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inysia [295]
2 years ago
13

A triangle can have sides whose lengths are 9cm, 16 cm, and

Mathematics
1 answer:
vodka [1.7K]2 years ago
5 0

Answer: (B) 8 cm

Step-by-step explanation:

A triangle has 3 sides.

In order for it to be a triangle, the sum of sides a and b must be greater than the length of side c.

Testing (A) 7cm:

7 + 9 > 16

16 > 16

False

Testing (B) 8cm:

8+9 > 16

17 > 16

True

Testing (C) 25cm:

9 + 16 > 25

25 > 25

False

Testing (D) 26cm:

9 + 16 > 26

25 > 26

False

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Dubnium-262 has a half-life of 34 s. How long will it take for 500.0 grams to
dmitriy555 [2]

Answer:

the time taken for the radioactive element to decay to 1 g is 304.8 s.

Step-by-step explanation:

Given;

half-life of the given Dubnium = 34 s

initial mass of the given Dubnium, m₀ = 500 grams

final mass of the element, mf = 1 g

The time taken for the radioactive element to decay to its final mass is calculated as follows;

1 = 500 (0.5)^{\frac{t}{34}} \\\\\frac{1}{500} =  (0.5)^{\frac{t}{34}}\\\\log(\frac{1}{500}) = log [(0.5)^{\frac{t}{34}}]\\\\log(\frac{1}{500})  = \frac{t}{34} log(0.5)\\\\-2.699 = \frac{t}{34} (-0.301)\\\\t = \frac{2.699 \times 34}{0.301} \\\\t = 304.8 \ s

Therefore, the time taken for the radioactive element to decay to 1 g is 304.8 s.

4 0
3 years ago
Can someone help with geometry?
grin007 [14]

Answer:

SSS

Step-by-step explanation:

bc all the sides are congruent.

6 0
3 years ago
"Quadrilateral BCDE is inscribed in circle A as shown in the picture.<br> What is m∠E? "
Veseljchak [2.6K]
96 Degrees.You need to find the length of the arcs outside by multiplying the inside angles by 2. That number must go with the arcs that are enclosed.
7 0
3 years ago
Read 2 more answers
F(x) = 22c2 – 10x + 11<br><br> What is f(4)?<br><br> A) 3<br><br> B) 99<br><br> C)128<br><br> D) 52
Sidana [21]

Answer:

A

Step-by-step explanation:

B IS WRONG

C IS WRONG

7 0
3 years ago
Given that −4i is a zero, factor the following polynomial function completely. Use the Conjugate Roots Theorem, if applicable. f
GalinKa [24]

Answer:

\large \boxed{\sf \bf \ \ f(x)=(x-4i)(x+4i)(x+3)(x-5) \ \ }

Step-by-step explanation:

Hello, the Conjugate Roots Theorem states that if a complex number is a zero of real polynomial its conjugate is a zero too. It means that (x-4i)(x+4i) are factors of f(x).

\text{Meaning that } (x-4i)(x+4i) =x^2-(4i)^2=x^2+16 \text{ is a factor of f(x).}

The coefficient of the leading term is 1 and the constant term is -240 = 16 * (-15), so we a re looking for a real number such that.

f(x)=x^4-2x^3+x^2-32x-240\\\\ =(x^2+16)(x^2+ax-15)\\\\ =x^4+ax^3-15x^2+16x^2+16ax-240

We identify the coefficients for the like terms, it comes

a = -2 and 16a = -32 (which is equivalent). So, we can write in \mathbb{R}.

\\f(x)=(x^2+16)(x^2-2x-15)

The sum of the zeroes is 2=5-3 and their product is -15=-3*5, so we can factorise by (x-5)(x+3), which gives.

f(x)=(x^2+16)(x^2-2x-15)\\\\=(x^2+16)(x^2+3x-5x-15)\\\\=(x^2+16)(x(x+3)-5(x+3))\\\\=\boxed{(x^2+16)(x+3)(x-5)}

And we can write in \mathbb{C}

f(x)=\boxed{(x-4i)(x+4i)(x+3)(x-5)}

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

7 0
3 years ago
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