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Anestetic [448]
3 years ago
10

change the precent to a fraction 120% (simplify your answer) show work with this and steps on the side

Mathematics
1 answer:
storchak [24]3 years ago
4 0
Here are my steps 11/5= 2 1/5
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Please help this is due tomorrow morning
Minchanka [31]

Answer:

I can try to help you..uhh alright 1ll it's definitely multiplying by 4 on the bottom but adding 1 one the top. I'm not sure if i'll be correct but try 1 and 4 i'm so sorry if it doesn't help!

7 0
3 years ago
The Value of x in the system of equation at the right is -2 what is the value of a? use elimination to help you find the value.
sammy [17]

2x-4y=-16
ax+4y=6      +
--------------------
2x+ax=-10;     for x=-2, 
2(-2)+a(-2)=-10
-2a=-10+4
a=-6/-2
a=3
7 0
3 years ago
Nancy prepared 12 kilograms of dough after working 6 hours. How much dough did Nancy prepare if she worked for 10 hours? Solve u
damaskus [11]

Answer:

20 kilograms of dough

Step-by-step explanation:

Given that;

12 kilograms working 6 hours

Rate = 12/6 kilogram per hour = 2 kilogram per hour

So,

In 10 hours he would produce;

Amount of dough = Time × rate

Time = 10 hours

= 10 hours × 2 kilograms per hour

= 20 kilograms of dough

6 0
3 years ago
Read 2 more answers
How is the graph of y = |x| − 1 different from the parent function f(x) = |x|?
Jlenok [28]

Answer:

Step-by-step explanation:

the graph moves down one unit

if you add/subtract outside the parent function the graph moves up/down

3 0
3 years ago
Plz hurry!!!! thank you!!!!
Nikitich [7]

Answer:

Area of trapezium = 4.4132 R²

Step-by-step explanation:

Given, MNPK is a trapezoid

MN = PK and ∠NMK = 65°

OT = R.

⇒ ∠PKM = 65° and also ∠MNP = ∠KPN = x (say).

Now, sum of interior angles in a quadrilateral of 4 sides = 360°.

⇒ x + x + 65° + 65° = 360°

⇒ x = 115°.

Here, NS is a tangent to the circle and ∠NSO = 90°

consider triangle NOS;

line joining O and N bisects the angle ∠MNP

⇒ ∠ONS = \frac{115}{2} = 57.5°

Now, tan(57.5°) = \frac{OS}{SN}

⇒ 1.5697 = \frac{R}{SN}

⇒ SN = 0.637 R

⇒ NP = 2×SN = 2× 0.637 R = 1.274 R

Now, draw a line parallel to ST from N to line MK

let the intersection point be Q.

⇒ NQ = 2R

Consider triangle NQM,

tan(∠NMQ) = \frac{NQ}{QM}

⇒ tan65° = \frac{NQ}{QM}

⇒ QM = \frac{2R}{2.1445}

QM = 0.9326 R .

⇒ MT = MQ + QT

          = 0.9326 R + 0.637 R  (as QT = SN)

⇒ MT = 1.5696 R

⇒ MK = 2×MT = 2×1.5696 R = 3.1392 R

Now, area of trapezium is (sum of parallel sides/ 2)×(distance between them).

⇒ A = (\frac{NP + MK}{2}) × (ST)

       = (\frac{1.274 R + 3.1392 R}{2}) × 2 R

       = 4.4132 R²

⇒ Area of trapezium = 4.4132 R²

5 0
3 years ago
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