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Airida [17]
3 years ago
15

(a) Write a pair of negative integers whose difference gives 8.

Mathematics
1 answer:
elixir [45]3 years ago
7 0
(a) -2 and -10
(b) -1 and 5
(c) 1 and -4
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Solve for c.<br><br> a(c−b)=d
tatiyna

Answer:

c= ab + d/ a

Step-by-step explanation:

Let's solve for c.

a(c-b)=d

Step 1: Add ab to both sides.

-ab + ac + ab = d+ ab

ac = ab + d

Step 2: Divide both sides by a.

ac / a = ab + d / a

c = ab + d/ a

Answer:

c= ab + d/ a

Hope this helps ☝️☝☝

3 0
3 years ago
What is the total cost of 0.5 pound of peaches selling for $0.80 per pound and 0.7 pound of oranges selling for $0.90 pet pound?
kari74 [83]
The total cost for peaches is $0.40 you get that by (0.5)(0.80) as for oranges (0.7)(.90)=$0.63 so $0.40+$0.63=$1.03
3 0
2 years ago
| x − 3 | &gt; 7 solve inequality
Cerrena [4.2K]
yuhhh get into ittttttt luv
6 0
3 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%28x%20-%201%29%28%20x%20-%20%28%20%5Cfrac%7B1%7D%7B2%7D%20%2B%203i%29%28x%20-%20%28%20%5Cfrac
abruzzese [7]

(X-1)(\frac{37}{4} -3ix+\frac{x}{2} )

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3 0
3 years ago
If two objects travel through space along two different curves, it’s often important to know whether they will collide. (Will a
galina1969 [7]

Answer:

<em>Both objects collide at t=3 in the point  <9,9,9></em>

Step-by-step explanation:

<em>Collision Of Moving Objects </em>

Two objects can describe different trajectories in the space. Those trajectories can intersect in one or more points but it doesn't mean they collide. Collision occurs if they are in the same position at the same time. If we know the positions as a function of time of each object, we could try so find if, for a given time, they are in the same position.

The positions of two object are given as

r1(t)=

r2(t)=

Let's find out if there is at least one value of t that makes both positions to be the same. We can try by equating one of the three coordinates and testing if the value of t make both have the same x,y,z coordinate. Let's try equating the x-components of both

t^2=4t-3

Rearranging

t^2-4t+3=0

Factoring

(t-1)(t-3)=0

We found two solutions

t=1,\ t=3

for t=1 the x-coordinates are

x1=t^2=1

x2=4t-3=1

For t=3

x1=t^2=9

x2=4t-3=9

Now we'll test both values in the y-coordinates

y1=7t-12

y2=t^2

For t=1

y1=-5

y2=1

Thus they don't collide at t=1. Let's try t=3

y1=7(3)-12=9

y2=3^2=9

Now let's try the z-coordinate for t=3

z1=t^2=9

z2=5t-6=9

Since the three coordinates match, we can say both objects collide at t=3 in the point  <9,9,9>

6 0
3 years ago
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