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3 years ago

Find the missing length of the triangle.

Mathematics

1 answer:

Margaret [11]3 years ago

4 0

Answer:

B=5.44 i think hopefully

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Please help me out :)

Tomtit [17]

(f+g)(x)=5x-6+x^2-4x-8
A.(f+g)(x)=x^2+x-14 is correct:)

5 0

3 years ago

The reduce feature on a copy machine is set at 84%. The picture being copied is 4.5 inches wide. What will be the width of yhe r

snow_tiger [21]

Answer:

The answer is 3.825 inches, the reduced picture will be 84% of the size of the original

8 0

3 years ago

Read 2 more answers

Can i have some help

Andrews [41]

Answer: g(x) = 3/2 x + 9

Step-by-step explanation:

h(x) = -2/3 x - 1

perpendicular lines always have the opposite sign, reciprocal slope

so, the slope of the perpendicular line would be: m = 3/2

y = mx + b

y = 3/2 x + b

plug in (-4, 3) to find b

3 = 3/2 (-4) + b

3 = -6 + b

b = 9

y = 3/2 x + 9

g(x) = 3/2 x + 9

5 0

2 years ago

1.901 x 10^-7 in standard notation

WITCHER [35]

Answer:

In standard form, it would be 0.0000001901!

Step-by-step explanation:

As you can see, if the exponent is negative, the answer would be a decimal so moving the decimal place 7 times to the left, that would be your answer.

7 0

3 years ago

In Exercises 5-7, find all the exact t-values for which the given statement is true,

bearhunter [10]

Answer: See Below

Step-by-step explanation:

NOTE: You need the Unit Circle to answer these (attached)

5) cos (t) = 1

Where on the Unit Circle does cos = 1?

Answer: at 0π (0°) and all rotations of 2π (360°)

In radians: t = 0π + 2πn

In degrees: t = 0° + 360n

******************************************************************************

$6)\quad sin (t) = \dfrac{1}{2}$

Where on the Unit Circle does $sin = \dfrac{1}{2}$

Hint: sin is only positive in Quadrants I and II

$\text{Answer: at}\ \dfrac{\pi}{6}\ (30^o)\ \text{and at}\ \dfrac{5\pi}{6}\ (150^o)\ \text{and all rotations of}\ 2\pi \ (360^o)$

$\text{In radians:}\ t = \dfrac{\pi}{6} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{6} + 2\pi n$

In degrees: t = 30° + 360n and 150° + 360n

******************************************************************************

$7)\quad tan (t) = -\sqrt3$

Where on the Unit Circle does $\dfrac{sin}{cos} = \dfrac{-\sqrt3}{1}\ or\ \dfrac{\sqrt3}{-1}\quad \rightarrow \quad (1,-\sqrt3)\ or\ (-1, \sqrt3)$

Hint: sin and cos are only opposite signs in Quadrants II and IV

$\text{Answer: at}\ \dfrac{2\pi}{3}\ (120^o)\ \text{and at}\ \dfrac{5\pi}{3}\ (300^o)\ \text{and all rotations of}\ 2\pi \ (360^o)$

$\text{In radians:}\ t = \dfrac{2\pi}{3} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{3} + 2\pi n$

In degrees: t = 120° + 360n and 300° + 360n

4 0

3 years ago