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3 years ago

Solve this equation for a: 6a+3=45

Mathematics

1 answer:

fenix001 [56]3 years ago

3 0

Answer: 7

Step-by-step explanation:

Step 1: Subtract 3 from both sides.

6a+3−3=45−3

6a=42

Step 2: Divide both sides by 6.

6a /6 = 42 /6

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What is the answer and how to do this?

zhenek [66]

Answer:

Step-by-step explanation:

Given :

⇒ P (Sumit) = 1/2

⇒ P (Sujan) = 1/3

⇒ P (Rakesh) = 1/a

⇒ P (total) = 3/4

============================================================

Solving :

⇒ 1/2 × 1/3 × 1/a = 3/4

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2 years ago

Point G is the centroid of the right △ABC with m∠C=90° and m∠B=30°. Find AG if CG=4 ft.

o-na [289]

Answer: $\text{Length of AG=}\frac{2\sqrt{63}}{3}$

Explanation:

Please follow the diagram in attachment.

As we know median from vertex C to hypotenuse is CM

$\therefore CM=\frac{1}{2}AB$

We are given length of CG=4

Median divide by centroid 2:1

CG:GM=2:1

Where, CG=4

$\therefore GM=2$ ft

Length of CM=4+2= 6 ft

$\therefore CM=\frac{1}{2}AB\Rightarrow AB=12$

In $\triangle ABC, \angle C=90^0$

Using trigonometry ratio identities

$AC=AB\sin 30^0\Rightarrow AC=6$ ft

$BC=AB\cos 30^0\Rightarrow BC=6\sqrt{3}$ ft

$CN=\frac{1}{2}BC\Rightarrow CN=3\sqrt{3}$ ft

In $\triangle CAN, \angle C=90^0$

Using pythagoreous theorem

$AN=\sqrt{6^2+(3\sqrt{3})^2\Rightarrow \sqrt{63}$

Length of AG=2/3 AN

$\text{Length of AG=}\frac{2\sqrt{63}}{3}$ ft

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3 years ago

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