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Sunny_sXe [5.5K]

3 years ago

7

30 points and mark as brainly

1 answer:

Alex17521 [72]3 years ago

3 0

Answer:

16. B

17. B

18. B

19. B

20. B

I am not sure if they are all correct but I hope I helped you a tiny bit

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If the mass of your father is 70kg, what is his weight (N)?

Lana71 [14]

Answer:

686.7 N

Step-by-step explanation:

mass the person is 70 kg , which is constant for the entire universe.Although weight of the person is depends upon the gravity of the planet on which he resides

W= mg

m= mass W= wight and g= Acceleration due to gravity of the planet which 9.81 m/sec^2 on earth.

therefore weight W = 70×9.81 = 686.7 N

Hence the weight of the father will be 686.7 N

8 0

3 years ago

21) Jaquan earns $5 per hour raking leaves. He needs at least $170 to buy a new deo game system. Write an inequality that descri

Anna007 [38]

5x>/170
(WORK SHOWN BELOW)

5 0

3 years ago

Draw a line from each rectangular prism to its matching group of face shapes

IrinaK [193]

Can you put more information into this question

6 0

4 years ago

Read 2 more answers

How do I complete these questions a bit confused . (extra credit work )

otez555 [7]

Answer:

(1)

$a = \frac{3\sqrt 3}{2}$

$b = \frac{3}{2}$

(2)

$a = \sqrt 6$

$b = \sqrt 2$

Step-by-step explanation:

Solving (1):

Considering

$\theta = 60^o$

We have:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

This gives:

$\sin(60^o) = \frac{a}{3}$

Solve for a

$a = 3 * \sin(60^o)$

$\sin(60^o) = \frac{\sqrt 3}{2}$

So:

$a = 3 * \frac{\sqrt 3}{2}$

$a = \frac{3\sqrt 3}{2}$

To solve for b, we make use of Pythagoras theorem

$3^2 = a^2 + b^2$

This gives

$3^2 = (\frac{3\sqrt 3}{2})^2 + b^2$

$9 = \frac{9*3}{4} + b^2$

$9 = \frac{27}{4} + b^2$

Collect like terms

$b^2 = 9 - \frac{27}{4}$

Take LCM and solve

$b^2 = \frac{36 - 27}{4}$

$b^2 = \frac{9}{4}$

Take square roots

$b = \frac{3}{2}$

Solving (2):

Considering

$\theta = 60^o$

We have:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

This gives:

$\sin(60^o) = \frac{a}{2\sqrt 2}$

Solve for a

$a = 2\sqrt 2 * \sin(60^o)$

$\sin(60^o) = \frac{\sqrt 3}{2}$

So:

$a = 2\sqrt 2 * \frac{\sqrt 3}{2}$

$a = \sqrt 2 * \sqrt 3$

$a = \sqrt 6$

To solve for b, we make use of Pythagoras theorem

$(2\sqrt 2)^2 = a^2 + b^2$

This gives

$(2\sqrt 2)^2 = (\sqrt 6)^2 + b^2$

$8 = 6 + b^2$

Collect like terms

$b^2 = 8 - 6$

$b^2 = 2$

Take square roots

$b = \sqrt 2$

3 0

3 years ago

The domain for f(x) and g(x) is the set of all real numbers.

Tom [10]

Answer:

Option B

Step-by-step explanation:

=> $f(x) = 3x+5$

=> $g(x) = x^2$

Subtracting both

=> $(f-g)(x) = 3x+5-x^2$

Assembling

=> $(f-g)(x) = -x^2+3x+5$

3 0

3 years ago