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katrin [286]
3 years ago
15

Need help ASAP. I don't know what the answer is. Please help asap

Mathematics
2 answers:
IrinaVladis [17]3 years ago
7 0

Answer:

43.25 degrees   B.

Step-by-step explanation:

This is an isosceles triangle (DF = EF) and so the 2 base angles are euql .

So m < x = (180 - 93.5) / 2

=  43.25 degrees

svetoff [14.1K]3 years ago
5 0

Yeah it is D i hope this helps u

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Solve for x by simplifying both sides of the equation, then isolating the variable.

x = 10

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Which expression is equivalent to the area of square A, in square inches?
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Answer:

length of 1 side of A, using the Pyth. Thm. and the dimensions of the other two squares: (side of A)^2 = (10 in)^2 + (24 in)^2. Then:

(side of A)^2 = 100+ 576 in^2 = 676 in^2.

Here I have not bothered to solve for the length of the side of A, since we want the area of square A. But if you do want the side length, find it: sqrt(676) = 26 in. Then the area of A is (26 in)^2 = 676 in^2.

Then the area of square A is (26 in)^2 =

Read more on Brainly.com - brainly.com/question/10676137#readmore

Step-by-step explanation:

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4 years ago
Which function represents the relationship between x and y?
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The correct answer is c
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3 years ago
Find the angle between u =the square root of 5i-8j and v =the square root of 5i+j.
fenix001 [56]

Answer:

The angle between vector \vec{u} = 5\, \vec{i} - 8\, \vec{j} and \vec{v} = 5\, \vec{i} + \, \vec{j} is approximately 1.21 radians, which is equivalent to approximately 69.3^\circ.

Step-by-step explanation:

The angle between two vectors can be found from the ratio between:

  • their dot products, and
  • the product of their lengths.

To be precise, if \theta denotes the angle between \vec{u} and \vec{v} (assume that 0^\circ \le \theta < 180^\circ or equivalently 0 \le \theta < \pi,) then:

\displaystyle \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|}.

<h3>Dot product of the two vectors</h3>

The first component of \vec{u} is 5 and the first component of \vec{v} is also

The second component of \vec{u} is (-8) while the second component of \vec{v} is 1. The product of these two second components is (-8) \times 1= (-8).

The dot product of \vec{u} and \vec{v} will thus be:

\begin{aligned} \vec{u} \cdot \vec{v} = 5 \times 5 + (-8) \times1 = 17 \end{aligned}.

<h3>Lengths of the two vectors</h3>

Apply the Pythagorean Theorem to both \vec{u} and \vec{v}:

  • \| u \| = \sqrt{5^2 + (-8)^2} = \sqrt{89}.
  • \| v \| = \sqrt{5^2 + 1^2} = \sqrt{26}.

<h3>Angle between the two vectors</h3>

Let \theta represent the angle between \vec{u} and \vec{v}. Apply the formula\displaystyle \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|} to find the cosine of this angle:

\begin{aligned} \cos(\theta)&= \frac{\vec{u} \cdot \vec{v}}{\| u \| \cdot \| v \|} = \frac{17}{\sqrt{89}\cdot \sqrt{26}}\end{aligned}.

Since \theta is the angle between two vectors, its value should be between 0\; \rm radians and \pi \; \rm radians (0^\circ and 180^\circ.) That is: 0 \le \theta < \pi and 0^\circ \le \theta < 180^\circ. Apply the arccosine function (the inverse of the cosine function) to find the value of \theta:

\displaystyle \cos^{-1}\left(\frac{17}{\sqrt{89}\cdot \sqrt{26}}\right) \approx 1.21 \;\rm radians \approx 69.3^\circ .

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