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Savatey [412]
3 years ago
6

These lines are perpendicular. Is this statement true or false?

Mathematics
1 answer:
umka2103 [35]3 years ago
5 0
False they are not exact recipricals



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-73 greater than or equal to 15 + 11x
Alecsey [184]
-73 ≥ 15 + 11x 
so the first thing you should do is to add -15 on both sides of the equation so you will have:
-73 -15 ≥ 11x 
- 88 ≥ 11x 
so the answer is:
-88/11 ≥ x or -8 ≥ x :)))
i hope this is helpful
have a nice day 
5 0
3 years ago
Read 2 more answers
I’m not sure how to solve “system of equations”
natka813 [3]

Hey there! :)

Answer:

(2, -2)

Step-by-step explanation:

-2x + y = -6

4x + 3y = 2

We can begin by setting the first equation equal to y:

-2x + y = -6

Add 2x to both sides:

y = 2x - 6

Plug this equation for y into the second equation:

4x + 3(2x - 6) = 2

Distribute:

4x + 6x - 18 = 2

Combine like terms:

10x = 20

x = 2

Plug the value of 'x' into an equation to solve for 'y':

-2(2) + y = -6

-4 + y = -6

y = -2

Therefore, the solution is (2, -2)

7 0
3 years ago
The students in Mr. Wilson's Physics class are making golf ball catapults. The
Mnenie [13.5K]

Answer:

Part a) About 48.6 feet

Part b) About 8.3 feet

Part c) The domain is 0 \leq x \leq 48.6\ ft and the range is 0 \leq y \leq 8.3\ ft

Step-by-step explanation:

we have

y=-0.014x^{2} +0.68x

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

where

x is the ball's  distance from the catapult in feet

y is the flight of the balls in feet

Part a)  How far did the ball fly?

Find the x-intercepts or the roots of the quadratic equation

Remember that

The x-intercept is the value of x when the value of y is equal to zero

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

-0.014x^{2} +0.68x=0

so

a=-0.014\\b=0.68\\c=0

substitute in the formula

x=\frac{-0.68(+/-)\sqrt{0.68^{2}-4(-0.014)(0)}} {2(-0.014)}

x=\frac{-0.68(+/-)0.68} {(-0.028)}

x=\frac{-0.68(+)0.68} {(-0.028)}=0

x=\frac{-0.68(-)0.68} {(-0.028)}=48.6\ ft

therefore

The ball flew about 48.6 feet

Part b) How high above the ground did the ball fly?

Find the maximum (vertex)

y=-0.014x^{2} +0.68x

Find out the derivative and equate to zero

0=-0.028x +0.68

Solve for x

0.028x=0.68

x=24.3

<em>Alternative method</em>

To determine the x-coordinate of the vertex, find out the midpoint  between the x-intercepts

x=(0+48.6)/2=24.3\ ft

To determine the y-coordinate of the vertex substitute the value of x in the quadratic equation and solve for y

y=-0.014(24.3)^{2} +0.68(24.3)

y=8.3\ ft

the vertex is the point (24.3,8.3)

therefore

The ball flew above the ground about 8.3 feet

Part c) What is a reasonable domain and range for this function?

we know that

A  reasonable domain is the distance between the two x-intercepts

so

0 \leq x \leq 48.6\ ft

All real numbers greater than or equal to 0 feet and less than or equal to 48.6 feet

A  reasonable range is all real numbers greater than or equal to zero and less than or equal to the y-coordinate of the vertex

so

we have the interval -----> [0,8.3]

0 \leq y \leq 8.3\ ft

All real numbers greater than or equal to 0 feet and less than or equal to 8.3 feet

8 0
3 years ago
----------------------------------------------------------------------------------------<br> 10x10?
a_sh-v [17]
10x10=100
------------------------------------ :))
3 0
3 years ago
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Sophia needs $47 to go on a field trip. She has saved $11.50. She earns $6.50 per hour cleaning her neighbor's garden, and she e
laiz [17]
<span>a) Yes, because the total will be $48
hope i helped</span>
6 0
3 years ago
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