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Mademuasel [1]
3 years ago
7

Move (not take away) ONE stick to make the equation true

Mathematics
1 answer:
omeli [17]3 years ago
3 0
Move the middle line of the +
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15 POINTS PLEASE HELP
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What is the slope of the line that passes through (-3,7) and (5,-1)
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To find the slope(m), use the slope formula:

m=\frac{y_2-y_1}{x_2-x_1}     And plug in the two points on the line

(-3, 7) = (x₁, y₁)

(5, -1) = (x₂, y₂)

m=\frac{y_2-y_1}{x_2-x_1}

m=\frac{-1-7}{5-(-3)}  (two negative signs cancel each other out and become positive)

m=\frac{-1-7}{5+3}

m=\frac{-8}{8}     Simplify the fraction

m = -1

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In a telephone poll, 11 people said they like shopping and 9 people said they do not like shopping. What is the ratio of the num
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9:11

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A $75 fund is available for a holiday party. If 75% of the available money is spent for food and beverages, how much is left for
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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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