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3 years ago

Find each missing length

Mathematics

2 answers:

Llana [10]3 years ago

5 0

Answer:

you didnt provide any info over the question

STatiana [176]3 years ago

3 0

Cant see photo needs to show photo in order for me to answer

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Maximize the objective function P = 2x + 1.5y for the feasible region shown. State the maximum value for P and the ordered pair

umka2103 [35]

Incomplete question. However, let's assume this are feasible regions to consider:

Points:

- (0, 100)

- (0, 125)

- (0, 325)

- (1, 200)

Answer:

Maximum value occurs at 325 at the point (0, 325)

Step-by-step explanation:

Remember, we substitute the points value for x, y in the objective function P = 2x + 1.5y.

- For point (0, 100): P= 2(0) + 1.5 (100) =150

- For point (0, 125): P= 2(0) + 1.5 (125) =187.5

For point (0, 325): P= 2(0) + 1.5 (325) = 487.5

For point (1, 200): P= 2(1) + 1.5 (200) = 302

Therefore, we could notice from the above solutions that at point (0,325) we attain the maximum value of P.

7 0

3 years ago

Please help Combine like terms. 9y + 5y - 3 = [? ]y + [ ]

almond37 [142]

Answer:

= 14y + ( – 3 )

I hope I helped you^_^

7 0

2 years ago

What is the measure of angle 1?

frutty [35]

Sum of all the interior angles = (5-2)x180 = 540°
540 - 100 - 120 - 100 - 90 = 130°
So the last unknown angle in the polygon is 130°

Sum of all the adjacent angles on a straight line is 180°
Angle 1 = 180 - 130 = 50°

3 0

3 years ago

16) Please help with question. WILL MARK BRAINLIEST + 10 POINTS.

Katyanochek1 [597]

We will use the sine and cosine of the sum of two angles, the sine and consine of $\frac{\pi}{2}$ , and the relation of the tangent with the sine and cosine:

$\sin (\alpha+\beta)=\sin \alpha\cdot\cos\beta + \cos\alpha\cdot\sin\beta

\cos(\alpha+\beta)=\cos\alpha\cdot\cos\beta-\sin\alpha\cdot\sin\beta$

$\sin\dfrac{\pi}{2}=1,\ \cos\dfrac{\pi}{2}=0$

$\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha}$

If you use those identities, for $\alpha=x,\ \beta=\dfrac{\pi}{2}$ , you get:

$\sin\left(x+\dfrac{\pi}{2}\right) = \sin x\cdot\cos\dfrac{\pi}{2} + \cos x\cdot\sin\dfrac{\pi}{2} = \sin x\cdot0 + \cos x \cdot 1 = \cos x$

$\cos\left(x+\dfrac{\pi}{2}\right) = \cos x \cdot \cos\dfrac{\pi}{2} - \sin x\cdot\sin\dfrac{\pi}{2} = \cos x \cdot 0 - \sin x \cdot 1 = -\sin x$

Hence:

$\tan \left(x+\dfrac{\pi}{2}\right) = \dfrac{\sin\left(x+\dfrac{\pi}{2}\right)}{\cos\left(x+\dfrac{\pi}{2}\right)} = \dfrac{\cos x}{-\sin x} = -\cot x$

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3 years ago

-21 = 7(3 - x) Please help. You have to use distributive property to solve.

OlgaM077 [116]

The answer is 6

hope this helps !!

4 0

3 years ago

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