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2 years ago

Find the value of x from the following trigonometrical equation :

Mathematics

1 answer:

Yuliya22 [10]2 years ago

6 0

Answer:

x = 2/3

Step-by-step explanation:

sin²30° + cos²30° + x.tan²60° = cot²30°

=> (1/2)² + (√3/2)² + x(√3)² = (√3)²

=> 1/4 + 3/4 + 3x = 3

=> 3x = 2

=> x = 2/3

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I NEED HELP WITH MY FINAL PLEASEE

madreJ [45]

Answer:

y=(x-1)(x-4) & 100°

Step-by-step explanation:

<u>Graph</u>

1) there are x-intercepts at 1 and 4so you set those equal to x

ex. x=1 x=4

2) then subtract them over to equal 0 and set it equal to y

ex. y=(x-1)(x-4)

<u>Angle</u>

*IMPORTANT* Straight lines are equal to 180°

∠CBD = 80°

1) So you just subtract ∠CBD from 180°

ex. 180° - 80° = 100°

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3 years ago

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3 years ago

When designing an experiment a benefit of larger sample sizes would be more

NARA [144]

The correct answer is B. Representative Data. This is because if you consult more people pertaining to your experiment, this is basically collecting more data for your experiment. This relationship is illustrated by answer choice B.

5 0

3 years ago

You are Looking at two area rugs for your bedroom. One is a square with a side length of 6.3 feet the other is rectangle that is

Ray Of Light [21]

Answer:

Part a) The square rug cover the greater area

Part b) The square rug cover 0.33 square feet more than the rectangular rug

Step-by-step explanation:

step 1

Find the area of the square rug

The area of a square is

$A=b^2$

where

b is the length of the square

we have

$b=6.3\ ft$

substitute

$A=(6.3)^2=39.69\ ft^2$

step 2

Find the area of the rectangular rug

The area of the rectangle is

$A=LW$

we have

$L=4.8\ ft\\W=8.2\ ft$

substitute

$A=4.8(8.2)=39.36\ ft^2$

step 3

Compare the areas

$39.69 > 39.36$

The square rug cover the greater area

step 4

Find the difference

$39.69-39.36=0.33\ ft^2$

The square rug cover 0.33 square feet more than the rectangular rug

8 0

3 years ago

Two buildings on opposites sides of a highway are 3x^3- x^2 + 7x +100 feet apart. One building is 2x^2 + 7x feet from the highwa

iVinArrow [24]

Given:

Distance between two buildings = $3x^3- x^2 + 7x +100$ feet apart.

Distance between highway and one building = $2x^2 + 7x$ feet.

Distance between highway and second building = $x^3 + 2x^2 - 18$ feet.

To find:

The standard form of the polynomial representing the width of the highway between the two building.

Solution:

We know that,

Width of the highway = Distance between two buildings - Distance of both buildings from highway.

Using the above formula, we get the polynomial for width (W) of the highway.

$W=3x^3- x^2 + 7x +100-(2x^2 + 7x)-(x^3 + 2x^2 - 18)$

$W=3x^3- x^2 + 7x +100-2x^2-7x-x^3 -2x^2+18$

Combining like terms, we get

$W=(3x^3-x^3)+(- x^2 -2x^2-2x^2)+ (7x -7x)+(100 +18)$

$W=2x^3-5x^2+0+118$

$W=2x^3-5x^2+118$

Therefore, the width point highway is $2x^3-5x^2+118$ .

8 0

3 years ago