All categories

3 years ago

Please help with this

Mathematics

1 answer:

solong [7]3 years ago

3 0

Could you retake the picture it’s hard to read

You might be interested in

Part B

solmaris [256]

The length of the other leg of the right triangle will be 13.86 cm.

The complete question is given below.

The triangle BCD is a right triangle. The length of the hypotenuse is 19 centimeters. The length of one of the legs is 13 centimeters.

What is the length of the other leg?

<h3>What is a Pythagoras theorem?</h3>

The Pythagoras theorem states that the sum of two squares equals the squared of the longest side.

The Pythagoras theorem formula is given as

H² = P² + B²

Let the unknown sides be x. Then we have,

19² = 13² + x²

361 = 169 + x²

x² = 361 – 169

x² = 192

x = 13.856 ≈ 13.86 cm

More about the Pythagoras theorem link is given below.

brainly.com/question/343682

#SPJ1

5 0

2 years ago

The measure of an angle is 2°. What is the measure of its complementary angle?

MAVERICK [17]

Answer:

88°

Step-by-step explanation:

Complementary angles have a sum of 90°, just do 90 - 2 and there's your answer.

5 0

3 years ago

The box plot represents the number on minutes it takes for the student in a class to run a mile. WHAT IS THE MINIMUM OF THE DATA

tatyana61 [14]

B. 5

the minimum is where the end of the left tail is. the first quartile is where the beginning of the box is, the mid is in the middle of the box, the 3rd quartile is at the end of the box and the maximum is the end of the right tail.

7 0

3 years ago

When you read an article, how can you tell which sentence states the author’s claim?

Neko [114]

Answer:

The claim states a “truth” that two reasonable people might disagree about.

Step-by-step explanation:

ive done this quiz

7 0

2 years ago

Integrate dx/3sinx+4cosx

german

$\displaystyle\int\frac{\mathrm dx}{3\sin x+4\cos x}$

A standard approach would be the tangent half-angle substitution:

$t=\tan\dfrac x2\implies\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx$

Then

$\sin x=2\sin\dfrac x2\cos\dfrac x2\implies\sin x=\dfrac{2t}{1+t^2}$

$\cos x=\cos^2\dfrac x2-\sin^2\dfrac x2\implies\cos x=\dfrac{1-t^2}{1+t^2}$

from which we get

$\mathrm dx=\dfrac2{1+t^2}\,\mathrm dt$

So the integral becomes

$\displaystyle\int\frac{\frac2{1+t^2}}{\frac{6t}{1+t^2}+\frac{4(1-t^2)}{1+t^2}}\,\mathrm dt=\int\frac{\mathrm dt}{3t+2(1-t^2)}=-\int\frac{\mathrm dt}{2t^2-3t-2}$