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olga nikolaevna [1]

3 years ago

8

Solve for theta

olve \: for \: \alpha ." alt="solve \: for \: \alpha ." align="absmiddle" class="latex-formula">
, if Cos30 = 0.45

1 answer:

nordsb [41]3 years ago

6 0

I don’t know i’m sorry

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Can someone pls help me

max2010maxim [7]

Answer:third graph (open circle on 518, colored to the right)

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3 years ago

ANSWER THIS QUICK QUESTION FOR 16 POINTS!

denpristay [2]

If you flipped the picture, it would be 60 degrees. But since its not flipped, it could possibly be 120 degrees since all sides are equal as the sides are parallel to each other.

Hope this helps.

4 0

3 years ago

Read 2 more answers

Solve the inequality for u<br>-5u+27 [less than or equal to] 2

baherus [9]

Answer:

5≤u

Step-by-step explanation:

Put the words and numbers into an equation:

-5u+27≤2

Put all like terms on one side:

27-2≤5u ---> Do you see what I did there? Now you don't have to work with negatives. :)

Simplify:

25≤5u

Solve:

5≤u

:)

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3 years ago

Reyna's health insurance plan requires that she have a physician who manages all of her health care. Reyna, most likely, has wha

faust18 [17]

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3 years ago

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The green triangle is a dilation of the red triangle with a scale factor of s=13 and the center of dilation is at the point (4,2

Arada [10]

Given:

Scale factor $s=\dfrac{1}{3}$

Center of dilation = (4,2)

To find:

The coordinates of the points C' and A.

Solution:

We know that, if a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

$(x,y)\to (k(x-a)+a,k(y-b)+b)$

The scale factor is $\dfrac{1}{3}$ and the center of dilation is at (4,2).

$(x,y)\to (\dfrac{1}{3}(x-4)+4,\dfrac{1}{3}(y-2)+2)$ ...(i)

Suppose the vertices of red triangle are A(m,n), B(10,14) and C(-2,11).

Using rule (i), we get

$C(-2,11)\to C'(\dfrac{1}{3}(-2-4)+4,\dfrac{1}{3}(11-2)+2)$

$C(-2,11)\to C'(\dfrac{1}{3}(-6)+4,\dfrac{1}{3}(9)+2)$

$C(-2,11)\to C'(-2+4,3+2)$

$C(-2,11)\to C'(2,5)$

Hence, the coordinates of Point C' are C'(2,5).

Let us assume that point A is A(m,n).

Using rule (i), we get

$A(m,n)\to A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)$

From the given figure it is clear that the image of point A is (8,4).

$A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)=A'(8,4)$

On comparing both sides, we get

$\dfrac{1}{3}(m-4)+4=8$

$\dfrac{1}{3}(m-4)=8-4$

$(m-4)=3(4)$

$m=12+4$

$m=16$

And,

$\dfrac{1}{3}(n-2)+2=4$

$\dfrac{1}{3}(n-2)=4-2$

$(n-2)=3(2)$

$n=6+2$

$n=8$

Therefore, the coordinates of point A are (16,8).

4 0

3 years ago