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3 years ago

Help me please I don’t know the answer

Mathematics

1 answer:

lorasvet [3.4K]3 years ago

6 0

Answer:

Step-by-step explanation:

The picture is kinda blurry for me, and I can't see the exact numbers, but with what I could see, I believe it's B

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Will give brainliest

telo118 [61]

Answer:

f(x) = x^(1/2) + 3

Step-by-step explanation:

Translating to the right 3 units would change these:

0^(1/2) = 0 /// 3 units to the right would be (0,3)

1^(1/2) = 1 //// 3 units to the right would be (1,4)

4^(1/2) = 2 //// 3 units to the right would be (4,5) etc

4 0

2 years ago

Suppose that Prudie earns the same each day.write an inequality to find how much she needs to earn each day

Sedbober [7]

let the amount of money earned by Prudie be p

since she earns the same amount of money each day after two days she has earned 2p

therefore the amount of money she needs to earn each day equals p $\geq$ 0

3 0

4 years ago

Factor y2 + 5y +6. (y + 1)(y + 6) (y + 2)(y + 3) (y + 2)(y + 4)

Natali5045456 [20]

Factor y2 + 5y +6.

Multiply all of the multiple choices

(y+1)(y+6)

(y*y)(y*6)(+1*y)(+1)(+6)

y^2+6y+y+6

=y^2+7y+6

(y+2)(y+3)

(y*y)(y*3)(2*y)(2*3)

y^2+3y+2y+6

=y^2+5y+6

Answer: (y + 2)(y + 3)

5 0

3 years ago

The gradient of the curve

AleksAgata [21]

Answer:

Coordinate of P is (-2, -3)

Step-by-step explanation:

Gradient is the derivative of the curve:

$y = {3x}^{2} + 11x + 7 \\ \\ \frac{dy}{dx} = 6x + 11$

but dy/dx is -1 :

$6x + 11 = - 1 \\ 6x = - 12 \\ x = - 2$

substitute for x :

$y = 3( - 2) {}^{2} + 11( - 2) + 7 \\ y = - 3$

3 0

3 years ago

I need help with finding the answer to a) and b). Thank you!

shtirl [24]

Answer:

$\displaystyle \sin\Big(\frac{x}{2}\Big) = \frac{7\sqrt{58} }{ 58 }$

$\displaystyle \cos\Big(\frac{x}{2}\Big)=-\frac{3 \sqrt{58}}{58}$

$\displaystyle \tan\Big(\frac{x}{2}\Big)=-\frac{7}{3}$

Step-by-step explanation:

We are given that:

$\displaystyle \sin(x)=-\frac{21}{29}$

Where x is in QIII.

First, recall that sine is the ratio of the opposite side to the hypotenuse. Therefore, the adjacent side is:

$a=\sqrt{29^2-21^2}=20$

So, with respect to x, the opposite side is 21, the adjacent side is 20, and the hypotenuse is 29.

Since x is in QIII, sine is negative, cosine is negative, and tangent is also negative.

And if x is in QIII, this means that:

$180$

So:

$\displaystyle 90 < \frac{x}{2} < 135$

Thus, x/2 will be in QII, where sine is positive, cosine is negative, and tangent is negative.

Recall that:

$\displaystyle \sin\Big(\frac{x}{2}\Big)=\pm\sqrt{\frac{1 - \cos(x)}{2}}$

Since x/2 is in QII, this will be positive.

Using the above information, cos(x) is -20/29. Therefore:

$\displaystyle \sin\Big(\frac{x}{2}\Big)=\sqrt{\frac{1 + 20/29}{2}$

Simplify:

$\displaystyle \sin\Big(\frac{x}{2}\Big)=\sqrt{\frac{49/29}{2}}=\sqrt{\frac{49}{58}}=\frac{7}{\sqrt{58}}=\frac{7\sqrt{58}}{58}$

Likewise:

$\displaystyle \cos \Big( \frac{x}{2} \Big) =\pm \sqrt{ \frac{1+\cos(x)}{2} }$

Since x/2 is in QII, this will be negative.

Using the above information, cos(x) is -20/29. Therefore:

$\displaystyle \cos \Big( \frac{x}{2} \Big) =-\sqrt{ \frac{1- 20/29}{2} }$

Simplify:

$\displaystyle \cos\Big(\frac{x}{2}\Big)=-\sqrt{\frac{9/29}{2}}=-\sqrt{\frac{9}{58}}=-\frac{3}{\sqrt{58}}=-\frac{3\sqrt{58}}{58}$

Finally:

$\displaystyle \tan\Big(\frac{x}{2}\Big) = \frac{\sin(x/2)}{\cos(x/2)}$

Therefore:

$\displaystyle \tan\Big(\frac{x}{2}\Big)=\frac{7\sqrt{58}/58}{-3\sqrt{58}/58}=-\frac{7}{3}$

5 0

3 years ago