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3 years ago

WILL MARK THE BRAINILEST What relationship do the ratios of sin x° and cos yº share?

Mathematics

2 answers:

joja [24]3 years ago

8 0

Sin x = 4/5
cos y = 4/5
What relationship do these share?

mariarad [96]3 years ago

4 0

Answer:

The sine of the measure of an angle in a right triangle, is the ratio of the opposite side of that angle, to the hypotenuse.

The cosine of the measure of an angle in a right triangle, is the ratio of the adjacent side of that angle, to the hypotenuse.

According to these, we have:

$\displaystyle{ \sin(x^{\circ})= \frac{6}{10}$

and

$\displaystyle{ \sin(y^{\circ})= \frac{6}{10} .$

Thus, the ratios of both are identical $( \frac{6}{10} \ and \ \frac{6}{10} )$

Step-by-step explanation:

Please give me brainliest

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In 2010, the world's largest pumpkin weighed 1,810 kilograms. An average-sized pumpkin weighs 5,000 grams.

jekas [21]

Answer:

1,805 kg.

Step-by-step explanation:

We have been given that in 2010, the world's largest pumpkin weighed 1,810 kilograms. An average-sized pumpkin weighs 5,000 grams. We are asked to find the how much the world's largest pumpkin weighs than an average pumpkin.

First of all, we will convert the weight of average pumpkin in kilograms by dividing 5,000 by 1000 as 1 kg equals 1,000 gm.

$\text{The weight of average pumpkin}=\frac{5000\text{ gm}}{\frac{\text{1000 gm}}{\text{ 1 kg}}}$

$\text{The weight of average pumpkin}=5000\text{ gm}\times \frac{\text{ 1 kg}}{\text{1000 gm}}$

$\text{The weight of average pumpkin}=5\text{ kg}$

Now, we will subtract the weight of average pumpkin from world's largest pumpkin's weight.

$1,810\text{ kilograms}-5\text{ kilograms}=1,805\text{ kilograms}$

Therefore, the 2010 world-record pumpkin weighs 1,805 kilograms more than an average-sized pumpkin.

3 0

3 years ago

Read 2 more answers

Segment BC is a midsegment of triangle TUV. What is the length of segment VC? (A)a (B)2b (C)b (D)c

Ganezh [65]

It would be twice whatever the midsegment equals.

7 0

3 years ago

A satellite dish has cross-sections shaped like parabolas. The receiver is located 13 inches from the base along the axis of sym

horsena [70]

Answer:

Depth = 3.3 inches

Step-by-step explanation:

Given that the shape of the satellite looks like a parabola

The equation of parabola is given as follows

$x^2=4\times a\times y$

Where

a= 13

Therefore

$x^2=4\times 13\times y$

$x^2=52\times y$

Lets take (13 , y) is a

Now by putting the values in the above equation we get

$13^2=52\times y$

$y=\dfrac{13^2}{52}=3.25$

y=3.25 in

Therefore the depth of the satellite at the nearest integer will be 3.3 inches.

Depth = 3.3 inches

7 0

3 years ago

Changing Bases to Evaluate Logarithms In Exercise, use the change-of-base formula and a calculator to evaluate the logarithm.

Ymorist [56]

Answer:

$\frac{3}{2}$

Step-by-step explanation:

Changing Bases to Evaluate Logarithms

$log_{16}(64)$

Apply change of base formula'

$log_b(a)= \frac{log a}{log b}$

log term should be the numerator and denominator is the log base

$log_{16}(64)$

$log_{16}(a)= \frac{log 64}{log 16}$

64 is 4^3 and 16 is 4^2

$log_{16}(a)= \frac{log 4^3}{log 4^2}$

Move the exponent before log

$\frac{3log 4}{2log 4}$

top and bottom has same log so cancel it out

$\frac{3}{2}$

3 0

3 years ago

How do i do this......

andrew11 [14]

Answer:

b is correct answer

Step-by-step explanation:

hope it helped

5 0

3 years ago