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3 years ago

If you help me answer this question i will give you brainiest!!

Mathematics

2 answers:

coldgirl [10]3 years ago

5 0

Angle BDC is given as 38° but not drawn in the figure. Amateurs.

We calculate angle DBC

DBC = 180° - 96° - 38° = 180° - 134°

We won't bother subtracting because we're really after

angle ABD = 180° - DBC = 134°

Now we have a Law of Cosines situation,

AD² = AB² + BD² - 2(AB)(BD) cos ABD

AD = √( 5.8² + 27.3² - 2(5.8)(27.3) cos 134°)

AD ≈ 31.61 m

Answer: 31.6 m

NikAS [45]3 years ago

4 0

Answer:

31.6m

Step-by-step explanation:

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What is the equation of a line perpendicular to -x +2=y that intersects the line at (-1.5, 3.5)?

Nina [5.8K]

Answer:

y=x+5

Step-by-step explanation:

y-y1=m(x-x1)

y=-x+2

slope of y=-x+2 is -1, perpendicular slope is 1

y-3.5=1(x-(-1.5))

y-3.5=x+1.5

y=x+5

6 0

3 years ago

Rationalise the denominator.

grin007 [14]

Let's see what to do buddy...

_________________________________

To do this, we have to multiply the face and denominator of the fraction by the denominator conjunction, which is :

$(5 -\sqrt{2})$

°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°

Reminder :

$(a + b)(a - b) = {a}^{2} - {b}^{2} \\$

°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°

So we have :

$\frac{1}{ 5-\sqrt{2} } \times \frac{ 5+\sqrt{2} }{ 5 +\sqrt{2} } = \frac{1 \times ( 5 +\sqrt{2}) }{(5- \sqrt{2} )( 5 +\sqrt{2}) } = \\$

The other is in the photo.

_________________________________

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

7 0

3 years ago

17. Find the length of the hypotenuse of an isosceles right triangle whose legs measure

defon

Answer:

4.950 m

Step-by-step explanation:

3.5^2+3.5^2=hypotenuse^2

3 0

3 years ago

Whats .4 (repeating) divided by 3/8??

hichkok12 [17]

.444444 divided by 3/8 = 0.01666666666

6 0

3 years ago

Read 2 more answers

PLS HELP ME ON THIS QUESTION I WILL MARK YOU AS BRAINLIEST IF YOU KNOW THE ANSWER!!

nata0808 [166]

Answer:

Step-by-step explanation:

Survey is the easiest, best, and fastest way o collect data.

(especially from if it's large)

<h2>I hope this helped!!</h2>

6 0

3 years ago