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2 years ago

The _____ of an angle is the ratio of the adjacent leg length to the hypotenuse length.

Mathematics

2 answers:

Savatey [412]2 years ago

8 0

Answer:

Cosine

Step-by-step explanation:

We have to fill the given blank space .

We know that

$sin\theta=\frac{Opposite\;side}{Hypotenuse}$

$cos\theta=\frac{Adjacent\;side}{Hypotenuse}$

$Tan\theta=\frac{Opposite\;side}{adjacent\;side}$

Therefore, the cosine of an angle is the ratio of the adjacent leg length to the hypotenuse length.

Answer: Cosine

nlexa [21]2 years ago

4 0

Answer:cosine ,just answered in apex

Step-by-step explanation:

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Gnesinka [82]

Answer:

y= -l×+4l-1

Step-by-step explanation:

To find x-intercept/zero, substitute y=0

0= -l×+4l-1

Sole the equation for x

x∈∅

Since there is no solution for y=0, there is no x-intercept/zero

No x-intercept/Zero

5 0

3 years ago

Read 2 more answers

Find the missing angles

atroni [7]

Answer:

y=55

x=70

Step-by-step explanation:

6 0

2 years ago

Simplify 8^-7 x 8^5 please show work

Pie

When you are multiplying exponents with the same base, you can add the exponents. This is represented as:

$x^a \cdot x^b = x^{(a +b)}$

In our case we are multiplying exponents with the same base 8 and exponents -7 and 5. The sum of -7 and 5 is -2, so our answer is $\boxed{8^{-2}}$ .

7 0

2 years ago

1. Explain why n root a^n= |a| when n is even, but not when n is odd.

kogti [31]

Step-by-step explanation:

In the expression a^n, for integer values of n greater than 1, there are n factors. For example, a^2 = a * 2 (2 factors), a^3 = a * a * a (3 factors), etc.

For a non-negative value of a, a^n is non-negative for all values of n.

If a is negative, and n is even, then a^n is non-negative.

If a is negative, and n is odd, then a^n is negative.

|a| is non-negative for all values of a.

sqrt_n(a^n) is negative for negative a and odd n, but |a| is always non-negative, so sqrtn(a^n) cannot equal |a| for odd n.

8 0

3 years ago

A water tank in the shape of an inverted circular cone has a base radius of 3 m and height of 7 m . If water is being pumped int

Wittaler [7]

The water level increases by 0.608 meters per minute when the water is 3.5 m deep

<h3>How to determine the rate?</h3>

The given parameters are:

Radius, r = 3
Height, h = 7
Rate in, V' = 4.3m^3/min

The relationship between the radius and height is:

r/h = 3/7

Make r the subject

r = 3h/7

The volume of a cone is;

$V = \frac 13\pi r^2h$

This gives

$V = \frac 13\pi (\frac{3h}{7})^2h$

Expand

$V = \frac{3h^3}{49}\pi$

Differentiate

$V' = \frac{9h^2}{49}\pi h'$

Make h' the subject

$h' = \frac{49}{9\pi h^2}V'$

When the water level is 3.5.

We have:

$h' = \frac{49}{9\pi * 3.5^2}V'$

Also, we have:

V' = 4.3

So, the equation becomes

$h' = \frac{49}{9\pi * 3.5^2} * 4.3$

Evaluate the products

$h' = \frac{210.7}{346.36}$

Evaluate the quotient

h' = 0.608

Hence, the water level increases by 0.608 meters per minute