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2 years ago

{2x + 3y = 5}{ y= 5 x -4}

Mathematics

1 answer:

Natasha_Volkova [10]2 years ago

8 0

Solve the system using substitutions:

$\begin{gathered} 2x+3y=5 \\ 2x+3(5x-4)=5 \\ 2x+15x-12=5 \\ 17x=5+12 \\ 17x=17 \\ x=\frac{17}{17} \\ x=1 \end{gathered}$

x has a value of 1. Use this value to find the value of y.

$\begin{gathered} y=5x-4 \\ y=5\cdot1-4 \\ y=5-4 \\ y=1 \end{gathered}$

y has a value of 1.

x=1

y=1

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Find an equation equivalent to r = 10sin ø in rectangular coordinates.

kherson [118]

Answer: x^2 + y^2 -10y = 0

Step-by-step explanation:

Cartesian coordinates, also called the Rectangular coordinates, isdefined in terms of x and y. So, for the problem θ has to be eliminated or converted using basic foundations that are described by the unit circle and the right triangle trigonometry.

r= 10sin(θ)

Remember that:

x= r × cos(θ)

y= r × sin(θ)

r^2= x^2 + y^2

Multiply both sides of the equation by r. This will give:

r × r = 10r × sin(θ)

r^2 = 10r × sin(θ)

x^2 + y^2= 10r × sin(θ)

Because y= r × sin(θ), we can make a substitution. This will be:

x^2 + y^2= 10y

x^2 + y^2 -10y = 0

The above equation is the Rectangular coordinate equivalent to the given equation.

4 0

4 years ago

The top and bottom margins of a poster are each 12 cm and the side margins are each 8 cm. The area of printed material on the po

grin007 [14]

Answer:

Dimensions of printed poster are

length is 32 cm

width is 48 cm

Step-by-step explanation:

Let's assume

length of printed poster is x cm

width of printed poster is y cm

now, we can find area of printed poster

so, area of printed poster is

$=xy$

we are given that area as 1536

so, we can set it to 1536

$xy=1536$

now, we can solve for y

$y=\frac{1536}{x}$

now, we are given

The top and bottom margins of a poster are each 12 cm and the side margins are each 8 cm

so, total area of poster is

$A=(8+x+8)\times (12+y+12)$

$A=(x+16)\times (y+24)$

now, we can plug back y

$A=(x+16)\times (\frac{1536}{x}+24)$

now, we have to minimize A

so, we will find derivative

$A'=\frac{d}{dx}\left(\left(x+16\right)\left(\frac{1536}{x}+24\right)\right)$

we can use product rule

$A'=\frac{d}{dx}\left(x+16\right)\left(\frac{1536}{x}+24\right)+\frac{d}{dx}\left(\frac{1536}{x}+24\right)\left(x+16\right)$

$=\frac{d}{dx}\left(x+16\right)\left(\frac{1536}{x}+24\right)+\frac{d}{dx}\left(\frac{1536}{x}+24\right)\left(x+16\right)$

now, we can simplify it

$A'=-\frac{24576}{x^2}+24$

now, we can set it to 0

and then we can solve for x

$A'=-\frac{24576}{x^2}+24=0$

$-\frac{24576}{x^2}x^2+24x^2=0\cdot \:x^2$

$-24576+24x^2=0$

$x=32,\:x=-32$

Since, x is dimension

and dimension can never be negative

so, we will only consider positive value

$x=32$

now, we can solve for y

$y=\frac{1536}{32}$

$y=48$

so, dimensions of printed poster are

length is 32 cm

width is 48 cm

5 0

3 years ago

10. Polygon WXYZ has vertices W(-1.5, 1.5),

vovikov84 [41]

Answer:

See explanation

Step-by-step explanation:

Polygon WXYZ has vertices W(-1.5, 1.5), X(6, 1.5), Y(6, -4.5), and Z(-1.5, -4.5).

Find the slopes of all sides:

$WX:\ \dfrac{1.5-1.5}{6-(-1.5)}=\dfrac{0}{7.5}=0$ - horizontal line

$XY:\ \dfrac{-4.5-1.5}{6-6}=\dfrac{-6}{0}$ - slope undefined, so the line is vertical

$YZ:\ \dfrac{-4.5-(-4.5)}{-1.5-6}=\dfrac{0}{-7.5}=0$ - horizontal line

$ZW:\ \dfrac{1.5-(-4.5)}{-1.5-(-1.5)}=\dfrac{6}{0}$ - slope undefined, so the line is vertical

Vertical and horizontal lines are perpendicular, so all angles W, X, Y and Z are right angles. Hence, quadrilateral WXYZ is a rectangle

6 0

4 years ago

Tomas drew a rectangle with an area of 6 square centimeters. What is the greatest possible perimeter for this rectangle.

Delvig [45]

6 by 1.

This is the greatest because it makes 14 (6+6+1+1), which is greater than 3 by 2, which is only 10. (3+3+2+2)

Hope this helps!

8 0

4 years ago

Find the simplified product b-5/2b x b^2+3b/b-5

White raven [17]

Answer:

The product $\frac{b-5}{2b}\times\frac{b^2+3b}{b-5}=\frac{b+3}{2}$

Step-by-step explanation:

Given expression $\frac{b-5}{2b}$ and $\frac{b^2+3b}{b-5}$

We have to find the product of $\frac{b-5}{2b}\times\frac{b^2+3b}{b-5}$

Consider the given expression $\frac{b-5}{2b}\times\frac{b^2+3b}{b-5}$

Multiply fractions, we have,

$\frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d}$

$=\frac{\left(b-5\right)\left(b^2+3b\right)}{2b\left(b-5\right)}$

Cancel common factor ( b - 5 )

we have, $=\frac{b^2+3b}{2b}$

Apply exponent rule,

$\:a^{b+c}=a^ba^c$

$b^2=bb$

$=bb+3b=b(b+3)$

$=\frac{b\left(b+3\right)}{2b}$

Cancel common factor b , we have,

$=\frac{b+3}{2}$

Thus, the product $\frac{b-5}{2b}\times\frac{b^2+3b}{b-5}=\frac{b+3}{2}$

8 0

4 years ago

Read 2 more answers