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Nikitich [7]
4 years ago
9

Add 15,7,.5 step by step please I'll mark u brainlest PLEASE

Mathematics
2 answers:
Korvikt [17]4 years ago
6 0

Answer:

27

Step-by-step explanation:

15+5=20

20+7=27

Vladimir [108]4 years ago
5 0

Answer: 27

Step-by-step explanation:

15+5=20

20+7=27

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3 0
3 years ago
Find the center that eliminates the linear terms in the translation of 4x^2 - y^2 + 24x + 4y + 28 = 0.(-3, 2)(-3,- 2)(4, 0)
baherus [9]

Step 1

Given;

4x^2-y^2+24x+4y+28=0

Required; To find the center that eliminates the linear terms

Step 2

\begin{gathered} 4x^2-y^2+24x+4y=-28 \\ 4x^2+24x-y^2+4y=-28 \\ Complete\text{ the square }; \\ 4x^2+24x \\ \text{use the form ax}^2+bx\text{ +c} \\ \text{where} \\ a=4 \\ b=24 \\ c=0 \end{gathered}\begin{gathered} consider\text{ the vertex }form\text{ of a }parabola \\ a(x+d)^2+e \\ d=\frac{b}{2a} \\ d=\frac{24}{2\times4} \\ d=\frac{24}{8} \\ d=3 \end{gathered}\begin{gathered} Find\text{ the value of e using }e=c-\frac{b^2}{4a} \\ e=0-\frac{24^2}{4\times4} \\ e=0-\frac{576}{16}=-36 \end{gathered}

Step 3

Substitute a,d,e into the vertex form

\begin{gathered} a(x+d)^2+e \\ 4(x+_{}3)^2-36 \end{gathered}\begin{gathered} 4(x+3)^2-36-y^2+4y=-28 \\ 4(x+3)^2-y^2+4y=\text{ -28+36} \\  \\  \end{gathered}

Step 4

Completing the square for -y²+4y

\begin{gathered} \text{use the form ax}^2+bx\text{ +c} \\ \text{where} \\ a=-1 \\ b=4 \\ c=0 \end{gathered}\begin{gathered} consider\text{ the vertex }form\text{ of a }parabola \\ a(x+d)^2+e \\ d=\frac{b}{2a} \\ d=\text{ }\frac{4}{2\times-1} \\ d=\frac{4}{-2} \\ d=-2 \end{gathered}\begin{gathered} Find\text{ the value of e using }e=c-\frac{b^2}{4a} \\ e=0-\frac{4^2}{4\times(-1)} \\  \\ e=0-\frac{16}{-4} \\ e=4 \end{gathered}

Step 5

Substitute a,d,e into the vertex form

\begin{gathered} a(y+d)^2+e \\ =-1(y+(-2))^2+4 \\ =-(y-2)^2+4 \end{gathered}

Step 6

\begin{gathered} 4(x+3)^2-y^2+4y=\text{ -28+36} \\ 4(x+3)^2-(y-2)^2+4=-28+36 \\ 4(x+3)^2-(y-2)^2=-28+36-4 \\ 4(x+3)^2-(y-2)^2=4 \\ \frac{4(x+3)^2}{4}-\frac{(y-2)^2}{4}=\frac{4}{4} \\ (x+3)^2-\frac{(y-2)^2}{2^2}=1 \end{gathered}

Step 7

\begin{gathered} \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1 \\ \text{This is the }form\text{ of a hyperbola.} \\ \text{From here } \\ a=1 \\ b=2 \\ k=2 \\ h=-3 \end{gathered}

Hence the answer is (-3,2)

4 0
1 year ago
Which of the following equations matches the graph above?
shtirl [24]
It’s 4x -4 because at at x=0 the y = -4 so it has -4 and then it goes up by 4 each x
6 0
3 years ago
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