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4 years ago

3+2-2-1+19+1536-12+163+5268

Mathematics

2 answers:

timurjin [86]4 years ago

6 0

6976 - simply plug it into a calculator

Fed [463]4 years ago

3 0

3+2-2-1+19+1536-12+163+5268=6966

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PLEASE HELP!! What is the value for y? Enter your answer in the box. y =

Mamont248 [21]

Answer:

y=28

Step-by-step explanation:

1. As ABC is an isosceles triangle, that means that angle A is congruent to angle B, which means angle B measures 34˚.

2. The sum of the angles of a triangle add up to 180˚. So, make an equation:

A+B+C=180

As we already know A and B, plug them in:

34+34+C=180

Then solve:

C=112˚

3. As we know that C=4y, we can solve for y: 112/4=28

6 0

4 years ago

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PLEASE PLEASE PLEASE HELP. This is a geometry worksheet.

Vinil7 [7]

Answer:

where's the question?

3 0

3 years ago

Bonjour , je n'arrive pas a faire ces calcules quelqu'un pourrais m'aidez Voici les calculs : 1 sur 4 - 3 sur 4 et calculer un p

Gre4nikov [31]

Bonjour!
90% de 2300=2070
112% de 80= 100% de 80 + 12% de 80 = 89.6km/h
:)

3 0

3 years ago

Describe the steps required to determine the equation of a quadratic function given its zeros and a point.

faltersainse [42]

Answer:

Procedure:

1) Form a system of 3 linear equations based on the two zeroes and a point.

2) Solve the resulting system by analytical methods.

3) Substitute all coefficients.

Step-by-step explanation:

A quadratic function is a polynomial of the form:

$y = a\cdot x^{2}+b\cdot x + c$ (1)

Where:

$x$ - Independent variable.

$y$ - Dependent variable.

$a$ , $b$ , $c$ - Coefficients.

A value of $x$ is a zero of the quadratic function if and only if $y = 0$ . By Fundamental Theorem of Algebra, quadratic functions with real coefficients may have two real solutions. We know the following three points: $A(x,y) = (r_{1}, 0)$ , $B(x,y) = (r_{2},0)$ and $C(x,y) = (x,y)$

Based on such information, we form the following system of linear equations:

$a\cdot r_{1}^{2}+b\cdot r_{1} + c = 0$ (2)

$a\cdot r_{2}^{2}+b\cdot r_{2} + c = 0$ (3)

$a\cdot x^{2} + b\cdot x + c = y$ (4)

There are several forms of solving the system of equations. We decide to solve for all coefficients by determinants:

$a = \frac{\left|\begin{array}{ccc}0&r_{1}&1\\0&r_{2}&1\\y&x&1\end{array}\right| }{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&1\\r_{2}^{2}&r_{2}&1\\x^{2}&x&1\end{array}\right| }$

$a = \frac{y\cdot r_{1}-y\cdot r_{2}}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x+x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}$

$a = \frac{y\cdot (r_{1}-r_{2})}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x +x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}$

$b = \frac{\left|\begin{array}{ccc}r_{1}^{2}&0&1\\r_{2}^{2}&0&1\\x^{2}&y&1\end{array}\right| }{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&1\\r_{2}^{2}&r_{2}&1\\x^{2}&x&1\end{array}\right| }$

$b = \frac{(r_{2}^{2}-r_{1}^{2})\cdot y}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x +x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}$

$c = \frac{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&0\\r_{2}^{2}&r_{2}&0\\x^{2}&x&y\end{array}\right| }{\left|\begin{array}{ccc}r_{1}^{2}&r_{1}&1\\r_{2}^{2}&r_{2}&1\\x^{2}&x&1\end{array}\right| }$

$c = \frac{(r_{1}^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1})\cdot y}{r_{1}^{2}\cdot r_{2}+r_{2}^{2}\cdot x + x^{2}\cdot r_{1}-x^{2}\cdot r_{2}-r_{2}^{2}\cdot r_{1}-r_{1}^{2}\cdot x}$

And finally we obtain the equation of the quadratic function given two zeroes and a point.

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3 years ago

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ehidna [41]

I don't know about square inches, I can't convert 128 inches to square inches is there anyway Google or something can help you?

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3 years ago

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