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3 years ago

5

Which of the following can be the side lengths of a triangle?

1 answer:

dolphi86 [110]3 years ago

3 0

To solve this question, we need to know the triangle inequality property, which states that the sum of two legs of a triangle is always <em>greater </em>(it can't be equal to) than the third leg. Using this property, we can deduct that the only answer that works is A, 6,6, and 5.

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H(t) = 3t2 + 4t

Anarel [89]

Answer:

first we need to substitute g(t) value in h(t) value and then evaluate the expression as follows :

h(g(t))=3(t-1)²+4(t-1).

by solving

h(g(t))=3(t²+1-2t)+4t-4

h(g(t))=3t²-2t-1 will.be the final answer

7 0

3 years ago

40 POINTS IF YOU ANSWER CORRECTLY!

noname [10]

Answer:

Step-by-step explanation:

y is 15.49

Formula is b = √c^2-a^2

You could use sine [ opposite/hypotenuse ],

Cosine [adjacent/hypotenuse]

or Tangent [ opposite/adjacent ]

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6 0

3 years ago

he high temperature on Friday afternoon was 82°F. The temperature that evening was 66°F. How much did the temperature change? A)

Sunny_sXe [5.5K]

66-82

-16

The temperature decreased

A Fell 16F

3 0

3 years ago

Read 2 more answers

Find the slope for each of the order pair. (-2,-1) and (8,-3)

kondaur [170]

Answer:

-0.2

Step-by-step explanation:

Slope (m) =ΔY/ΔX= 8-(-2)/-3-(-1)

= -1/5

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4 0

3 years ago

Sinx = 1/2, cosy = sqrt2/2, and angle x and angle y are both in the first quadrant.

Leviafan [203]

Answer:

Option D. 3.73

Step-by-step explanation:

we know that

$tan(x+y)=\frac{tan(x)+tan(y)}{1-tan(x)tan(y)}$

and

$sin^{2}(\alpha)+cos^{2}(\alpha)=1$

step 1

Find cos(X)

we have

$sin(x)=\frac{1}{2}$

we know that

$sin^{2}(x)+cos^{2}(x)=1$

substitute

$(\frac{1}{2})^{2}+cos^{2}(x)=1$

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

$cos^{2}(x)=\frac{3}{4}$

$cos(x)=\frac{\sqrt{3}}{2}$

step 2

Find tan(x)

$tan(x)=sin(x)/cos(x)$

substitute

$tan(x)=1/\sqrt{3}$

step 3

Find sin(y)

we have

$cos(y)=\frac{\sqrt{2}}{2}$

we know that

$sin^{2}(y)+cos^{2}(y)=1$

substitute

$sin^{2}(y)+(\frac{\sqrt{2}}{2})^{2}=1$

$sin^{2}(y)=1-\frac{2}{4}$

$sin^{2}(y)=\frac{2}{4}$

$sin(y)=\frac{\sqrt{2}}{2}$

step 4

Find tan(y)

$tan(y)=sin(y)/cos(y)$

substitute

$tan(y)=1$

step 5

Find tan(x+y)

$tan(x+y)=\frac{tan(x)+tan(y)}{1-tan(x)tan(y)}$

substitute

$tan(x+y)=[1/\sqrt{3}+1}]/[{1-1/\sqrt{3}}]=3.73$

7 0

3 years ago