40% of 70% may seem difficult, but parsing the expression's different parts is essential to being able to understand this
"Of" is an indicator of multiplication. Knowing this, the expressions changes to \(40\%*70\%\).
\(40\%*70\%\) | A number in a percentage is the same as the number over 100. |
\(\frac{40}{100}*\frac{70}{100}\) | Simplify both fractions with its GCF. |
\(\frac{2}{5}*\frac{7}{10}\) | Now, multiply both the fractions. |
\(\frac{14}{50}=0.28=28\%\) | |
Hello, equations with absolute value bars can be mysterious sometimes, so I'll help you out. The original equation is \(\frac{1}{3}\left|4p-11\right|=p+4\).
\(\frac{1}{3}\left|4p-11\right|=p+4\) | Multiply by 3 on both sides. | ||
\(|4p-11|=3p+12\) | The absolute value symbol can be replaced with a plus-minus symbol, which creates 2 separate equations. | ||
\(\pm(4p-11)=3p+12\) | Now, split this up into 2 equations. | ||
| Now, solve both equations. | ||
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These could be extraneous solutions, so we must check to ensure that both are indeed valid solutions.
\(\frac{1}{3}\left|4*23-11\right|=23+4\) | |
\(\frac{1}{3}\left|81\right|=27\) | Multiply by 3 on both sides. |
\(|81|=81\) | |
\(81=81\) | This is true, so \(p_1=23\) is a valid solution. |
Now, let's check \(p=-\frac{1}{7}\):
\(\frac{1}{3}\left|4*-\frac{1}{7}-11\right|=-\frac{1}{7}+4\) | Do the same simplification as last time. |
\(\frac{1}{3}\left|\frac{-4}{7}+\frac{-77}{7}\right|=\frac{-1}{7}+\frac{28}{7}\) | |
\(\frac{1}{3}|\frac{-81}{7}|=\frac{27}{7}\) | Muliply by 3 on both sides. |
\(|\frac{-81}{7}|=\frac{81}{7}\) | |
\(\frac{81}{7}=\frac{81}{7}\) | |
Therefore, both solutions are valid solutions.
Before we begin solving, let's insert \(\overline{PU}\) such that the segments extends perpendicularly from \(\overline{OS}\) , and \(U\) lies on \(\overline{QT}\).
We know from the given information that parallelogram \(PQRS\) has a perimeter of 24. Of course, opposite sides of a parallelogram are congruent, so \(\overline{RS}\cong\overline{QP}\), which means that \(RS=QP=5\). Now, we can solve for the remaining sides of the parallelogram (\(\overline{PS}\) and \(\overline{QR}\))
\(5+5+PS+QR=24\) | Simplify the left-hand side of the equation. |
\(10+PS+QR=24\) | Subtract 10 on both sides of the equation. |
\(PS+QR=14\) | Since \(PS=QR\) because of the property of a parallelogram, we can actually solve this equation. |
\(PS+PS=14\) | |
\(2PS=14\) | Divide by 2 on both sides. |
\(PS=7\) | |
\(PS=QR=7\) |
To recap, we now know that \(RS=QP=5\) and \(QR=PS=7\). Now, let's worry about that segment I drew in since the beginning of the problem.
\(\overline{PU}\) is an example of an altitude (perpendicula height that extends from a vertex). The area of the parallelogram is given to be \(28units^2\). Let's review the formula for the area of a parallelogram. It is the following:
\(A=bh\)
b = length of the base
h = length of the altitude (\(\overline{PU}\))
Although it is true that we do not know the length of the altitude, we know the length of the base (7) and the total area of the parallelogram (28). We can now solve for the length of the altitude.
\(A=bh\) | Substitute in the known values and solve for the unknown, in this case h. |
\(28=7h\) | Divide by 7 on both sides of the equation. |
\(4=h\) | |
We know that \(PU=OQ=ST=4\). Now, we must solve for the final missing lengths. They are \(\overline{RT}\) and \(\overline{PO}\). We can use Pythagorean's Theorem for this as we know the lengths of the one of the legs and the hypotenuse.
\(a^2+b^2=c^2\) | This is the equation for the Pythagorean Theorem. Solve for the unknown. |
\(RT^2+4^2=5^2\) | Simplify. |
\(RT^2+16=25\) | Subtract 16 on both sides. |
\(RT^2=9\) | Take the square root of both sides to eliminate the exponent of 2. |
\(RT=\sqrt{9}=3\) | Of course, we can only have positive lengths in the context of geometry, so you should immediately reject a negative side length. |
\(RT=OP=3\) |
We now know all the sides of the parallelogram, and we can solve.
\(OP+PS+ST+TR+RQ+QO=P\) | P, in this case, stands for perimeter, in units, as no specific unit is specified in the problem. |
\(\textcolor{red}{3+7}+4+\textcolor{red}{3+7}+4=P\) | Since this is all addition, the addition can be computed in any order. In red, I suggest doing that first as it results in 10, which is a friendly number computationally speaking. |
\(10+4+10+4=P\) | |
\(28=P\) | |
Therefore, the perimeter of rectangle \(QOST\) is \(28units\)
.First, we must understand what standard form of a line is. Standard form of a line is written like \(Ax+By=C\) such that A,B, and C are all integers, and A must be positive. First, we must calculate the slope of the line that passes through theses coordinates.
\(m=\frac{y_2-y_1}{x_2-x_1}\) | As a refresher, this is the equation to figure out the slope of two coordinates. |
\(m=\frac{-8-(-3)}{4-7}\) | Now, we just simplify the numerator and denominator. |
\(m=\frac{-8+3}{4-7}\) | |
\(m=\frac{-5}{-3}=\frac{5}{3}\) | |
The next step is to utilize point-slope form, which is \(y-y_1=m(x-x_1)\) where \((x_1,y_1)\) is a point on the line. Of course, we already know that (7,-3) and (4,-8) both lie of the line. Therefore, plug in one fot he coordinates. Once converted into point-slope, we must then convert into standard form. This is what is demonstrated in the next step.
\(y+3=\frac{5}{3}(x-7)\) | Let's multiply all sides by 3 to get rid of the fraction early. |
\(3y+9=5(x-7)\) | Distribute the 5 to both terms in the parentheses. |
\(3y+9=5x-35\) | Subtract 9 from both sides. |
\(3y=5x-44\) | Subtract 5x on both sides. |
\(-5x+3y=-44\) | We aren't done yet! The coefficient of the x-term must be positive. Therefore, divide by -1 on both sides. |
\(5x-3y=44\) | This is standard form now, so we are done! |
I'm assuming that the question is asking for the roots of the expression \((9x+1)(x-1)\).
Finding the roots simply means to set the expression equal to zero and solve for any variables. Let's do that!
\((9x+1)(x-1)=0\) | Set both factors equal to 0 and solve. | ||
| Move the constant term over the righthand side. | ||
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Therefore, the roots are the following:
\(x_1=-\frac{1}{9}\)
\(x_2=1\)
.Yes, assuming it is a circle (which I believe it is because "circumference" is used in the title), the area is indeed 0\(114.91ft^2\). Apparently, being ill does not obstruct your computational abilities. I hope you get better soon!
Of course, \(r=\frac{19}{\pi}\), and the area of a circle is \(\pi r^2\). Plugging what we know for r, we ge tthe following:
\(\pi*\left(\frac{19}{\pi}\right)^2\) | Distribute the exponent to both the numerator and denominator. |
\(\left(\frac{19}{\pi}\right)^2=\frac{19^2}{\pi^2}=\frac{361}{\pi^2}\) | |
\(\frac{\pi}{1}*\frac{361}{\pi^2}\) | Before multiplying the fractions together, notice that there is a common factor of pi in both the numerator of one fraction anf the denominator in another. |
\(\frac{361}{\pi}\approx114.91ft^2\) | |
In other words, good job!