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4 years ago

Help ppppppppppppllllllllllllllllsssssssss

Mathematics

1 answer:

iVinArrow [24]4 years ago

5 0

Answer:

The correct option is D ....

Step-by-step explanation:

Angle is ending at the point = (-3, -4)

Signs of both x and y are negative therefore it is in the third quadrant.

To find the the tangent of an angle we have to find the ratio of the length of the opposite side to the length of the adjacent side.

In this question the opposite side of the angle is of 4 units and the adjacent side of the angle is 3 units.

tanθ = y/x

tanθ = -4/-3

tanθ = 4/3

Thus the correct option is D ....

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make skateboards 1/4 of a mile and 1/2 of an hour at this rate how far will she skateboard in an hour

Taya2010 [7]

Answer:

1/2 a mile

Step-by-step explanation:

because if she travels 1/4 miles in 1/2 an hour that all you have to do is multiply 1/2 by 2 to get 1 hour so you do the same thing to 1/4 to get 2/4 simplified equals 1/2 so your answer is she travels 1/2 a mile in 1 hour.

And also please mark brainliest.

4 0

3 years ago

What is the area of TriangleABC if AD = 20 and BC = 25? А B

scoray [572]

Answer:

where's the picture at??

6 0

3 years ago

The reaction of the body to a dose of medicine can sometimes be represented by an equation of the formR = M^2 (C/2- M/3),where C

mr Goodwill [35]

Answer:

The derivative is $\frac{dR}{dM} = CM - M^2$

Step-by-step explanation:

From the question we are told that

The equation representing the reaction is $R = M^2 (\frac{C}{2} - \frac{M}{3} )$

Generally this equation can be represented as

$R = \frac{C M^2}{2} - \frac{M^3}{3}$

Generally $\frac{dR}{dM}$ as a function of M is mathematically represented as

$\frac{dR}{dM} = 2 * \frac{C M^{2 - 1 }}{2} + 3 * \frac{M^{3-1}}{3}$

=> $\frac{dR}{dM} = CM - M^2$

8 0

4 years ago

Eric, Shenna, and Yael are playing a game. Eric has half as many points as Shenna. Shenna has three fewer points than Yael. Alto

Anon25 [30]

Using algebraic equations, the number of points Yael has is calculated as: 29 points.

<h3>How to Use Algebraic Equations to Solve Word Problems?</h3>

In the given world problem, we known the following:

E = Eric

S = Shenna

Y = Yael

Their total points would be: E + S + Y = 68

S = 2(E) (Shenna has twice the points of Erik)

Y = S + 3 (Yael has three points more than Shenna)

We would therefore have the following algebraic equation:

E + 2E + (S + 3) = 68

Substitute S with 2E

E + 2E + 2E + 3 = 68

Solve for E

5E + 3 = 68

5E = 68 - 3

5E = 65

5E/5 = 65/5

E = 13

Eric's points is: 13

Yael's points = Y = S + 3

Y = S + 3 = 2E + 3

Plug in the value of E

Y = 2(13) + 3

Y = 26 + 3

Y = 29

Yael has 29 points.

Learn more about algebraic equation on:

brainly.com/question/2164351

#SPJ1

5 0

1 year ago

For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4

Elina [12.6K]

Answer:

C. $(5-\frac{1}{2})^6$

Step-by-step explanation:

Given

$15(5)^2(-\frac{1}{2})^4$

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

$Sum = 2 + 4$

$Sum = 6$

Each term of a binomial expansion are always of the form:

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

Where n = the sum above

$n = 6$

Compare $15(5)^2(-\frac{1}{2})^4$ to the above general form of binomial expansion

$(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......$

Substitute 6 for n

$(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......$

[Next is to solve for a and b]

From the above expression, the power of (5) is 2

Express 2 as 6 - 4

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

By direct comparison of

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

and

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

We have;

$^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4$

Further comparison gives

$^nC_r = 15$

$a^{n-r} =(5)^{6-4}$

$b^r= (-\frac{1}{2})^4$

[Solving for a]

By direct comparison of $a^{n-r} =(5)^{6-4}$

$a = 5$

$n = 6$

$r = 4$

[Solving for b]

By direct comparison of $b^r= (-\frac{1}{2})^4$

$r = 4$

$b = \frac{-1}{2}$

Substitute values for a, b, n and r in

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

$(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

Solve for $^6C_4$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......$

Check the list of options for the expression on the left hand side

The correct answer is $(5-\frac{1}{2})^6$

3 0

3 years ago