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2 years ago

Adam plans to choose a video game from a section of the store where everything is

Mathematics

1 answer:

alexira [117]2 years ago

4 0

Original price of video game minus the sale price (-.75d) = the percent of original price Adam will pay (.25d)

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Which ordered pair is a solution to the system of linear equation? 2x+ 3y=6 <br> -3x+5y=10

allochka39001 [22]

Answer:

Step-by-step explanation:

2x + 3y = 6________( 1 )

-3x + 5y = 10_______( 2 )

( 1 ) × 3 ---- 6x + 9y = 18_____( 3 )

( 2 ) × 2 ---- -6x + 10y = 20_____( 4 )

$(4) + (3) \\ 6x + 9y + ( - 6x + 10y) = 20 - 18 \\ 6x + 9y - 6x - 10y = 2 \\ y = 2$

y = 2 ,

$2x + 3y = 6 \\ 2x + 3(2) = 2 \\ 2x + 6 = 2 \\ 2x = 2 - 6 \\ 2x = - 4 \\x = \frac{ - 4}{2} \\ x = ( - 2) \\$

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2 years ago

Idk help me ong , plzzz help meeeeeeeeee

krek1111 [17]

Answer:6\

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

What values for q (0 ≤q≤2π)<br> satisfy the equation?<br><br> 22√sin q + 2 = 0

Vesna [10]

Answer:
$\frac{3 \pi }{4}$ , $\frac{7 \pi }{4}$

Explanation:
2√2 sin(q) + 2 = 0
2√2 sin(q) = -2
sin(q) = $\frac{-2}{2 \sqrt{2} }$
sin(q) = $\frac{- \sqrt{2} }{2}$

Now, we know that:
sin (45) = $\frac{ \sqrt{2} }{2}$

From the ASTC rule, we know that the sine function is negative in the third and fourth quadrant.
This means that:
either q = 90 + 45 = 135° which is equivalent to $\frac{3 \pi }{4}$
or q = 270 + 45 = 315° which is equivalent to $\frac{7 \pi }{4}$

Hope this helps :)

6 0

3 years ago

Read 2 more answers

Y=99.99+23.75(x-1) as a Y=ax+b equation

Hatshy [7]

Answer:

$y = 23.75x + 76.24$

Step-by-step explanation:

We want to rewrite

$y = 99.99 + 23.75(x - 1)$

in the form

$y = ax + b$

We expand to get:

$y = 99.99 + 23.75x - 23.75$

Regroup similar terms:

$y = 23.75x - 23.75 + 99.99$

Simplify:

$y = 23.75x + 76.24$

This is now of the form

$y = ax + b$

where a=23.75 and b=76.24

6 0

3 years ago

Arrange the cones in order from lease volume to greatest volume

bearhunter [10]

Answer:

Volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

cone with DIAMETER of 18 & height of 10

cone with RADIUS of 10 & height of 9

cone with RADIUS of 11 & height of 9

cone with DIAMETER of 20 & height of 12

Step-by-step explanation:

Let $V_{2}. V_{3}. and\ V_{4}.$ be the volume of the cone.

Let d, r and h be the diameter, radius and height of the cone.

Given:

$d_{1} = 20\ and\ h_{1}=12$

$d_{2} = 18\ and\ h_{2}=10$

$r_{3} = 10\ and\ h_{3}=9$

$r_{4} = 11\ and\ h_{14}=9$

Arrange the cones in order from lease volume to greatest volume.

Solution:

The volume of the cone is given below.

$V=\pi r^{2} \frac{h}{3}$ ----------------(1)

where: r is radius of the base of cone.

and h is height of the cone.

The volume of the cone for $d_{1} = 20\ and\ h_{1}=12$

$r_{1} = \frac{d_{1}}{2}$

$r_{1} = \frac{20}{2}=10\ units$

$V_{1}=\pi (r_{1})^{2} \frac{h_{1}}{3}$

$V_{1}=\pi (10)^{2} \frac{12}{3}$

$V_{1}=\pi\times 100\times 4$

$V_{1}=400\pi\ units^{3}$

Similarly, for volume of the cone for $d_{2} = 18\ and\ h_{2}=10$

$r_{2} = \frac{d_{2}}{2}$

$r_{2} = \frac{18}{2}=9\ units$

$V_{2}=\pi (r_{2})^{2} \frac{h_{2}}{3}$

$V_{2}=\pi (9)^{2} \frac{10}{3}$

$V_{2}=\pi\times 81\times \frac{10}{3}$

$V_{2}=\pi\times 27\times 10$

$V_{2}=270\pi\ units^{3}$

Similarly, for volume of the cone for $r_{3} = 10\ and\ h_{3}=9$

$V_{3}=\pi (r_{3})^{2} \frac{h_{3}}{3}$

$V_{3}=\pi (10)^{2} \frac{9}{3}$

$V_{3}=\pi\times 100\times 3$

$V_{3}=\pi\times 300$

$V_{3}=300\pi\ units^{3}$

Similarly, for volume of the cone for $r_{4} = 11\ and\ h_{4}=9$

$V_{4}=\pi (r_{4})^{2} \frac{h_{4}}{3}$

$V_{4}=\pi (11)^{2} \frac{9}{3}$

$V_{4}=\pi\times 121\times 3$

$V_{4}=\pi\times 363$

$V_{4}=363\pi\ units^{3}$

So, the volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

7 0

2 years ago