(x^2+5x+39\right)=x^2+5x+39-\left(x^2+5x+39\right)

Assuming you know how to simplify this, I will not show the steps but can add them later on upon request
x^2-2x-46=0

Lets use the quadratic formula now
\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}
x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:} a=1,\:b=-2,\:c=-46: x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:1\left(-46\right)}}{2\cdot \:1}

\frac{-\left(-2\right)+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \  \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \  \frac{2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}

Multiply the numbers 2 * 1 = 2
\frac{2+\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}

2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \  \sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}

\mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \  \sqrt{\left(-2\right)^2+1\cdot \:4\cdot \:46} \ \textgreater \  \left(-2\right)^2=2^2, 2^2 = 4

\mathrm{Multiply\:the\:numbers:}\:4\cdot \:1\cdot \:46=184 \ \textgreater \  \sqrt{4+184} \ \textgreater \  \sqrt{188} \ \textgreater \  2 + \sqrt{188}
\frac{2+\sqrt{188}}{2} \ \textgreater \  Prime\;factorize\;188 \ \textgreater \  2^2\cdot \:47 \ \textgreater \  \sqrt{2^2\cdot \:47}

\mathrm{Apply\:radical\:rule}: \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b} \ \textgreater \  \sqrt{47}\sqrt{2^2}

\mathrm{Apply\:radical\:rule}: \sqrt[n]{a^n}=a \ \textgreater \  \sqrt{2^2}=2 \ \textgreater \  2\sqrt{47} \ \textgreater \  \frac{2+2\sqrt{47}}{2}

Factor\;2+2\sqrt{47} \ \textgreater \  Rewrite\;as\;1\cdot \:2+2\sqrt{47}
\mathrm{Factor\:out\:common\:term\:}2 \ \textgreater \  2\left(1+\sqrt{47}\right) \ \textgreater \  \frac{2\left(1+\sqrt{47}\right)}{2}

\mathrm{Divide\:the\:numbers:}\:\frac{2}{2}=1 \ \textgreater \  1+\sqrt{47}

Moving on, I will do the second part excluding the extra details that I had shown previously as from the first portion of the quadratic you can easily see what to do for the second part.

\frac{-\left(-2\right)-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \  \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \  \frac{2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}

\frac{2-\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}

2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \  2-\sqrt{188} \ \textgreater \  \frac{2-\sqrt{188}}{2}

\sqrt{188} = 2\sqrt{47} \ \textgreater \  \frac{2-2\sqrt{47}}{2}

2-2\sqrt{47} \ \textgreater \  2\left(1-\sqrt{47}\right) \ \textgreater \  \frac{2\left(1-\sqrt{47}\right)}{2} \ \textgreater \  1-\sqrt{47}

Therefore our final solutions are
x=1+\sqrt{47},\:x=1-\sqrt{47}

Hope this helps!
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\bf \begin{array}{cccccclllll}
\textit{something}&&\textit{varies directly to}&&\textit{something else}\\ \quad \\
\textit{something}&=&{{ \textit{some value}}}&\cdot &\textit{something else}\\ \quad \\
y&=&{{ k}}&\cdot&x
&&  y={{ k }}x
\end{array}\\ \quad \\


and also

\bf \begin{array}{llllll}
\textit{something}&&\textit{varies inversely to}&\textit{something else}\\ \quad \\
\textit{something}&=&\cfrac{{{\textit{some value}}}}{}&\cfrac{}{\textit{something else}}\\ \quad \\
y&=&\cfrac{{{\textit{k}}}}{}&\cfrac{}{x}
&&y=\cfrac{{{  k}}}{x}
\end{array}


now, we know that V varies directly to T and inversely to P simultaneously
thus\bf V=T\cdot \cfrac{k}{P}

so     \bf V=T\cdot \cfrac{k}{P}\qquad 
\begin{cases}
V=42\\
T=84\\
P=8
\end{cases}\implies 42=\cfrac{84k}{8}\implies 4=k
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\begin{cases}
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