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3 years ago

Solve for x.

1 answer:

3 0

the answer would be x=1 because if u do the math u get 3/2 and that simplified is 1 so this how i got the answer not 100% sure though but thats what i got

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Find the image of A(-2,4) under a translation (x,y)->(x-2,y-3)

vichka [17]

Answer:

Point A would be at the coordinates, (-4, 1). Hope this helps!

4 0

2 years ago

Adrian earns $3600 per month. he spends 1/4 of his income on rent, 2/5 on food, 1/9 on activities, and he saves the rest. how mu

murzikaleks [220]

he saves $860 per month

calculate each of his outgoings

rent = $\frac{1}{4}$ × $3600 = $\frac{3600}{4}$ = $900

food = $\frac{2}{5}$ × $3600 = $\frac{2(3600)}{5}$ = $1440

activities = $\frac{1}{9}$ × $3600 = $\frac{3600}{9}$ = $400

savings = $3600 - ( 900 + 1440 + 400 ) = $860

4 0

3 years ago

Read 2 more answers

Can someone please help solve 2-7!!! I will give brainliest!!!

patriot [66]

Question 2 is 19.4 I hope I helped!

8 0

3 years ago

Read 2 more answers

An equation is given.

svp [43]

Answer:

option A is correct answer

divide both sides with 2

hope it helps

6 0

2 years ago

Read 2 more answers

In a lab experiment, a student is trying to apply the conservation of momentum. Two identical balls, each with a mass of 1.0 kg.

muminat

Answer:

Trial- 2 shows the conservation of momentum in a closed system.

Step-by-step explanation:

Given: Mass of balls are $m= 1.0\ kg$

Conservation of momentum in a closed system occurs when momentum before collision is equal to momentum after collision.

Let initial velocity of ball $A\ is\ u_1$
Initial velocity of ball $B\ is\ u_2$
Final velocity of ball $A\ is\ v_1$
Final velocity of ball $B\ is\ v_2$
Momentum before collision $= mu_1+mu_2$
Momentum after collision $=mv_1+mv_2$

Now, According to conservation of momentum.

Momentum before collision = Momentum after collision

$mu_1+mu_2=mv_1+mv_2$

We will plug each trial to this equation.

Trial 1

$mu_1+mu_2=mv_1+mv_2\\1.0(1)+1.0(-2)=1.0(-2)+1.0(-1)\\1-2=-2-1\\-1=-3$

Trial 2

$mu_1+mu_2=mv_1+mv_2\\1.0(.5)+1.0(-1.5)=1.0(-.5)+1.0(-\.5)\\.5-1.5=-.5-.5\\-1=-1$

Trial 3

$mu_1+mu_2=mv_1+mv_2\\1.0(2)+1.0(1)=1.0(1)+1.0(-2)\\2+1=1-2\\3=-1$

Trial 4

$mu_1+mu_2=mv_1+mv_2\\1.0(.5)+1.0(-1)=1.0(1.5)+1.0(-1.5)\\.5-1=1.5-1.5\\-.5=0$

We can see only Trial 2 satisfies the princple of conservation of momentum. That is momentum before collison should equal to momentum after collision.

5 0

3 years ago