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S_A_V [24]
3 years ago
5

Please find the missing number for the surface area. I will mark brainiest if correct. 2.

Mathematics
1 answer:
Paha777 [63]3 years ago
5 0

It is 18 because it is a square so all the sides have to be the same so that would make it 18

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Dvinal [7]

just look it up on google


6 0
3 years ago
Question is in the picture
Basile [38]

Answer:

(6, 5)

Step-by-step explanation:

-3 + 9 = 6

15 - 10 = 5

4 0
3 years ago
We need help finding the means of the sides of this figure.<br> (Photo attached)<br><br> Thank you
VikaD [51]

Answer:

3) 5

4) 8.2

5) 6.8

6)FG = 6.4

7) EF = 3.6

8)DF = 4.8

Step-by-step explanation:

From the attached triangle, using trigonometric ratio we can find ∠G as; tan^(-1) (6/8)

Thus, ∠G = 36.87°

Still using trigonometric ratios;

FG/8 = cos 36.87°

FG = 8 cos 36.87°

FG = 6.4

3) EF + FG = EG

Since EG is 10, mean of EF and FG = 10/2 = 5

4) mean of EG and FG = (10 + 6.4)/2 = 8.2

5) EF + FG = EG

Thus; EF = EG - FG

EF = 10 - 6.4

EF = 3.6

Mean of EG and EF = (10 + 3.6)/2 = 6.8

6) FG = 6.4

7) EF = 3.6

8) Using trigonometric ratio;

DF/8 = sin 36.87

DF = 8 × 0.6

DF = 4.8

8 0
2 years ago
What is the value of x?
STatiana [176]
I thought it was 5 because the 55 degrees given is equal to the angle above it. If you look, there are 4, 90 degree angles so the 8x-5 has to equal 35 to make 90 but idk
3 0
3 years ago
You are given the parametric equations x=2cos(θ),y=sin(2θ). (a) List all of the points (x,y) where the tangent line is horizonta
vladimir1956 [14]

Answer:

The solutions listed from the smallest to the greatest are:

x:  -\sqrt{2}   -\sqrt{2}  \sqrt{2}  \sqrt{2}

y:      -1         1     -1     1

Step-by-step explanation:

The slope of the tangent line at a point of the curve is:

m = \frac{\frac{dy}{dt} }{\frac{dx}{dt} }

m = -\frac{\cos 2\theta}{\sin \theta}

The tangent line is horizontal when m = 0. Then:

\cos 2\theta = 0

2\theta = \cos^{-1}0

\theta = \frac{1}{2}\cdot \cos^{-1} 0

\theta = \frac{1}{2}\cdot \left(\frac{\pi}{2}+i\cdot \pi \right), for all i \in \mathbb{N}_{O}

\theta = \frac{\pi}{4} + i\cdot \frac{\pi}{2}, for all i \in \mathbb{N}_{O}

The first four solutions are:

x:   \sqrt{2}   -\sqrt{2}  -\sqrt{2}  \sqrt{2}

y:     1        -1        1     -1

The solutions listed from the smallest to the greatest are:

x:  -\sqrt{2}   -\sqrt{2}  \sqrt{2}  \sqrt{2}

y:      -1         1     -1     1

6 0
3 years ago
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