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3 years ago

A, B & C form a triangle where Z BAC = 90°.

Mathematics

1 answer:

zhannawk [14.2K]3 years ago

8 0

Answer:

BC ≈ 8.6

Step-by-step explanation:

Using Pythagoras' identity in the right triangle , that is

BC² = AB² + AC² = 7.3² + 4.6² = 53.29 + 21.16 = 74.45 ( square root both sides )

BC = $\sqrt{74.45}$ ≈ 8.6 ( to 1 dec. place )

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Each of 3 bags contains x pencils. Tyler adds 2 more pencils to each bag. Altogether, the gift bags contain 12 pencils.

WARRIOR [948]

Answer:

Below

Step-by-step explanation:

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3 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

5 0

3 years ago

Of angle JEK is complementary to angle DFH, then

siniylev [52]

I would say the answer is C

6 0

3 years ago

Which function is the inverse of f(x) = 2x+37

WINSTONCH [101]

Answer:

$f^{-1} (x)=\frac{x-3}{2}$

Step-by-step explanation:

y = f(x)

y = 2x+3

You need to switch the places of x and y, into this:

x = 2y + 3

Then solve for y:

2y = x-3

$y=\frac{x-3}{2}$

$f^{-1} (x)=\frac{x-3}{2}$

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3 years ago

Help anyone can help me do this question,I will mark brainlest.

astraxan [27]

d =√((x_2-x_1)²+(y_2-y_1)²)

This is the formula of distance between

So, I hope u will get all answer of the questions by using this formula...

Thanks...

8 0

3 years ago