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2 years ago

10

25 POINTS PLEASE HELP ME!!! IM SO CLOSE TO FAILING RN ITS NOT FUNNY

2 answers:

Vilka [71]2 years ago

8 0

We can’t see the picture :( so we can’t do it

Arte-miy333 [17]2 years ago

7 0

I can’t see the image

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I need help. ASAP!<br> Divide.<br><br><br> (3b^3 – 10b^2 + 4) ÷ (3b – 1)

viktelen [127]

Step-by-step explanation:

try surfing how to divide a function by the long division method

3 0

3 years ago

Perform the indicated operation. (36m 2 - 60m + 25) ÷ (6m - 5)

9966 [12]

The correct answer for this is 6m - 5
That is the final expression.
Mark brainliest :)

6 0

3 years ago

Which of the following is not equal to sin⁡(-230°)?

xenn [34]

From trigonometry we know that:

if $sin(\theta)=sin(\alpha)$

then, $\theta=n\pi+(-1)^n\alpha$ (where $n$ is an integer)

This can be rewritten in degrees as:

$\theta=n(180^{\circ})+(-1)^n\alpha$ .............(Equation 1)

Now, in our case, $\alpha=-230^{\circ}$

Therefore, (Equation 1) can be written as:

$\theta=n(180^{\circ})+(-1)^n(-230^{\circ})$ ..........(Equation 2)

Now, to find the correct options all that we have to do is replace n by relevant integers and find the values of $\theta$ that match.

For n=2, (Equation 2) gives us: $\theta=2\times 180^{\circ}+(-1)^2(-230^{\circ})=360^{\circ}-230^{\circ}=130^{\circ}$ .

Thus, $sin(230^{\circ})=sin(130^{\circ})$

Now, we know that: $-sin(-50^{\circ})=sin(50^{\circ})$

Let n=-1, then:

$\theta=(-1)\times 180^{\circ}+(-1)^{-1}(-230^{\circ})=-180^{\circ}+230^{\circ}=50^{\circ}$

Thus, $sin(-230^{\circ})=-sin(-50^{\circ})$

Likewise, $sin(-230^{\circ})=sin(50^{\circ})$

Only the last option $sin(-50^{\circ})$ will never match $sin(-230^{\circ})$ because no integral value of $n$ will ever give $\theta=-50^{\circ}$

Thus the last option is the correct option.

3 0

3 years ago

Read 2 more answers

I need help with this one

Anika [276]

Answer:

B

Step-by-step explanation:

if you look closely, you can see that the side of it goes up 2, then 2, then 1.

7 0

3 years ago

Read 2 more answers

Two smaller triangle have areas of 75 m square and 12 m squared. find the ratio of their perimeters

Paul [167]

we know that two similar triangles will have ratio corresponding side is same.

ratio triangle 1 is 75m^2.

and triangle 2 is 12m^2.

$75:12=\frac{75}{12} =\frac{25}{4} .$

so we have ratio of area $25:4$ .

ratio of perimeter will be $\frac{25}{4} =\frac{50}{8} =\frac{25}{12}$ .

from these perimeter ratio also $\frac{25}{4} =25:4$ .

5 0

3 years ago