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ZanzabumX [31]
3 years ago
11

I've been stuck can someone help me​

Mathematics
2 answers:
NISA [10]3 years ago
4 0

3.5,

since 3350/670 = 5, we know that the 2345.0/670 = 3.5

Helen [10]3 years ago
4 0

Answer:

3.5

Step-by-step explanation:

1. divide 3350/670=5

2. place the 5 in the spot above, which is right after the decimal

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aivan3 [116]
Again using the Pythagorean Theorem:

h^2=x^2+y^2, where h is the hypotenuse of a right triangle and x and y are the side lengths.  You are given that the hypotenuse is 34 ft and the base side is 16ft. Let h=34, x=16, and y=height, then you have:

34^2=16^2+y^2

y^2=34^2-16^2

y^2=1156-256

y^2=900

y=√900

y=30 ft

So the ladder will reach a spot 30 feet high on the building. 
8 0
3 years ago
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kondaur [170]
Choice b would be correct since the slope of the line is negative !
4 0
3 years ago
Read 2 more answers
Where should the point P be chosen on line segment AB so as to maximize the angle θ? (Assume a = 4 units, b = 5 units, and c = 9
taurus [48]
From the figure, let the distance of point P from point A on line segment AB be x and let the angle opposite side a be M and the angle opposite side c be N.

Using pythagoras theorem,
\tan M= \frac{a}{b-x} \\ \\ M=\tan^{-1}\left(\frac{a}{b-x}\right)
and
\tan N= \frac{c}{x} \\ \\ N=\tan^{-1}\left(\frac{c}{x}\right)

Angle θ is given by
\theta=180-M-N \\  \\ =180-\tan^{-1}\left(\frac{a}{b-x}\right)-\tan^{-1}\left(\frac{c}{x}\right)

Given that a = 4 units, b = 5 units, and c = 9 units, thus
\theta=180-\tan^{-1}\left(\frac{4}{5-x}\right)-\tan^{-1}\left(\frac{9}{x}\right)

To maximixe angle θ, the differentiation of <span>θ with respect to x must be equal to zero.
i.e.
\frac{d\theta}{dx} = -\frac{4}{x^2-10x+41} + \frac{9}{x^2+81} =0 \\  \\ -4(x^2+81)+9(x^2-10x+41)=0 \\  \\ -4x^2-324+9x^2-90x+369=0 \\  \\ 5x^2-90x+45=0 \\  \\ x^2-18x+9=0 \\  \\ x=9\pm6 \sqrt{2}

Given that x is a point on line segment AB, this means that x is a positive number less than 5.

Thus
x=9-6 \sqrt{2}=0.5147

Therefore, The distance from A of point P, so that </span>angle θ is maximum is 0.51 to two decimal places.
6 0
3 years ago
Given the graph of a function f. Identify the function by name. Then Graph, state domain &amp; range in set notation:A) f(x) +2B
WARRIOR [948]

The function in the graph has the name of square function.

The domain of a function is all values of x the function can have. The domain of this function is all real numbers:

\mleft\lbrace x\in\R\mright\rbrace

The range of a function is all values of y the function can have. The range of this function is all positive numbers, including zero:

\mleft\lbrace y\in\R\mright|y\ge0\}

In order to graph f(x) + 2, we just need to translate the graph 2 units up. To find the new points, we need to increase all y-coordinates by 2:

(-2, 6), (-1, 3), (0, 2), (1, 3), (2, 6)

Domain: {x ∈ ℝ}

Range: {y ∈ ℝ | y ≥ 2}

Then, in order to graph f(x) - 2, we just need to translate the graph 2 units down. To find the new points, we need to decrease all y-coordinates by 2:

(-2, 2), (-1, -1), (0, -2), (1, -1), (2, 2)

Domain: {x ∈ ℝ}

Range: {y ∈ ℝ | y ≥ -2}

4 0
1 year ago
Let f(x) = tan(x) - 2/x. Let g(x) = x^2 + 8. What is f(x)*g(y)?
Tema [17]

Answer:

f(x)\times g(y)=y^2tan(x)+8tan(x)-\frac{2y^2}{x}-\frac{16}{x}

Step-by-step explanation:

We are given that

f(x)=tan(x)-\frac{2}{x}

g(x)=x^2+8

We have to find f(x)\times g(y)

To find the value of f(x)\times g(y) we will multiply f(x) by g(y)

g(y)=y^2+8

Now,

f(x)\times g(y)=(tanx-\frac{2}{x})(y^2+8)

f(x)\times g(y)=tan(x)(y^2+8)-\frac{2}{x}(y^2+8)

f(x)\times g(y)=y^2tan(x)+8tan(x)-\frac{2y^2}{x}-\frac{16}{x}

Hence,

f(x)\times g(y)=y^2tan(x)+8tan(x)-\frac{2y^2}{x}-\frac{16}{x}

5 0
3 years ago
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