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lakkis [162]
2 years ago
11

Bob is standing 25 feet from a lamppost that is to his left and 30 feet from a lamppost that is to his right. The distance betwe

en the two lampposts is 20 feet. What is the measure of the angle formed from the line from each lamppost to Bob? Approximate to the nearest degree. 1. 202 = 252 302 − 2(25)(30)cos(A) 2. 400 = 625 900 − (1500)cos(A) 3. 400 = 1525 − (1500)cos(A) 4. −1125 = −(1500)cos(A) degrees.
Mathematics
1 answer:
irina1246 [14]2 years ago
3 0

Answer:

The answer is 41

Step-by-step explanation:

400 = 900 + 625 - 1500 cos x

400 - 1525 = -1500 cos x

-1125 = -1500 cos

cos x = -1125/-1500

x = cos⁻¹ 0.75

x = 41.409622109

x ≈ 41°

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IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. If a certain statistician has an IQ of 112
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Step-by-step explanation:

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3 years ago
Name a pair of overlapping congruent triangles in each diagram. State whether the triangles are congruent by SSS, SAS, ASA, AAS,
MariettaO [177]

Answer:

ΔPTS≅ΔRTA by AAS axiom of congruency

Step-by-step explanation:

Consider ΔPQA and ΔRQS

∠PQA=∠RQS     (Vertically Opposite Angles)

∠QAP=∠QSR     (Complementary of two equal angles, ∠RAT and∠PST)

Due to angle sum property of a triangle, we come to the conclusion that

∠APQ=∠SRQ

Consider ΔPTS and ΔRTA

TA=TS  (Given)

∠RAT=∠PST(Given)

∠APQ=∠SRQ (Proved above)

Therefore, ΔPTS≅ΔRTA by AAS axiom of congruency.

8 0
3 years ago
Can someone please help me on number 16-ABC
melomori [17]

Answer:

Please check the explanation.

Step-by-step explanation:

Given the inequality

-2x < 10

-6 < -2x

<u>Part a) Is x = 0 a solution to both inequalities</u>

FOR  -2x < 10

substituting x = 0 in -2x < 10

-2x < 10

-3(0) < 10

0 < 10

TRUE!

Thus, x = 0 satisfies the inequality -2x < 10.

∴ x = 0 is the solution to the inequality -2x < 10.

FOR  -6 < -2x

substituting x = 0 in -6 < -2x

-6 < -2x

-6 < -2(0)

-6 < 0

TRUE!

Thus, x = 0 satisfies the inequality -6 < -2x

∴ x = 0 is the solution to the inequality -6 < -2x

Conclusion:

x = 0 is a solution to both inequalites.

<u>Part b) Is x = 4 a solution to both inequalities</u>

FOR  -2x < 10

substituting x = 4 in -2x < 10

-2x < 10

-3(4) < 10

-12 < 10

TRUE!

Thus, x = 4 satisfies the inequality -2x < 10.

∴ x = 4 is the solution to the inequality -2x < 10.

FOR  -6 < -2x

substituting x = 4 in -6 < -2x

-6 < -2x

-6 < -2(4)

-6 < -8

FALSE!

Thus, x = 4 does not satisfiy the inequality -6 < -2x

∴ x = 4 is the NOT a solution to the inequality -6 < -2x.

Conclusion:

x = 4 is NOT a solution to both inequalites.

Part c) Find another value of x that is a solution to both inequalities.

<u>solving -2x < 10</u>

-2x\:

Multiply both sides by -1 (reverses the inequality)

\left(-2x\right)\left(-1\right)>10\left(-1\right)

Simplify

2x>-10

Divide both sides by 2

\frac{2x}{2}>\frac{-10}{2}

x>-5

-2x-5\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-5,\:\infty \:\right)\end{bmatrix}

<u>solving -6 < -2x</u>

-6 < -2x

switch sides

-2x>-6

Multiply both sides by -1 (reverses the inequality)

\left(-2x\right)\left(-1\right)

Simplify

2x

Divide both sides by 2

\frac{2x}{2}

x

-6

Thus, the two intervals:

\left(-\infty \:,\:3\right)

\left(-5,\:\infty \:\right)

The intersection of these two intervals would be the solution to both inequalities.

\left(-\infty \:,\:3\right)  and \left(-5,\:\infty \:\right)

As x = 1 is included in both intervals.

so x = 1 would be another solution common to both inequalities.

<h3>SUBSTITUTING x = 1</h3>

FOR  -2x < 10

substituting x = 1 in -2x < 10

-2x < 10

-3(1) < 10

-3 < 10

TRUE!

Thus, x = 1 satisfies the inequality -2x < 10.

∴ x = 1 is the solution to the inequality -2x < 10.

FOR  -6 < -2x

substituting x = 1 in -6 < -2x

-6 < -2x

-6 < -2(1)

-6 < -2

TRUE!

Thus, x = 1 satisfies the inequality -6 < -2x

∴ x = 1 is the solution to the inequality -6 < -2x.

Conclusion:

x = 1 is a solution common to both inequalites.

7 0
3 years ago
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