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ololo11 [35]
3 years ago
5

I need help again I used to love math but I’m not good at it no more

Mathematics
2 answers:
zavuch27 [327]3 years ago
4 0

Answer:

<h2>A    </h2>

x-38.45=17.71

Step-by-step explanation:

stiks02 [169]3 years ago
3 0

Even if you're not that good, you can still love it!

Well x stands for the money she had in her pocket when she first started.

38.45 is the price of a new shirt.

17.71 is what's left in her pocket after spending the price on the shirt.

So that means you have to subtract 38.45 from "x" to get 17.71:

x - 38.45 = 17.71

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Factor completely. x2−8x+15
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Transversal / cuts lines a, b, c, and d.<br> Which two lines are parallel?
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An aircraft jet engine has a thrust of I80N. What impulse is produced<br> in 30 S?
neonofarm [45]

The impulse that was produced by this aircraft jet engine is 5,400 Ns.

<u>Given the following data:</u>

  • Thrust = 180 Newton.
  • Time = 30 seconds.

To determine the impulse that was produced by this aircraft jet engine:

<h3>How to calculate impulse.</h3>

Mathematically, impulse is given by this formula:

I = Ft

<u>Where:</u>

  • F is the force or thrust.
  • t is the time measured in seconds.

Substituting the given parameters into the formula, we have;

I = 180 \times 30

Impulse, I = 5,400 Ns.

Read more on impulse here: brainly.com/question/16750406

8 0
2 years ago
EASY TRIG - 15 POINTS
poizon [28]

Answer: The answers are (a) 40 cm and (b) \sin^{-1}\dfrac{23}{62}.


Step-by-step explanation: The calculations are as follows:

(a) See the figure (a). As given in the question, A circle with centre 'O' circumscribes a triangle ABC with BC = 20 cm and ∠BAC = 30°. We need to find the diameter DC of the circle.

Let us draw BD. Now, ∠BAC and ∠BDC are angles on the same arc BC, so we have

∠BAC = ∠BDC = 30°.

Also, ∠CBD = 90°, since it stands on the diameter DC. So, ΔBCD will be a right angled triangle.

We can write

\sin \angle BDC=\dfrac{BC}{DC}\\\\\\ \Rightarrow \sin 30^\circ=\dfrac{20}{DC}\\\\\\\Rightarrow \dfrac{1}{2}=\dfrac{20}{DC}\\\\\\\Rightarrow DC=40.

Thus, the diameter of the circle = 40 cm.


(b) See the figure (b).

As given in the question, A circle with centre 'O'' circumscribes a triangle DEF with EF = 4.6 inches and diameter GF = 12.4 in.. We need to find the angle EDF.

Let us draw GE. Now, ∠EGF and ∠EDF are angles on the same arc EF, so we have

∠EGF = ∠EDF = ?

Also, ∠GEF = 90°, since it stands on the diameter GF. So, ΔGEF will be a right angled triangle.

We can write

\sin \angle EGF=\dfrac{EF}{GF}\\\\\\ \Rightarrow \sin \angle EGF=\dfrac{4.6}{12.4}\\\\\\\Rightarrow \sin \angle EGF=\dfrac{23}{62}\\\\\\\Rightarrow \angle EGF=\sin^{-1}\dfrac{23}{62}.

Thus,

\angle EDF=\angle EGF=\sin^{-1}\dfrac{23}{62}.

4 0
3 years ago
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