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Oliga [24]
4 years ago
11

What is the peremeter of this tile 3in 3in 3in 3in I ready

Mathematics
1 answer:
Leokris [45]4 years ago
8 0

Answer:

12 in.

Step-by-step explanation:

The perimeter is calculated by adding all up the sides of a shape.  In this case, the shape is a square, considering that all the sides are the same measurements.  3+3+3+3 can be simplified to 3x4.  This gives you 12 in. as the perimeter.

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Dmitriy789 [7]
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v = <u>2 + 4</u>    v = <u>2 - 4</u>
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v = <u>6</u>          v = <u>-2</u>
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4 0
3 years ago
When price increases, quantity supplied:
lara31 [8.8K]
It should increase but I’m between stays the same or increase, I’m not sure in a 100% sorry...
3 0
3 years ago
Law of sines: StartFraction sine (uppercase A) Over a EndFraction = StartFraction sine (uppercase B) Over b EndFraction = StartF
Vsevolod [243]

First of all, this problem is properly done with the Law of Cosines, which tells us

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

giving us a quadratic equation for b we can solve.  But let's do it with the Law of Sines as asked.

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

We have c,a,A so the Law of Sines gives us sin C

\sin C = \dfrac{c \sin A}{a} = \dfrac{5.4 \sin 20^\circ}{3.3} = 0.5597

There are two possible triangle angles with this sine, supplementary angles, one acute, one obtuse:

C_a = \arcsin(.5597)  = 34.033^\circ

C_o = 180^\circ - C_a = 145.967^\circ

Both of these make a valid triangle with A=20°.   They give respective B's:

B_a = 180^\circ - A - C_a = 125.967^\circ

B_o = 180^\circ - A - C_o = 14.033^\circ

So we get two possibilities for b:

b = \dfrac{a \sin B}{\sin A}

b_a = \dfrac{3.3 \sin 125.967^\circ}{\sin 20^\circ} = 7.8

b_o = \dfrac{3.3 \sin 14.033^\circ}{\sin 20^\circ} = 2.3

Answer: 2.3 units and 7.8 units

Let's check it with the Law of Cosines:

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

0 = b^2 - (2 c \cos A)b + (c^2-a^2)

There's a shortcut for the quadratic formula when the middle term is 'even.'

b = c \cos A \pm \sqrt{c^2 \cos^2 A - (c^2-a^2)}

b = c \cos A \pm \sqrt{c^2( \cos^2 A - 1) + a^2}

b = 5.4 \cos 20 \pm \sqrt{5.4^2(\cos^2 20 -1) + 3.3^2}

b = 2.33958 \textrm{ or } 7.80910 \quad\checkmark

Looks good.

6 0
3 years ago
Read 2 more answers
In ΔJKL, the measure of ∠L=90°, KL = 22 feet, and JK = 54 feet. Find the measure of ∠J to the nearest degree.
nydimaria [60]

We have been given that in ΔJKL, the measure of ∠L=90°, KL = 22 feet, and JK = 54 feet. We are asked to find the measure of angle J to nearest degree.

First of all, we will draw a triangle as shown in the attachment.

We can see from our attachment that side KL is opposite side to angle J and side JK is hypotenuse of right triangle.

We know that sine relates opposite side of right triangle to hypotenuse.

\text{sin}=\frac{\text{Opposite}}{\text{Hypotenuse}}

\text{sin}(\angle J)=\frac{22}{54}

Using inverse sine or arcsin, we will get:

\angle J=\text{sin}^{-1}(\frac{22}{54})

\angle J=24.042075905756^{\circ}

Upon rounding to nearest degree, we will get:

\angle J\approx 24^{\circ}

Therefore, the measure of angle J is approximately 24 degrees.

8 0
4 years ago
The low temperatures in degrees Fahrenheit in Marsh City during a week in January were: -3, -5, 3, 7, -2, -2, -5 What was the av
miskamm [114]

Answer:

-1

Step-by-step explanation:

You can find the average my finding the sum of all temperatures and then dividing it by the number of temperatures ( in this case 7)

So you get (-3-5+3+7-2-2-5)/7=-7/7=-1

4 0
4 years ago
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