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Maru [420]
3 years ago
11

How do you subtract this

Mathematics
1 answer:
Maurinko [17]3 years ago
8 0

To subtract fractions, you need to obtain a common denominator. We can achieve this, in this case, by multiplying:

(\frac{4}{4} * \frac{22}{13}) - (\frac{13}{13}*\frac{21}{4})

Simplify this:

\frac{88}{52} - \frac{286}{52}

Now we have common denominators, so we can easily just subtract the numerators, and we have the answer:

-\frac{198}{52}

Notice that this is divisible by 2:

-\frac{99}{26}

That's your answer!

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For number 6, evaluate the definite integral.
maks197457 [2]
\bf \displaystyle \int\limits_{0}^{28}\ \cfrac{1}{\sqrt[3]{(8+2x)^2}}\cdot dx\impliedby \textit{now, let's do some substitution}\\\\
-------------------------------\\\\
u=8+2x\implies \cfrac{du}{dx}=2\implies \cfrac{du}{2}=dx\\\\
-------------------------------\\\\

\bf \displaystyle \int\limits_{0}^{28}\ \cfrac{1}{\sqrt[3]{u^2}}\cdot \cfrac{du}{2}\implies \cfrac{1}{2}\int\limits_{0}^{28}\ u^{-\frac{2}{3}}\cdot du\impliedby 
\begin{array}{llll}
\textit{now let's change the bounds}\\
\textit{by using } u(x)
\end{array}\\\\
-------------------------------\\\\
u(0)=8+2(0)\implies u(0)=8
\\\\\\
u(28)=8+2(28)\implies u(28)=64

\bf \\\\
-------------------------------\\\\
\displaystyle  \cfrac{1}{2}\int\limits_{8}^{64}\ u^{-\frac{2}{3}}\cdot du\implies \cfrac{1}{2}\cdot \cfrac{u^{\frac{1}{3}}}{\frac{1}{3}}\implies \left. \cfrac{3\sqrt[3]{u}}{2} \right]_8^{64}
\\\\\\
\left[ \cfrac{3\sqrt[3]{(2^2)^3}}{2} \right]-\left[ \cfrac{3\sqrt[3]{2^3}}{2}  \right]\implies \cfrac{12}{2}-\cfrac{6}{2}\implies 6-3\implies 3
3 0
3 years ago
Graph the function 2(x+2)(x+6)
SOVA2 [1]
The answer is 32 :))
7 0
3 years ago
Read 2 more answers
The table shows the relationship between the number of Calories Lucie burns while jumping rope and the number of minutes she jum
iragen [17]

Answer:

The amount of calories Lucie will burn by jumping for 1 minute is 8 calories

Step-by-step explanation:

The given data values are;

Minutes {}                  Calories Burned

40                {}            320

80        {}                    640

120 {}                          960

160 {}                          1280

The value of calories burnt and the number of minutes of jump robe are seen to be directly proportional, such that we have;

40 minutes of jump rope will yield 320 calories burnt

Therefore we have;

1 minute of jump rope will yield 320/40 = 8 calories burnt

The amount of calories Lucie will burn by jumping for 1 minute = 8 calories.

8 0
3 years ago
Write an equation for a parabola with a focus of (1,-2) and a directrix of y=6
FinnZ [79.3K]

Answer:

y = - \frac{1}{16}(x - 1)² + 2

Step-by-step explanation:

Any point (x, y) on the parabola is equidistant from the focus and the directrix.

Using the distance formula

\sqrt{(x-1)^2+(y+2)^2^} = | y - 6 |

Square both sides

(x - 1)² + (y + 2)² = (y - 6)² ( expand the factors in y )

(x - 1)² + y² + 4y + 4 = y² - 12y + 36 ( subtract y² - 12y from both sides )

(x - 1)² + 16y + 4 = 36 ( subtract 4 from both sides )

(x - 1)² + 16y = 32 ← subtract (x - 1)² from both sides )

16y = - (x - 1)² + 32 ( divide all terms by 16 )

y = - \frac{1}{16} (x - 1)² + 2

8 0
3 years ago
For what value of c is the function defined below continuous on (-\infty,\infty)?
kozerog [31]
f(x)= \left \{ {{x^2-c^2,x \ \textless \  4} \atop {cx+20},x \geq 4} \right


It's clear that for x not equal to 4 this function is continuous. So the only question is what happens at 4.
<span>A function, f, is continuous at x = 4 if 
</span><span>\lim_{x \rightarrow 4} \  f(x) = f(4)

</span><span>In notation we write respectively
</span>\lim_{x \rightarrow 4-} f(x) \ \ \ \text{ and } \ \ \ \lim_{x \rightarrow 4+} f(x)

Now the second of these is easy, because for x > 4, f(x) = cx + 20. Hence limit as x --> 4+ (i.e., from above, from the right) of f(x) is just <span>4c + 20.
</span>
On the other hand, for x < 4, f(x) = x^2 - c^2. Hence 
\lim_{x \rightarrow 4-} f(x) = \lim_{x \rightarrow 4-} (x^2 - c^2) = 16 - c^2

Thus these two limits, the one from above and below are equal if and only if
 4c + 20 = 16 - c²<span> 
 Or in other words, the limit as x --> 4 of f(x) exists if and only if
 4c + 20 = 16 - c</span>²

c^2+4c+4=0&#10;\\(c+2)^2=0&#10;\\c=-2

That is to say, if c = -2, f(x) is continuous at x = 4. 

Because f is continuous for all over values of x, it now follows that f is continuous for all real nubmers (-\infty, +\infty)

4 0
3 years ago
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