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allochka39001 [22]
2 years ago
14

80% of the students in 4-6 building at LVA are virtual.If we have 1200 students TOTAL in the 4-6 building, how many students are

virtual?
Mathematics
1 answer:
wariber [46]2 years ago
7 0
960 students are virtual !
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It is always helpful to rewrite a quadratic equation in standard form as a first step in solving it.
kotykmax [81]
THE ANSWER IS (B) true
5 0
2 years ago
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Triangle ABC has the coordinates A(-3, -2), B(2, 4) and C(2,-2). What is the area of triangle ABC?
Snezhnost [94]

Answer:   The area is 15 square units

Step-by-step explanation:  It is helpful to sketch the coordinates on a graph to get an idea of what the triangle looks like so you can see which side will be the base, and where to get the height. You will see that AB is the hypotenuse, it is a right triangle, so AC can be the base and BC is the height.

Use the differences in x-values of points C & A to get the length of the base:   2-(-3) is 5

Use the differences in y-values of points B & C to get the height:

4-(-2) = 6

Area = bh/2

A = 5×6/2 = 30/2

A = 15 square unts

3 0
2 years ago
How do I solve this problem? ​
natulia [17]

Step-by-step explanation:

you know the formulas ?

the area of a circle is : pi×r²

the circumference of a circle is : 2×pi×r

with r being the radius, which is half of the diameter.

so, here, r = 10/2 = 5 m

area = pi×r² = pi×5² = pi×25 or 25×pi m²

circumference = 2×pi×r = 2×pi×5 = 10×pi m

for the approximation we make the calculations with pi on the calculator and round to the nearest hundredth.

area = 25×pi ≈ 78.54 m²

circumference = 10×pi ≈ 31.42 m

5 0
3 years ago
Write the expression in factored form: m²-n²
mash [69]
This expression is a difference of squares: x^2 - y^2 = (x + y)(x - y)

m^2 - n^2 = (m + n)(m - n)
4 0
3 years ago
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Could someone help me rnnn?
GuDViN [60]

Answer:

vertex = (0, -4)

equation of the parabola:  y=3x^2-4

Step-by-step explanation:

Given:

  • y-intercept of parabola: -4
  • parabola passes through points: (-2, 8) and (1, -1)

Vertex form of a parabola:  y=a(x-h)^2+k

(where (h, k) is the vertex and a is some constant)

Substitute point (0, -4) into the equation:

\begin{aligned}\textsf{At}\:(0,-4) \implies a(0-h)^2+k &=-4\\ah^2+k &=-4\end{aligned}

Substitute point (-2, 8) and ah^2+k=-4 into the equation:

\begin{aligned}\textsf{At}\:(-2,8) \implies a(-2-h)^2+k &=8\\a(4+4h+h^2)+k &=8\\4a+4ah+ah^2+k &=8\\\implies 4a+4ah-4&=8\\4a(1+h)&=12\\a(1+h)&=3\end{aligned}

Substitute point (1, -1) and ah^2+k=-4 into the equation:

\begin{aligned}\textsf{At}\:(1.-1) \implies a(1-h)^2+k &=-1\\a(1-2h+h^2)+k &=-1\\a-2ah+ah^2+k &=-1\\\implies a-2ah-4&=-1\\a(1-2h)&=3\end{aligned}

Equate to find h:

\begin{aligned}\implies a(1+h) &=a(1-2h)\\1+h &=1-2h\\3h &=0\\h &=0\end{aligned}

Substitute found value of h into one of the equations to find a:

\begin{aligned}\implies a(1+0) &=3\\a &=3\end{aligned}

Substitute found values of h and a to find k:

\begin{aligned}\implies ah^2+k&=-4\\(3)(0)^2+k &=-4\\k &=-4\end{aligned}

Therefore, the equation of the parabola in vertex form is:

\implies y=3(x-0)^2-4=3x^2-4

So the vertex of the parabola is (0, -4)

5 0
2 years ago
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