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3 years ago

Write sin 33 in terms of cosine a. cos 33 b. cos 66 c. cos 57 d. cos 147

Mathematics

1 answer:

zlopas [31]3 years ago

4 0

Answer:

Step-by-step explanation:

a a cosine and sine have to add up to 90. so subtracting 33 from 90 gives you 57 your answer would be cosine 57

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Hello Kids! Its Joe giving out a free brainliest to whoever answers this first with the abc's backwards!

sveticcg [70]

Answer:

Z, Y, X,

W, V,

U, T, S, R, Q, P, O, N,

M, L, K,

J, I, H,

G, F, E, D,

C, B, and A;

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Identify the slope of the line for the equation y= -2/3× + 1/3

natita [175]

Answer:

$\textbf{The slope of the parallel line is given by $ \frac{-2}{3} $}\\$

Step-by-step explanation:

$\textup{The slope of any two parallel lines are equal and they only differ in their $y- intercepts$. }\\\textup{Given $y = \frac{-2}{3}x + \frac{1}{3} $}\\\textup{This is comparable to $y = mx + c$, where $m$ is the slope of that line.}\\\textup{Therefore, the slope of the given line and any line parallel to it is $ \frac{-2}{3}$.}$

7 0

3 years ago

Kelly has 4 dimes and some nickels. The value of her coins is 2.25 how many

Semmy [17]

4 dimes=4 *10=40 cents

2.25-0.40=1.85

1.85=0.05 times number of nickles
times 100 both sides
185=5 times number of nickels
divide both sides by 5
37=number of nickles

has 37 nickles

8 0

3 years ago

(0,-4),(1,0),(2,2) find a quadratic function in standard form for each set of points.

Sphinxa [80]

We are given the points
<span>(0,-4),(1,0),(2,2)
The standard form a quadratic function (in terms of x) is
y = Ax2 + Bx + C
Subsitute the points
-4 = 0 + 0 + C
C = -4

0 = A + B - 4

2 = 4A + 2B - 4

Solve for A and B
A = -1
B = 5

The function is
y = -x2 + 5x - 4</span>

4 0

3 years ago

The point nearest to the origin on a line is at (4, -4). Find the standard form of the equation of the line.

Scrat [10]

Just did a specific one of these; let's do the general case.

The point nearest the origin is (a,b).

The line from the origin through the point is

$bx - ay = 0$

The line we seek is perpendicular to this one. We swap the coefficients on x and y, negating one, to get the perpendicular family of lines. We set the constant by plugging in the point (a,b):

$ax + by = a^2 + b^2$

$ax + by -( a^2 + b^2) = 0$

That's standard form; let's plug in the numbers:

$4 x - 4 y - 32 = 0$

$x - y - 8 = 0$

5 0

3 years ago