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3 years ago

Find the value of x. Round to the nearest tenth.

Mathematics

1 answer:

Aliun [14]3 years ago

7 0

Answer:

Step-by-step explanation:

Triangle XYZ is a right angle triangle.

From the given right angle triangle

XZ represents the hypotenuse of the right angle triangle.

With ∠34 as the reference angle,

YZ represents the adjacent side of the right angle triangle.

XY represents the opposite side of the right angle triangle.

To determine x, we would apply the Cosine trigonometric ratio

Cos θ = opposite side/hypotenuse. Therefore,

Cos 34 = x/28

x = 28Cos34 = 28 × 0.8290

x = 23.212

To the nearest tenth, it becomes

23.2

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Determine whether the pair of equations represent parallel lines, perpendicular lines, or neither.

Ivahew [28]

Answer:

Parallel

Step-by-step explanation:

Parallel lines have the same slope but different y-intercepts. If you multiply the top equation by 2, you get:

2(12x + 4y = 16)

24x + 8y = 32

This shows that both lines have the same slope, but then you find the y-intercepts, they are different:

1st equation y-int = 4

2nd equation y-int = 9/2 or 4.5

6 0

3 years ago

There are six balls in the bag. Two are basketballs, three are playground balls, and one is a soccer ball. What is the probabili

aleksandr82 [10.1K]

Answer:

$\frac{1}{6}=0.16$

Step-by-step explanation:

Given : There are six balls in bag

Basketballs=2

Playground balls =3

Soccer ball =1

To Find : the probability that if you pick one of the balls without looking, it will be a soccer ball

Solution :

Total no. of balls = 6

No. of soccer balls = 1

Probability = favorable outcomes / total outcomes

Since total outcomes = total no. of balls =6

And favorable outcome = no. of soccer balls = 1

Thus , Probability of getting soccer ball = $\frac{1}{6}$

Hence the probability that if you pick one of the balls without looking, it will be a soccer ball:

$\frac{1}{6}=0.16$

7 0

4 years ago

Val needs to find the area enclosed by the figure. The figure is made by attaching semicircles to each side of a 58-by-58 squar

Savatey [412]

Answer:

We need to find the area of the semicircles + the area of the square.

The area of a square is equal to the square of the lenght of one side.

As = L^2 = 58m^2 = 3,364 m^2

Now, each of the semicircles has a diameter of 58m, and we have that the area of a circle is equal to:

Ac = pi*(d/2)^2 = 3.14*(58m/2)^2 = 3.14(27m)^2 = 2,289.06m^2

And the area of a semicircle is half of that, so the area of each semicircle is:

a = (2,289.06m^2)/2 = 1,144.53m^2

And we have 4 of those, so the total area of the semicircles is:

4*a = 4* 1,144.53m^2 = 4578.12m^2

Now, we need to add the area of the square 3,364 m^2 + 4578.12m^2 = 7942.12m^2

This is nothing like the provided anwer of Val, so the numbers of val may be wrong.

4 0

3 years ago

Tell whether the fractions are equal or not equal to 6/8 and 3/4

yulyashka [42]

Answer:

they are equal

Step-by-step explanation:

3/4 can be converted it 6/8 by multiplying both denominator and nominator by two

5 0

3 years ago

What is the true solution to the equation below? 2 In e^In2x -In e^In10x = In 30

stealth61 [152]

$The\ domain.\\D:x\in\mathbb{R^+}\\\\2\ln e^{\ln2x}-\ln e^{\ln10x}=\ln30\ \ \ |use\ \log_ab^n=n\log_ab\\\\2\ln2x\ln e-\ln10x\ln e=\ln30\ \ \ \ |use\ \log_aa=1\\\\2\ln2x-\ln10x=\ln30\ \ \ |use\ n\log_ab^n=\log_ab^n\\\\\ln(2x)^2-\ln10x=\ln30\\\\\ln4x^2-\ln10x=\ln30\ \ \ |use\ \log_ab-\log_ac=\log_a\dfrac{b}{c}\\\\\ln\dfrac{4x^2}{10x}=\ln30\\\\\ln\dfrac{2x}{5}=\ln30\iff\dfrac{2x}{5}=30\ \ \ \ |\cdot5\\\\2x=150\ \ \ \ |:2\\\\x=75\in D$

Answer: x = 75

8 0

3 years ago

Read 2 more answers