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Charra [1.4K]
3 years ago
10

1/2 + 9/10 + 4/5 A. 2 1/5 B. 2 2/10C. 14/17D. 22/10

Mathematics
2 answers:
worty [1.4K]3 years ago
8 0
Here is the answer so yo you don't get confused

lora16 [44]3 years ago
5 0
I didn't state this earlier, but you should find the common denominator so in this case it is 10.
1/2=5/10, 9/10 stays the same, and 4/5=8/10.
Now you add them up.
5/10 + 9/10 + 8/10 = 22/10
22/10 = 2 2/10
2 2/10 = 2 1/5
Answer: A. 2 1/5
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What is the equation of a quadratic graph with a focus of (-4,17/8) and a detric of y=15/8 answer
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keeping in mind that the vertex is between the focus point and the directrix, in this cases we have the focus point above the directrix, meaning is a vertical parabola opening upwards, Check the picture below, which means the "x" is the squared variable.

now, the vertical distance from the focus point to the directrix is \bf \cfrac{17}{8}-\cfrac{15}{8}\implies \cfrac{2}{8} , which means the distance "p" is half that or 1/8, and is positive since it's opening upwards.

since the vertex is 1/8 above the directrix, that puts the vertex at \bf \cfrac{15}{8}+\stackrel{p}{\cfrac{1}{8}}\implies \cfrac{16}{8}\implies 2 , meaning the y-coordinate for the vertex is 2.

\bf \textit{vertical parabola vertex form with focus point distance} \\\\ 4p(y- k)=(x- h)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h,k+p)}\qquad \stackrel{directrix}{y=k-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\cap}\qquad \stackrel{"p"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill

\bf \begin{cases} h=-4\\ k=2\\ p=\frac{1}{8} \end{cases}\implies 4\left(\frac{1}{8} \right)(y-2)=[x-(-4)]^2\implies \cfrac{1}{2}(y-2)=(x+4)^2 \\\\\\ y-2=2(x+4)^2\implies \blacktriangleright y = 2(x+4)^2+2 \blacktriangleleft

7 0
3 years ago
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sleet_krkn [62]

Answer:

2x^2+3x is the expression which is equivalent to  x\left(2x+3\right).

In other words:

x\left(2x+3\right)=2x^2+3x

Step-by-step explanation:

Given the expression

x\left(2x+3\right)

solving

x\left(2x+3\right)

\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac

a=x,\:b=2x,\:c=3

=x\cdot \:2x+x\cdot \:3

=2xx+3x

=2x^2+3x            ∴  2xx=2x^2

Thus,

x\left(2x+3\right)=2x^2+3x

Therefore, 2x^2+3x is the expression which is equivalent to  x\left(2x+3\right).

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