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VikaD [51]
2 years ago
10

PLEASE ANSWER QUESTION

Mathematics
1 answer:
CaHeK987 [17]2 years ago
8 0

Answer:

60+80+x=180

Step-by-step explanation:

__ + __ + __ = __ fill in the blanks

if 60, 80, and y have a sum of 180, the equation is 60+80+x=180

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I need help with this question. Can someone help me with it? And show work!
kondor19780726 [428]

Answer:

B

Step-by-step explanation:

K=1/3=10

L=1/5=6

10+6=16

30-16=14

4 0
3 years ago
Read 2 more answers
Find the vertex I’d this parabola y=-2x^2-4x+4
mamaluj [8]

Answer:

  (-1, 6)

Step-by-step explanation:

A graphing calculator shows the vertex to be (x, y) = (-1, 6).

__

You can also find it by putting the equation into vertex form.

  y = -2(x^2 +2x) +4

  y = -2(x^2 +2x +1) +4 +2(1) . . . . . add 1 inside parens; the opposite outside

  y = -2(x +1)^2 +6

Compare to ...

  y = a(x -h)^2 +k

and you see a=-2, h=-1, k=6

The vertex is (h, k) = (-1, 6).

5 0
3 years ago
Please help to solve this please with steps.
Blababa [14]

Answer:

284cm^2

Step-by-step explanation:

first, we split up the shape into seperate sections that we can easily find the areas of.

i will draw vertical lines in the bottom left and right, leaving me with 2 seperate rectangles and 1 irregular pentagon.

we know that these rectangles are 4x8cm, so we do 4 * 8 which gives us 32.

there are 2 of these, so 32 x 2 = 64cm^2.

now, i chose to seperarte the pentagon into a rectangle and a triangle,

and i found the height and width of the rectangle to be (18 - (4+4)) x (8+7), or 10 x 15.

the area of the rectangle is 150cm^2.

now, for the triangle.

the line through the centre of th shape is 22cm long, but we only want the part in the triangle. luckily, there are mesurements that can help us with this.

8 + 7 = 15.

22 - 15 = 7.

now we know that the height of the triangle is 7 cm.

from earlier, we also know the base, which is 10cm.

7 x 10 = 70cm^2.

now we add all these together:

70 + 150 + 64 = 284cm^2

4 0
3 years ago
What is 5.316 - 1.942 btw (show ur work) :)
erastovalidia [21]

Answer:

3.374

Step-by-step explanation:

\mathrm{Write\:the\:numbers\:one\:under\:the\:other,\:line\:up\:the\:decimal\:points.}

\mathrm{Add\:trailing\:zeroes\:so\:the\:numbers\:have\:the\:same\:length.}

\begin{matrix}\:\:&5&.&3&1&6\\ -&1&.&9&4&2\end{matrix}

\mathrm{Subtract\:each\:column\:of\:digits,\:starting\:from\:the\:right\:and\:working\:left}

\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:6-2=4\\

\frac{\begin{matrix}\:\:&5&.&3&1&\textbf{6}\\ -&1&.&9&4&\textbf{2}\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\:\:&\textbf{4}\end{matrix}}

\frac{\begin{matrix}\:\:&5&.&3&\textbf{1}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}

\frac{\begin{matrix}\:\:&5&.&\textbf{3}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}

\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&10&\:\:&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&3&1&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{\:\:}&\:\:&\:\:&\:\:&4\end{matrix}}\\

\frac{\begin{matrix}\:\:&4&\:\:&\textbf{13}&\:\:&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{3}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}

\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&10&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}

\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}

\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{7}&4\end{matrix}}

\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&11&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&\linethrough{1}&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{3}&7&4\end{matrix}}

\frac{\begin{matrix}\:\:&4&\textbf{\:\:}&12&11&\:\:\\ \:\:&\linethrough{5}&\textbf{.}&\linethrough{13}&\linethrough{1}&6\\ -&1&\textbf{.}&9&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\textbf{.}&3&7&4\end{matrix}}

\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&12&11&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&\linethrough{13}&\linethrough{1}&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{3}&.&3&7&4\end{matrix}}

=3.374

6 0
2 years ago
Using the equation representing the height of the firework (h = -16t2 + v0t + h0), algebraically determine the extreme value of
lubasha [3.4K]
-16t^2 + v0t + h0

=  -16(t^2 - 1/16 v0t)  + h0
 = -16 [(t - 1/32v0)^2 - (1/32 v0)^2 ) + h0
 = -16(t - 1/32v0)^2 + 1/64 (v0)^2 + h0   which is the vertex form 

The vertex  is at the point (1/32v0, 1/64 (v0)^2)

1/32v0  represents the time  when the firework is at its maximum height and 1/64 (v0)^2  is this maximum height
6 0
3 years ago
Read 2 more answers
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