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3 years ago

Evaluate the algebraic expression for the given value(s) of the variable(s).

Mathematics

1 answer:

Tanya [424]3 years ago

6 0

Answer:

-18

Step-by-step explanation:

2x²-7xy+3y²

2(3²)-7(3)(4)+3(4²)

2(9)-84+3(16)

18-84+48

18+48-84

-18

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Problem 81

netineya [11]

it hasn't been specified which ticket the question's is asking but I'm guessing the price of the adult ones so $30.00

altogether that's two adults and 3 children who all paid a total of $105.00

let's simplify that to algebra

children tickets = c adults tickets = a

3c + 2a = $105.00

we know that 1c is 0.5a because each child ticket is half of an adults one as stated in the question

so in terms of (adults) a: half of 3 meaning 3×0.5= $1.5 meaning 1.5a

so now 3c = 1.5a

so we can substitute and have like terms

meaning 1.5a + 2a= $105

3.5a =$105.00

divide by 3.5 on both sides to get a by itself thus the value of a:

a = $30

so each adult tickets cost $30

8 0

2 years ago

PLEASE HELP THIS IS A MAJOR GRADE FOR ME!!!! 31 Triangle RST is translated 6 units to the left and 4 units up to create triangle

Sladkaya [172]

D is the answer- x-6, y+4

4 0

3 years ago

A woman bought 70 Cubes

KatRina [158]

Answer:

56 Sugar cubes

Step-by-step explanation:

75 x .25 = 18.75

75 - 18.75 = 56.25

Round down for the closest answer = 56

8 0

3 years ago

a line whose perpendicular distance from the origin is 4 units and the slope of perpendicular is 2÷3. Find the equation of the l

GrogVix [38]

Answer:

$\huge\boxed{y=\dfrac{2}{3}x-\dfrac{4\sqrt{13}}{3}\ \vee\ y=\dfrac{2}{3}x+\dfrac{4\sqrt{13}}{3}}$

Step-by-step explanation:

The equation of a line:

$y=mx+b$

We have

$m=\dfrac{2}{3}$

substitute:

$y=\dfrac{2}{3}x+b$

The formula of a distance between a point and a line:

General form of a line:

$Ax+By+C=0$

Point:

$(x_0,\ y_0)$

Distance:

$d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+b^2}}$

Convert the equation:

$y=\dfrac{2}{3}x+b$ |subtract $y$ from both sides

$\dfrac{2}{3}x-y+b=0$ |multiply both sides by 3

$2x-3y+3b=0\to A=2,\ B=-3,\ C=3b$

Coordinates of the point:

$(0,\ 0)\to x_0=0,\ y_0=0$

substitute:

$d=4$

$4=\dfrac{|2\cdot0+(-3)\cdot0+3b|}{\sqrt{2^2+(-3)^2}}\\\\4=\dfrac{|3b|}{\sqrt{4+9}}$

$4=\dfrac{|3b|}{\sqrt{13}}\qquad|$ |multiply both sides by $\sqrt{13}$

$4\sqrt{13}=|3b|\iff3b=-4\sqrt{13}\ \vee\ 3b=4\sqrt{13}$ |divide both sides by 3

$b=-\dfrac{4\sqrt{13}}{3}\ \vee\ b=\dfrac{4\sqrt{13}}{3}$

Finally:

$y=\dfrac{2}{3}x-\dfrac{4\sqrt{13}}{3}\ \vee\ y=\dfrac{4\sqrt{13}}{3}$

4 0

2 years ago

Which expressions are not polynomials?

Naddika [18.5K]

Answer:

The middle two expressions shown are not polynomials

Step-by-step explanation:

Polynomial terms have positive integer exponents on any variables. A 2/3 power is not an integer exponent. Division by y is equivalent to an exponent of -1, which is not a positive integer.

8 0

3 years ago