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Troyanec [42]
3 years ago
8

What is 83.9 rounded two decimal places

Mathematics
2 answers:
Effectus [21]3 years ago
8 0
<h2><em>Rounding it 2 decimal places will end up with 83.90.</em></h2>
Stels [109]3 years ago
7 0
Answer:

~ 83.90

2 decimal place is basically the hundredths place
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Here's another problem plz answer this correct 10 points and brainly!<br><br> NO LINKS NO SCAM LINKS
Dafna11 [192]
Lolz i don’t know I just want points:)
8 0
3 years ago
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If S is 20 when T is 4, then T is what, when S is 30.
Digiron [165]
When s is 20 t is 4
when s is 10 t is 2
when s is 30 t is 6
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3 years ago
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True or false I need help plzzz
Stels [109]

Answer:

i cant rlly see it

Step-by-step explanation:

3 0
3 years ago
Select the correct answer
Tomtit [17]

Answer:

Step-by-step explanation:

The area for a triangle is

A=\frac{1}{2}bh

Our base is given as x, the height is given as 1 more than 6 times the base which looks like this: 6x + 1. We know that the area is 13, so filling in everything:

13=\frac{1}{2} (6x+1)(x)

We can get rid of the denominator of 2 by multiplying it on both sides to get:

26=(6x+1)(x) and simplifying a bit:

26=6x^2+x

Now we can move over the 26 and put the expression into standard quadratic form by setting it equal to 0:

6x^2+x-26=0 which is choice C

3 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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