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3 years ago

Help find the perimeter! Please help! (30 points)

Mathematics

2 answers:

lord [1]3 years ago

5 0

The perimeter is 16, according to my calculations.

melamori03 [73]3 years ago

3 0

The perimeter of the triangle is 16 inches.

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How do i do this, my son is struggling and he needs help

expeople1 [14]

Answer:

Step-by-step explanation:

angle a is the outside angle. It creates a circle that will always equal 360 degrees. Since the outside angle is 302, the inside angle must be 58 degrees (360-302). The top square angle will be 90 degrees (a square angle is always 90 degrees). Since the angles in a triangle always add up to 180 degrees, we need to add 58 and 90, and then subtract it from 180, which equals 32. So, angle x equals 32.

4 0

2 years ago

Read 2 more answers

I need help with questions #7 and #8 plz

katen-ka-za [31]

Answer:

7. A = 40.8 deg; B = 60.6 deg; C = 78.6 deg

8. A = 20.7 deg; B = 127.2 deg; C = 32.1 deg

Step-by-step explanation:

Law of Cosines

$c^2 = a^2 + b^2 - 2ab \cos C$

You know the lengths of the sides, so you know a, b, and c. You can use the law of cosines to find C, the measure of angle C.

Then you can use the law of cosines again for each of the other angles. An easier way to solve for angles A and B is, after solving for C with the law of cosines, solve for either A or B with the law of sines and solve for the last angle by the fact that the sum of the measures of the angles of a triangle is 180 deg.

We use the law of cosines to find C.

$18^2 = 12^2 + 16^2 - 2(12)(16) \cos C$

$324 = 144 + 256 - 384 \cos C$

$-384 \cos C = -76$

$\cos C = 0.2$

$C = \cos^{-1} 0.2$

$C = 78.6^\circ$

Now we use the law of sines to find angle A.

Law of Sines

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$

We know c and C. We can solve for a.

$\dfrac{a}{\sin A} = \dfrac{c}{\sin C}$

$\dfrac{12}{\sin A} = \dfrac{18}{\sin 78.6^\circ}$

Cross multiply.

$18 \sin A = 12 \sin 78.6^\circ$

$\sin A = \dfrac{12 \sin 78.6^\circ}{18}$

$\sin A = 0.6535$

$A = \sin^{-1} 0.6535$

$A = 40.8^\circ$

To find B, we use

m<A + m<B + m<C = 180

40.8 + m<B + 78.6 = 180

m<B = 60.6 deg

I'll use the law of cosines 3 times here to solve for all the angles.

Law of Cosines

$a^2 = b^2 + c^2 - 2bc \cos A$

$b^2 = a^2 + c^2 - 2ac \cos B$

$c^2 = a^2 + b^2 - 2ab \cos C$

Find angle A:

$a^2 = b^2 + c^2 - 2bc \cos A$

$8^2 = 18^2 + 12^2 - 2(18)(12) \cos A$

$64 = 468 - 432 \cos A$

$\cos A = 0.9352$

$A = 20.7^\circ$

Find angle B:

$b^2 = a^2 + c^2 - 2ac \cos B$

$18^2 = 8^2 + 12^2 - 2(8)(12) \cos B$

$324 = 208 - 192 \cos A$

$\cos B = -0.6042$

$B = 127.2^\circ$

Find angle C:

$c^2 = a^2 + b^2 - 2ab \cos C$

$12^2 = 8^2 + 18^2 - 2(8)(18) \cos B$

$144 = 388 - 288 \cos A$

$\cos C = 0.8472$

$C = 32.1^\circ$

8 0

3 years ago

If angle ABC is one degree less than three times angle ABD and angle DBC=47 find each measure

Contact [7]

It has not been indicated whether the figure in the questions is a triangle or a quadrilateral. Irrespective of the shape, this can be solved. The two possible shapes and angles have been indicated in the attached image.

Now, from the information given we can infer that there is a line BD that cuts angle ABC in two parts: angle ABD and angle DBC

⇒ Angle ABC = Angle ABD + Angle DBC

Also, we know that angle ABC is 1 degree less than 3 times the angle ABD, and that angle DBC is 47 degree

Let angle ABD be x

⇒ Angle ABC = 3x-1

Also, Angle ABC = Angle ABD + Angle DBC

Substituting the values in the above equations

⇒ 3x-1 = x+47

⇒ 2x = 48

⇒ x = 24

So angle ABD = 24 degree, and angle ABC = 3(24)-1 = 71-1 = 71 degree

8 0

3 years ago

Two equations are given below:

stealth61 [152]

a is given to us so just plug a into the first equation:

b-3 - 3b =9

Add 3 to both sides:

b-3b=12

Combine like terms:

-2b=12

Divide by -2 to get b by itself:

b=-6

The only answer with b as -6 is the first one, (-9,-6)

8 0

3 years ago

Read 2 more answers

Alexander says that 3x + 4y is equivalent to (3)(4) + xy because of any order, any grouping. Is he correct why or why not

snow_lady [41]

Answer:

Alexander is incorrect because the expressions are not equivalent.

Step-by-step explanation:

If the expression is evaluated for any value of x, y; the result will not be same.

For instance, let assume x = 1 and y = 2

3x + 4y = 3 + 4 = 7

(3)(4) + xy = (3)(4) + (1 * 2) = 12 + 2 = 14

So, the expressions are not the same and Alexander is incorrect.

3 0

3 years ago