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viktelen [127]
3 years ago
7

What's the lateral area of the drawing?

Mathematics
1 answer:
sleet_krkn [62]3 years ago
6 0

Answer:

294^2

Step-by-step explanation:

It’s D

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Which is greater 2/5 or 3/4
Brut [27]
3/4 is greater than 2/5. 2/5 is .4 and 3/4 is .75
7 0
3 years ago
Which definition best describes vertical angles?
Artyom0805 [142]

Answer:

B is the answer

Step-by-step explanation:

Vertical angles are pair angles formed when two lines intersect. Vertical angles are sometimes referred to as vertically opposite angles because the angles are opposite to each other

4 0
3 years ago
HILP3.)
Pachacha [2.7K]

Answer:

Step-by-step explanation:

a-number of advance  tickets

s-number of same day tickets

a*40+s*25=2400

a      +s      =75

I solve  with the substitution method

s=75-a

40a+25*(75-a)=2400

40a+1875-25a=2400

15a=2400-1875

15a=525

a=525/15

a=35

so  s=75-35

s=40

5 0
3 years ago
Let f be the function given by f(x)= (x-1)(x^2-4)/x^2-a. For what positive values of a is f continuous for all real numbers x?
9966 [12]

Answer:

a =1 and a=4.

Step-by-step explanation:

The function is

f(x)=\frac{(x-1)(x^2-4)}{x^2-a}

If we want f(x) to be continuous the denominator needs to be different to 0, otherwise f(x) will be indeterminate.

Now, for a a positive real we have that x=\sqrt{a} will annulate the denominator, i.e

(\sqrt{a})^2-a = a-a = 0. But, if a = 1 we have:

f(x)=\frac{(x-1)(x^2-4)}{x^2-1} = \frac{(x-1)(x^2-4)}{(x-1)(x+1)}=\frac{(x^2-4)}{x+1}

so, the value x=\sqrt{a} = \sqrt{1} = 1 won't annulate the denominator.

Now, for a = 4 we have:

f(x)=\frac{(x-1)(x^2-4)}{x^2-4} = x-1

so, the value x=\sqrt{a} = \sqrt{4} = 2 won't annulate the denominator.

In conclusion, for a=1 or a=1, the function will be continuos for all real numbers, since the denominator will never be 0.

4 0
3 years ago
Pls help me I rlly don’t get it :(
zhenek [66]

Step-by-step explanation:

Use SOH-CAH-TOA.

Sine = Opposite / Hypotenuse

Cosine = Adjacent / Hypotenuse

Tangent = Opposite / Adjacent

Let's start with #12.  The hypotenuse is 18.  The side adjacent to ∠B is 6.  Since we have the adjacent side and hypotenuse, we should use cosine.

cos B = 6/18

Solving for B:

B = cos⁻¹(6/18)

Using a calculator:

B ≈ 70.5°

Now let's do #14.  The side adjacent to ∠B is 19, and the side opposite of ∠B is 22.  Since we have the adjacent side and opposite side, we should use tangent.

tan B = 22/19

Solving for B:

B = tan⁻¹(22/19)

Using a calculator:

B ≈ 49.2°

8 0
3 years ago
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