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2 years ago

Any negative integer less than -5

Mathematics

2 answers:

Elden [556K]2 years ago

8 0

-6
that’s exactly how it goes yk, that’s the right answer

timama [110]2 years ago

5 0

Answer:

-6

Step-by-step explanation:

Math, thats how it works.

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6 2/7 rational or irrational

Marrrta [24]

Answer:

Rational

Step-by-step explanation:

Give brainliest or at least a thanks please!

6 0

3 years ago

Read 2 more answers

Write an equation of the line, in point-slope form, that passes through the points (–7, 2) and (3, –2). Use (–7, 2) as the point

Veronika [31]

Answer:

first use the equation m = y2 - y1 / x2- x1 to get the slope

m = -2 - 2 = -4

m = 3 + 7 = 10 when there is a negative for x1 it turns positive

the slope is -4/10 or -2/5

now plus the info into the point slope equation

y - y1 = m (x-x1)

this is in point slope form y - 2 = -2/5 (x + 7)

This is how to get it in y-intercept form

y -2 = -2/5x + -2.8

now add -2 on both sides to the -2.8

y = -2/5x (Slope) + -0.8 y- intercept

Hopefully this is right becuase this is just what i learned

Step-by-step explanation:

3 0

3 years ago

A triangle has side lengths of 10cm, 24cm ,and 33cm. classify it as an acute, obtuse or right

Lemur [1.5K]

Hey there! :D

Plug it into the Pythagorean theorem.

a^2+b^2=c^2

10^2+24^2= 33^2

100+ 576= 1089

676= 1089

Since the hypotenuse is bigger than the sides, it is an obtuse triangle.

I hope this helps!
~kaikers

6 0

3 years ago

What is the largest number that can be used to divide 36 other than itself

defon

Answer:

18!

Step-by-step explanation:

36/1=36

36/2=18

36/3=12

36/4=9

36/6=6

36/9=4

36/12=3

4 0

3 years ago

Need help with trig problem in pic

Sidana [21]

Answer:

a) $cos(\alpha)=-\frac{3}{5}\\$

b) $\sin(\beta)= \frac{\sqrt{3} }{2}$

c) $\frac{4+3\sqrt{3} }{10}\\$

d) $\alpha\approx 53.1^o$

Step-by-step explanation:

a) The problem tells us that angle $\alpha$ is in the second quadrant. We know that in that quadrant the cosine is negative.

We can use the Pythagorean identity:

$tan^2(\alpha)+1=sec^2(\alpha)\\(-\frac{4}{3})^2 +1=sec^2(\alpha)\\sec^2(\alpha)=\frac{16}{9} +1\\sec^2(\alpha)=\frac{25}{9} \\sec(\alpha) =+/- \frac{5}{3}\\cos(\alpha)=+/- \frac{3}{5}$

Where we have used that the secant of an angle is the reciprocal of the cos of the angle.

Since we know that the cosine must be negative because the angle is in the second quadrant, then we take the negative answer:

$cos(\alpha)=-\frac{3}{5}$

b) This angle is in the first quadrant (where the sine function is positive. They give us the value of the cosine of the angle, so we can use the Pythagorean identity to find the value of the sine of that angle:

$cos (\beta)=\frac{1}{2} \\\\sin^2(\beta)=1-cos^2(\beta)\\sin^2(\beta)=1-\frac{1}{4} \\\\sin^2(\beta)=\frac{3}{4} \\sin(\beta)=+/- \frac{\sqrt{3} }{2} \\sin(\beta)= \frac{\sqrt{3} }{2}$

where we took the positive value, since we know that the angle is in the first quadrant.

c) We can now find $sin(\alpha -\beta)$ by using the identity:

$sin(\alpha -\beta)=sin(\alpha)\,cos(\beta)-cos(\alpha)\,sin(\beta)\\$

Notice that we need to find $sin(\alpha)$ , which we do via the Pythagorean identity and knowing the value of the cosine found in part a) above:

$sin(\alpha)=\sqrt{1-cos^2(\alpha)} \\sin(\alpha)=\sqrt{1-\frac{9}{25} )} \\sin(\alpha)=\sqrt{\frac{16}{25} )} \\sin(\alpha)=\frac{4}{5}$

Then:

$sin(\alpha -\beta)=\frac{4}{5}\,\frac{1}{2} -(-\frac{3}{5}) \,\frac{\sqrt{3} }{2} \\sin(\alpha -\beta)=\frac{2}{5}+\frac{3\sqrt{3} }{10}=\frac{4+3\sqrt{3} }{10}$

Since $sin(\alpha)=\frac{4}{5}$

then $\alpha=arcsin(\frac{4}{5} )\approx 53.1^o$

4 0

3 years ago