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DedPeter [7]
3 years ago
14

I need help ASAP???​

Mathematics
1 answer:
melamori03 [73]3 years ago
4 0

Answer:

Since this is a check-all-that-apply question it would be the first and second choices.

Hope This Helps!

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Lola measures three buttons for a shirt she’s making. One button is 1/8, 3/8 and 1/4 inch. The button hole is 2/6 of an inch. Wh
mel-nik [20]

Answer:

3/8

Step-by-step explanation:

That's the only one bigger than 2/6, any smaller will fall out as it's a button. You'd twist it sort of in half to get it in the hole, therefore, 3/8 is the only button that will fit.

I'm not sure if this is correct but this is how I'd go about it :)

7 0
3 years ago
2. A right triangle has a hypotenuse of 15 cm. What are possible lengths for the two legs of the triangle? Explain your reasonin
dolphi86 [110]

Answer:

Two possible lengths for the legs A and B are:

B = 1cm

A = 14.97cm

Or:

B = 9cm

A = 12cm

Step-by-step explanation:

For a triangle rectangle, Pythagorean's theorem says that the sum of the squares of the cathetus is equal to the hypotenuse squared.

Then if the two legs of the triangle are A and B, and the hypotenuse is H, we have:

A^2 + B^2 = H^2

If we know that H = 15cm, then:

A^2 + B^2 = (15cm)^2

Now, let's isolate one of the legs:

A = √( (15cm)^2 - B^2)

Now we can just input different values of B there, and then solve the value for the other leg.

Then if we have:

B = 1cm

A = √( (15cm)^2 - (1cm)^2) = 14.97

Then we could have:

B = 1cm

A = 14.97cm

Now let's try with another value of B:

if B = 9cm, then:

A = √( (15cm)^2 - (9cm)^2) = 12 cm

Then we could have:

B = 9cm

A = 12cm

So we just found two possible lengths for the two legs of the triangle.

4 0
3 years ago
What is the area of a trapozoid
user100 [1]

I hope this helped you tell me if you need help understanding

8 0
3 years ago
Please help me with this !!
Sergio [31]

Answer: AC = 7√3

tan 30° = \frac{BC}{AC}

=> \frac{BC}{AC}=\frac{1}{\sqrt{3} }\\\\\=> AC=BC\sqrt{3} =7\sqrt{3}

Step-by-step explanation:

3 0
3 years ago
A basketball gymnasium is 25 meters high, 80 meters wide and 200 meters long. We want to connect two strings, one from each of t
Rudiy27

Answer:

a) the centre is at (12.5 m, 40 m , 100 m ) with respect to our position

b) the length of the strings S will be 216.85 m

c) the angle that is formed by the strings is  1.23 rad

Step-by-step explanation:

assuming that we stand on one of the corners on the floor , so our coordinates are (0,0,0)  , then the coordinates of the center of the gymnasium   are found through

x center = (25 + 0)/2 = 12.5 m

y center = (80+ 0)/2 = 40 m

z center = (200+ 0)/2 = 100 m

then the centre is at (12.5 m, 40 m , 100 m ) with respect to our position

b) the length of the strings S will be the modulus of the vector that points from our position to the diagonally opposite corners

|S| = √(25²+80²+200²) = 216.85 m

c) the angle can be found through the dot product of the vectors that represent the strings S₁ and S₂

S₁ =(25,80,10)

S₂ =(-25,80,100)

then

S₁*S₂ = 25*(-25) +80*80 + 100*100 = 15775

but also

S₁*S₂ = |S₁||S₂| cos θ = |S|² * cos θ

S₁*S₂ =  |S|² * cos θ

cos θ= S₁*S₂/|S|²

θ= cos ⁻¹ ( S₁*S₂/|S|² ) = cos ⁻¹ [15775/(25²+80²+200²)] = 1.23 rad

7 0
3 years ago
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