All categories

2 years ago

Find the missing side to the triangle in the attached image.

Mathematics

2 answers:

kari74 [83]2 years ago

8 0

Answer:

x=34

Step-by-step explanation:

1. formula for Pythagorean Theorem for right trangles is <u>a^2+b^2=c^2</u>

2. a=30 and b=16, so (30)^2+(16)^2=c^2

3. (30)^2=900 and (16)^2=256

4. 900+256=1156

5. 1156=c^2

6. $\sqrt{1156} =\sqrt{c^2}$

7. the square root of 1156 is 34, so c=34

8. The missing side of the triangle, labeled as "x<em>"</em> is 34

Arte-miy333 [17]2 years ago

5 0

Answer:

Step-by-step explanation:

1. formula for Pythagorean Theorem for right trangles is a^2+b^2=c^2

2. a=30 and b=16, so (30)^2+(16)^2=c^2

3. (30)^2=900 and (16)^2=256

4. 900+256=1156

5. 1156=c^2

7. the square root of 1156 is 34, so c=34

8. The missing side of the triangle, labeled as "x" is 34

You might be interested in

What does the measure of B Equal?

defon

You have two triangles, ADC and ABC.
Sides AD and AB are congruent.
Sides DC and BC are congruent.
Side AC is congruent to itself.
By SSS, triangles ADC and ABC are congruent.

Corresponding parts of congruent triangles are congruent.
That means that angles DAC and BAC are congruent.
Angles DCA and BCA are congruent.

Since m<DAC = 32, then m<BAC = 32
Since m<DCA = 41, then m<BCA = 41.

Now you know the measures of two angles of triangle ABC.
The measures of the interior angles of a triangle add to 180.
You can find the measure of angle B.

m<BAC + m<B + m<BCA = 180

32 + m<B + 41 = 180

m<B + 73 = 180

m<B = 107

8 0

3 years ago

(cos2a *cos 4a+ sin 2a*sin 4a)/sin4a

lyudmila [28]

Answer:

Step-by-step explanation:

(cos 4a*cos 2a+sin 4a*sin 2a)/sin 4a

=[cos (4a-2a)]/sin 4a

=(cos 2a)/sin 4a

=(cos 2a) /(2 sin 2a cos 2a)

=1/(2 sin 2a)

=1/2 csc 2a

4 0

2 years ago

<img src="https://tex.z-dn.net/?f=%28x%5E%7B2%7D%20%2Bx-3%29%3A%20%28x%5E%7B2%7D%20-4%29%5Cgeq%201" id="TexFormula1" title="(x^{

Jlenok [28]

Answer:

$x>2$

Step-by-step explanation:

When given the following inequality;

$(x^2+x-3):(x^2-4)\geq1$

Rewrite in a fractional form so that it is easier to work with. Remember, a ratio is another way of expressing a fraction where the first term is the numerator (value over the fraction) and the second is the denominator(value under the fraction);

$\frac{x^2+x-3}{x^2-4}\geq1$

Now bring all of the terms to one side so that the other side is just a zero, use the idea of inverse operations to achieve this:

$\frac{x^2+x-3}{x^2-4}-1\geq0$

Convert the (1) to have the like denominator as the other term on the left side. Keep in mind, any term over itself is equal to (1);

$\frac{x^2+x-3}{x^2-4}-\frac{x^2-4}{x^2-4}\geq0$

Perform the operation on the other side distribute the negative sign and combine like terms;

$\frac{(x^2+x-3)-(x^2-4)}{x^2-4}\geq0\\\\\frac{x^2+x-3-x^2+4}{x^2-4}\geq0\\\\\frac{x+1}{x^2-4}\geq0$

Factor the equation so that one can find the intervales where the inequality is true;

$\frac{x+1}{(x-2)(x+2)}\geq0$

Solve to find the intervales when the equation is true. These intervales are the spaces between the zeros. The zeros of the inequality can be found using the zero product property (which states that any number times zero equals zero), these zeros are as follows;

$-1, 2, -2$

Therefore the intervales are the following, remember, the denominator cannot be zero, therefore some zeros are not included in the domain

$x\leq-2\\-2$

Substitute a value in these intervales to find out if the inequality is positive or negative, if it is positive then the interval is a solution, if it is negative then it is not a solution. This is because the inequality is greater than or equal to zero;

$x\leq-2$ -> negative