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Crazy boy [7]
3 years ago
14

I fixed the question need help​

Mathematics
1 answer:
STatiana [176]3 years ago
8 0

Answer:

D

Step-by-step explanation:

90-33=57

57-1=56

56/2=28

28

You might be interested in
Express 9680 in scientific notation
vodomira [7]
Scientific\ notation:a\times10^n\ where\ a\in[1;\ 10)\ and\ n\in\mathbb{Z}


9680=9\underbrace{680}_{\leftarrow 3}=9.680\times10^3=9.68\times10^3
5 0
4 years ago
Read 2 more answers
4. Jill math score on her test last year 7th grade was 650. Her math teacher told her that her test level went up by 2586. What
MrRa [10]
Dang she scored in the negative Jill slow
7 0
3 years ago
I need help, urgently.
Korvikt [17]

Answer:

y = -0.5(x - 4)^2 + 5.

y = -0.5x^2 + 4x - 3.

Step-by-step explanation:

The vertex is at (4, 5).  so we have:

f(x) = a(x - 4)^2 + 5

When x = 0 y = -3 so substituting in the above:

- 3 = a(0-4)^2 + 5

-8 = 16a

a = -0.5.

So the vertex form is y = -0.5(x - 4)^2 + 5.

Standard form:

y = -0.5(x^2 - 8x + 16) + 5

y = -0.5x^2 + 4x - 8 + 5

y = -0.5x^2 + 4x - 3.

5 0
3 years ago
Uestion
Stella [2.4K]

Check the picture below, so the park looks more or less like so, with the paths in red, so let's find those midpoints.

~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ J(\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad K(\stackrel{x_2}{1}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 1 -3}{2}~~~ ,~~~ \cfrac{ 3 +1}{2} \right) \implies \left(\cfrac{ -2 }{2}~~~ ,~~~ \cfrac{ 4 }{2} \right)\implies JK=(-1~~,~~2) \\\\[-0.35em] ~\dotfill

~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ L(\stackrel{x_1}{5}~,~\stackrel{y_1}{-1})\qquad M(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -1 +5}{2}~~~ ,~~~ \cfrac{ -3 -1}{2} \right) \implies \left(\cfrac{ 4 }{2}~~~ ,~~~ \cfrac{ -4 }{2} \right)\implies LM=(2~~,~~-2) \\\\[-0.35em] ~\dotfill

~~~~~~~~~~~~\textit{distance between 2 points} \\\\ JK(\stackrel{x_1}{-1}~,~\stackrel{y_1}{2})\qquad LM(\stackrel{x_2}{2}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ JKLM=\sqrt{(~~2 - (-1)~~)^2 + (~~-2 - 2~~)^2} \\\\\\ JKLM=\sqrt{(2 +1)^2 + (-2 - 2)^2} \implies JKLM=\sqrt{( 3 )^2 + ( -4 )^2} \\\\\\ JKLM=\sqrt{ 9 + 16 } \implies JKLM=\sqrt{ 25 }\implies \boxed{JKLM=5}

now, let's check the other path, JM and KL

~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ J(\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad M(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -1 -3}{2}~~~ ,~~~ \cfrac{ -3 +1}{2} \right) \implies \left(\cfrac{ -4 }{2}~~~ ,~~~ \cfrac{ -2 }{2} \right)\implies JM=(-2~~,~~-1) \\\\[-0.35em] ~\dotfill

~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ K(\stackrel{x_1}{1}~,~\stackrel{y_1}{3})\qquad L(\stackrel{x_2}{5}~,~\stackrel{y_2}{-1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 5 +1}{2}~~~ ,~~~ \cfrac{ -1 +3}{2} \right) \implies \left(\cfrac{ 6 }{2}~~~ ,~~~ \cfrac{ 2 }{2} \right)\implies KL=(3~~,~~1) \\\\[-0.35em] ~\dotfill

~~~~~~~~~~~~\textit{distance between 2 points} \\\\ JM(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-1})\qquad KL(\stackrel{x_2}{3}~,~\stackrel{y_2}{1})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ JMKL=\sqrt{(~~3 - (-2)~~)^2 + (~~1 - (-1)~~)^2} \\\\\\ JMKL=\sqrt{(3 +2)^2 + (1 +1)^2} \implies JMKL=\sqrt{( 5 )^2 + ( 2 )^2} \\\\\\ JMKL=\sqrt{ 25 + 4 } \implies \boxed{JMKL=\sqrt{ 29 }}

so the red path will be  5~~ + ~~\sqrt{29} ~~ \approx ~~ \blacksquare~~ 10 ~~\blacksquare

3 0
2 years ago
Which expression is equivalent to (16x^8y^-12)^1/2
MAVERICK [17]

Answer:

The equivalent expression is \Rightarrow \dfrac{4x^4}{y^6}

Step-by-step explanation:

Given: (16x^8y^{-12})^{1/2}

We need to simplify the expression. We will distribute radical with each term inside the parentheses.

\Rightarrow (16x^8y^{-12})^{1/2}

Distribute the radical

\Rightarrow 16^{1/2}\cdot x^{8\cdot 1/2\cdot y^{-12\cdot 1/2}

\Rightarrow 4\cdot x^4\cdot y^{-6}

\Rightarrow 4x^4y^{-6}

Now we will simplify the negative exponent.

\Rightarrow \dfrac{4x^4}{y^6}

Hence, The equivalent expression is \Rightarrow \dfrac{4x^4}{y^6}

8 0
3 years ago
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