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2 years ago

What is the sum or difference? 19x3 – 13x3

Mathematics

2 answers:

mojhsa [17]2 years ago

8 0

Answer:

Step-by-step explanation:

use pemdas. multiplication comes first and you go left to right. then you subtract the products of 19*3 and 13*3

alexdok [17]2 years ago

4 0

Answer:

the difference is 6x³

Step-by-step explanation:

19x3 – 13x3

=6x³

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yanalaym [24]

It’s 10 it has it near the dot

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3 years ago

Read 2 more answers

Mrs. Olga’s $250 investment earns 19% interest compounded annually. What will her balance be after 3 years?

Maurinko [17]

Answer:

The answer is A. $421.29

Step-by-step explanation:

Just keep finding 19% of the numbers next, and you add that all up and you get the answer.

250 x 19%=47.5. (250+47.5)=> 297.5x19%= 56.525. (297.5+56.525) =>354.025x19%=67.26475

354.025+67.26475=> 421.28975, which can round up to 421.29.

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3 years ago

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What is the value of k in the function ƒ(x) = 112 - kx if ƒ(-3) = 121?

DedPeter [7]

f(x) = 112 - kx
f(-3) = 112 - k(-3)
f(-3) = 112 - (-3k)
f(-3) = 112 + 3k
121 = 112 + 3k
- 112 - 112
      9 = 3k
      3    3
      3 = k

6 0

2 years ago

What's 1.8499 x 10^9 in standard notation

maxonik [38]

you just have to move the decimal point 9 places to the right thats an easy trick to remember 1849900000

5 0

3 years ago

Solve the equation on the interval [0,2π]

WARRIOR [948]

$\bf 16sin^5(x)+2sin(x)=12sin^3(x)
\\\\\\
16sin^5(x)+2sin(x)-12sin^3(x)=0
\\\\\\
\stackrel{common~factor}{2sin(x)}[8sin^4(x)+1-6sin^2(x)]=0\\\\
-------------------------------\\\\
2sin(x)=0\implies sin(x)=0\implies \measuredangle x=sin^{-1}(0)\implies \measuredangle x=
\begin{cases}
0\\
\pi \\
2\pi 
\end{cases}\\\\
-------------------------------$

$\bf 8sin^4(x)+1-6sin^2(x)=0\implies 8sin^4(x)-6sin^2(x)+1=0$

now, this is a quadratic equation, but the roots do not come out as integers, however it does have them, the discriminant, b² - 4ac, is positive, so it has 2 roots, so we'll plug it in the quadratic formula,

$\bf 8sin^4(x)-6sin^2(x)+1=0\implies 8[~[sin(x)]^2~]^2-6[sin(x)]^2+1=0
\\\\\\
~~~~~~~~~~~~\textit{quadratic formula}
\\\\
\begin{array}{lcccl}
& 8 sin^4& -6 sin^2(x)& +1\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array} 
\qquad \qquad 
sin(x)= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}
\\\\\\
sin(x)=\cfrac{-(-6)\pm\sqrt{(-6)^2-4(8)(1)}}{2(8)}\implies sin(x)=\cfrac{6\pm\sqrt{4}}{16}
\\\\\\
sin(x)=\cfrac{6\pm 2}{16}\implies sin(x)=
\begin{cases}
\frac{1}{2}\\\\
\frac{1}{4}
\end{cases}$

$\bf \measuredangle x=
\begin{cases}
sin^{-1}\left( \frac{1}{2} \right)
sin^{-1}\left( \frac{1}{4} \right)
\end{cases}\implies \measuredangle x=
\begin{cases}
\frac{\pi }{6}~,~\frac{5\pi }{6}\\
----------\\
\approx~0.252680~radians\\
\qquad or\\
\approx~14.47751~de grees\\
----------\\
\pi -0.252680\\
\approx 2.88891~radians\\
\qquad or\\
180-14.47751\\
\approx 165.52249~de grees
\end{cases}$

3 0

3 years ago