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iren2701 [21]
3 years ago
6

How many posters and picture frames does the store sell on that day?

Mathematics
1 answer:
USPshnik [31]3 years ago
3 0
What is the question
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8−5y=−2xin standerd form
vazorg [7]
The answer is 8=-2x+5y
7 0
3 years ago
Help with this please ..
abruzzese [7]

Answer:

the answer is 2.59x10^7

7 0
3 years ago
What value of y makes this equation true?<br><br> 2(−4y−7)+8=−2y+18+6y ?
rodikova [14]

Answer:

idhjjg54543

Step-by-step explanation-6790985

4 0
3 years ago
(-4, 7), (-6,-4)<br><br> find the slope of the line through each pair of points
HACTEHA [7]

Answer:

slope = 11/2

Step-by-step explanation:

If you are given two points, you can find the slope using the point-slope equation. The equation looks like this:

y₁ - y₂ = m(x₁ - x₂)

In this form, "m" represents the slope, "x₁" and "y₁" represent the values from one point, and "x₂" and "y₂" represent the values from the other point. You can plug the values from the points into the equation and simplify to find the slope.

Point 1: (-4, 7)                Point 2: (-6, -4)

x₁ = -4                              x₂ = -6

y₁ = 7                               y₂ = -4

y₁ - y₂ = m(x₁ - x₂)                               <----- Point-slope form

7 - (-4) = m(-4 - (-6))                            <----- Insert values

11 = m(2)                                             <----- Simplify

11/2 = m                                             <----- Divide both sides by 2

5 0
2 years ago
Read 2 more answers
The height H of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 9
mariarad [96]

Answer:

The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.

Step-by-step explanation:

Statement is incorrect. Correct form is presented below:

<em>The height </em>h(t)<em> of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 90 feet per second is given by equation </em>h(t) = 3 +90\cdot t -16\cdot t^{2}<em>, where </em>t<em> is time in seconds. After how many seconds will the ball be 84 feet above the ground. </em>

We equalize the kinematic formula to 84 feet and solve the resulting second-order polynomial by Quadratic Formula to determine the instants associated with such height:

3+90\cdot t -16\cdot t^{2} = 84

16\cdot t^{2}-90\cdot t +81 = 0 (1)

By Quadratic Formula:

t_{1,2} = \frac{90\pm \sqrt{(-90)^{2}-4\cdot (16)\cdot (81)}}{2\cdot (16)}

t_{1} = 4.5\,s, t_{2} = 1.125\,s

The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.

3 0
3 years ago
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